CBSE 10TH Maths

Maths Syllabus 2010 - CBSE X

Maths Course Structure

CLASS X

One Paper Time : 3 Hours Marks : 80

UNITS MARKS

I. NUMBER SYSTEMS 04

II. ALGEBRA 20

III. TRIGONOMETRY 12

IV. COORDINATE GEOMETRY 08

V. GEOMETRY 16

VI. MENSURATION 10

VII. STATISTICS AND PROBABILITY 10

TOTAL 80

UNIT I : NUMBER SYSTEMS

1. REAL NUMBERS (15) Periods

Euclid's division lemma, Fundamental Theorem of Arithmetic - statements after reviewing work done earlier

and after illustrating and motivating through examples, Proofs of results - irrationality of decimal

expansions of rational numbers in terms of terminating/non-terminating recurring decimals.

UNIT II : ALGEBRA

1. POLYNOMIALS (6) Periods

Zeros of a polynomial. Relationship between zeros and coefficients of a polynomial with particular reference

to quadratic polynomials. Statement and simple problems on division algorithm for polynomials with real

coefficients.

2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods

Pair of linear equations in two variables. Geometric representation of different possibilities of solutions/

inconsistency.

Algebraic conditions for number of solutions. Solution of pair of linear equations in two variables algebraically

- by substitution, by elimination and by cross multiplication. Simple situational problems must be included.

Simple problems on equations reducible to linear equations may be included.

3. QUADRATIC EQUATIONS (15) Periods

Standard form of a quadratic equation ax² + bx + c = 0, (a ≠ 0). Solution of the quadratic equations

(only real roots) by factorization and by completing the square, i.e. by using quadratic formula. Relationship

between discriminant and nature of roots.

Problems related to day to day activities to be incorporated.

4. ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms.

UNIT III : TRIGONOMETRY

1. INTRODUCTION TO TRIGONOMETRY (12) Periods

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined);

motivate the ratios, whichever are defined at 0^o & 90^o. Values (with proofs) of the trigonometric ratios of

30^o, 45^o & 60^o. Relationships between the ratios.

2. TRIGONOMETRIC IDENTITIES (16) Periods

Proof and applications of the identity sin² A + cos² A = 1. Only simple identities to be given. Trigonometric

ratios of complementary angles.

3. HEIGHTS AND DISTANCES (8) Periods

Simple and believable problems on heights and distances. Problems should not involve more than two right

triangles. Angles of elevation / depression should be only 30^o, 45^o, 60^o.

UNIT IV : COORDINATE GEOMETRY

1. LINES (In two-dimensions) (15) Periods

Review the concepts of coordinate geometry done earlier including graphs of linear equations. Awareness of

geometrical representation of quadratic polynomials. Distance between two points and section formula

(internal). Area of a triangle.

UNIT V : GEOMETRY

1. TRIANGLES (15) Periods

Definitions, examples, counter examples of similar triangles.

1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct

points, the other two sides are divided in the same ratio.

2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are

proportional and the triangles are similar.

4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are

equal and the two triangles are similar.

5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including

these angles are proportional, the two triangles are similar.

6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the

hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each

other.

7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares on their

corresponding sides.

8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other

two sides.

9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the

angles opposite to the first side is a right traingle.

2. CIRCLES (8) Periods

Tangents to a circle motivated by chords drawn from points coming closer and closer to the point.

1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. (Prove) The lengths of tangents drawn from an external point to circle are equal.

3. CONSTRUCTIONS (8) Periods

1. Division of a line segment in a given ratio (internally)

2. Tangent to a circle from a point outside it.

3. Construction of a triangle similar to a given triangle.

UNIT VI : MENSURATION

1. AREAS RELATED TO CIRCLES (12) Periods

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and

perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems

should be restricted to central angle of 60^o, 90^o & 120^o only. Plane figures involving triangles, simple

quadrilaterals and circle should be taken.)

2. SURFACE AREAS AND VOLUMES (12) Periods

(i) Problems on finding surface areas and volumes of combinations of any two of the following: cubes,

cuboids, spheres, hemispheres and right circular cylinders/cones. Frustum of a cone.

(ii) Problems involving converting one type of metallic solid into another and other mixed problems. (Problems

with combination of not more than two different solids be taken.)

UNIT VII : STATISTICS AND PROBABILITY

1. STATISTICS (15) Periods

Mean, median and mode of grouped data (bimodal situation to be avoided). Cumulative frequency graph.

2. PROBABILITY (10) Periods

Classical definition of probability. Connection with probability as given in Class IX. Simple problems on

single events, not using set notation.

INTERNAL ASSESSMENT 20 Marks

Evaluation of activities 10 Marks

Project Work 05 Marks

Continuous Evaluation 05 Marks

RECOMMENDED BOOKS

1. Mathematics - Textbook for class IX - NCERT Publication

2. Mathematics - Textbook for class X - NCERT Publication

3. Guidelines for Mathematics Laboratory in Schools, class IX- CBSE Publication

4. Guidelines for Mathematics Laboratory in Schools, class X - CBSE Publication

CBSE 10th Maths Chap 1 - Real Numbers

Chapter Summary

1. Euclid’s division lemma :

Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0 ≤ r < b.

2. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows:

Step 1 : Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b.

Step 2 : If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be

HCF (a, b). Also, HCF(a, b) = HCF(b, r).

3. The Fundamental Theorem of Arithmetic :

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

4. If p is a prime and p divides a2, then p divides q, where a is a positive integer.

5. To prove that 2, 3 are irrationals.

6. Let x be a rational number whose decimal expansion terminates. Then we can express x in the form p

q , where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.

7. Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2n5m,where n, m are non-negative integers. Then x has a decimal expansion which terminates.

8. Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which isnon-terminating repeating (recurring).

NCERT Book Solutions

CBSE 10TH Maths Chap 1 Ex 1.1 Q1

Question:

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Answer:

Question:

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q what can you say about the prime factors of q? (i) 43.123456789 (ii) 0.120120012000120000. . . (iii) 43.123456789

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Chapter Summary

1. Two linear equations in the same two variables are called a pair of linear equations in two variables. The most general form of a pair of linear equations is
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
where a₁, a₂, b₁, b₂, c₁, c₂ are real numbers, such that a₁²+b₁² ≠ 0, a₂²+b₂² ≠ 0.

2. A pair of linear equations in two variables can be represented, and solved, by the:
(i) graphical method
(ii) algebraic method

3. Graphical Method :
The graph of a pair of linear equations in two variables is represented by two lines.
(i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.
(ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent).
(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.

4. Algebraic Methods : We have discussed the following methods for finding the solution(s) of a pair of linear equations :
(i) Substitution Method
(ii) Elimination Method
(iii) Cross-multiplication Method

5. There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.

NCERT Book Solutions

CBSE 10th Maths Chap 4 - Quadratic Equations

CBSE 10th Maths Chap 5 - Arithmetic Progressions

CBSE 10th Maths Chap 5 - Arithmetic Progressions - P99 - Q1

Question:

In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

(iv)The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

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Answer:

(i) It can be observed that

Taxi fare for 1^st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 + 8 = 31

Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes $\f$\small \frac{1}{4 }\f$$ of air remaining in the cylinder at a time. In other words, after every stroke, only $\f$\small 1-\frac{1}{4}= \frac{3}{4}\f$$ part of air will remain.

Therefore, volumes will be $\f$\small v,\left(\frac{3v}{4} \right),\left(\frac{3}{4}v \right)^{2}\left(\frac{3}{4}v \right)^{3}\f$$

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
(iii) Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250

Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be $\f$\small p\left[1+\frac{r}{100} \right]^{n}\f$$ after n years.

Therefore, after every year, our money will be $\f$\small 1000\left[1+\frac{8}{100} \right],1000\left[1+\frac{8}{100} \right]^{2},1000\left[1+\frac{8}{100} \right]^{3},1000\left[1+\frac{8}{100} \right]^{4},...\f$$

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q3

Question:

In the following APs find the missing term in the boxes

I. 2,_,26

II._,13,_,3

III.5,_,_, $\f$9\frac{1}{2}\f$$

IV.-4,_,_,_,_,6

V._,38,_,_,_,-22

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Answer:

I. 2,_,26

For this A.P.,

a = 2

a₃ = 26

We know that, a_n = a + (n − 1) d

a₃ = 2 + (3 − 1) d

26 = 2 + 2d

24 = 2d

d = 12

a₂ = 2 + (2 − 1) 12

= 14

Therefore, 14 is the missing term.

II._,13,_,3

For this A.P.,

a₂ = 13 and

a4 = 3

We know that, a_n = a + (n − 1) d

a₂ = a + (2 − 1) d

13 = a + d (I)

a4 = a + (4 − 1) d

3 = a + 3d (II)

On subtracting (I) from (II), we obtain

−10 = 2d

d = −5

From equation (I), we obtain

13 = a + (−5)

a = 18

a₃ = 18 + (3 − 1) (−5)

= 18 + 2 (−5) = 18 − 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

III..5,_,_, $\f$9\frac{1}{2}\f$$

For this A.P.

a=5

$\f$a_{4}=9\frac{1}{2}=\frac{19}{2}\f$$

We know that,

$\f$a_{n}=a+(n-1)d\f$$

$\f$a_{4}=a+(4-1)d\f$$

$\f$\frac{19}{2}=5+3d\f$$

$\f$\frac{19}{2}-5=3d\f$$

$\f$\frac{9}{2}=3d\f$$

$\f$d=\frac{3}{2}\f$$

$\f$a_{2}=a+d=5+\frac{3}{2}=\frac{13}{2}\f$$

$\f$a_{3}=a+2d=5+2\left(\frac{3}{2} \right)=8\f$$

Therefore, the missing terms are $\f$\frac{13}{2}\f$$ and 8 respectively.

IV.-4,_,_,_,_,6

For this A.P.,

a = −4 and

a₆ = 6

We know that,

a_n = a + (n − 1) d

a₆ = a + (6 − 1) d

6 = − 4 + 5d

10 = 5d

d = 2

a₂ = a + d = − 4 + 2 = −2

a₃ = a + 2d = − 4 + 2 (2) = 0

a₄ = a + 3d = − 4 + 3 (2) = 2

a₅ = a + 4d = − 4 + 4 (2) = 4

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

V._,38,_,_,_,-22

For this A.P.,

a₂ = 38

a₆ = −22

We know that

a_n = a + (n − 1) d

a₂ = a + (2 − 1) d

38 = a + d (1)

a₆ = a + (6 − 1) d

−22 = a + 5d (2)

On subtracting equation (1) from (2), we obtain

− 22 − 38 = 4d

−60 = 4d

d = −15

a = a₂ − d = 38 − (−15) = 53

a₃ = a + 2d = 53 + 2 (−15) = 23

a₄ = a + 3d = 53 + 3 (−15) = 8

a₅ = a + 4d = 53 + 4 (−15) = −7

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P99 - Q2

Question:

Write first four terms of the A.P. when the first term a and the common difference d are given as follows

(i) a = 10, d = 10

(ii) a = − 2, d = 0

(iii) a = 4, d = − 3

(iv) a = − 1 d = $\f$\small \frac{1}{2}\f$$

(v) a = − 1.25, d = − 0.25

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Answer:

(i) a = 10, d = 10

Let the series be a1, a2, a3, a4, a5 …

a₁ = a = 10

a₂ = a₁ + d = 10 + 10 = 20

a₃ = a₂ + d = 20 + 10 = 30

a₄ = a₃ + d = 30 + 10 = 40

a₅ = a₄ + d = 40 + 10 = 50

Therefore, the series will be 10, 20, 30, 40, 50 …

First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = −2, d = 0

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = −2

a₂ = a₁ + d = − 2 + 0 = −2

a₃ = a₂ + d = − 2 + 0 = −2

a₄ = a₃ + d = − 2 + 0 = −2

Therefore, the series will be −2, −2, −2, −2 …

First four terms of this A.P. will be −2, −2, −2 and −2.

(iii) a = 4, d = −3

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = 4

a₂ = a₁ + d = 4 − 3 = 1

a₃ = a₂ + d = 1 − 3 = −2

a₄ = a₃ + d = − 2 − 3 = −5

Therefore, the series will be 4, 1, −2 −5 …

First four terms of this A.P. will be 4, 1, −2 and −5.

(iv) a = −1, d = $\f$\frac{1}{2}\f$$

Let the series be a₁, a₂, a₃, a₄ …

$\f$a_{1}=a= - 1\f$$

$\f$a_{2}=a_{1}+d=-1+\frac{1}{2}=-\frac{1}{2}\f$$

$\f$a_{3}=a_{2}+d=-\frac{1}{2}+\frac{1}{2}=0\f$$

$\f$a_{4}=a_{3}+d=0+\frac{1}{2}=\frac{1}{2}\f$$

Clearly, the series will be

$\f$-1,-\frac{1}{2},0,\frac{1}{2}\f$$

First four terms of this A.P. will be $\f$-1,-\frac{1}{2},0 and \frac{1}{2}\f$$

(v) a = −1.25, d = −0.25

Let the series be a₁, a₂, a₃, a₄ …

a₁ = a = −1.25

a₂ = a1 + d = − 1.25 − 0.25 = −1.50

a₃ = a₂ + d = − 1.50 − 0.25 = −1.75

a₄ = a₃ + d = − 1.75 − 0.25 = −2.00

Clearly, the series will be 1.25, −1.50, −1.75, −2.00 ……..

First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P99 - Q3

Question:

For the following A.P.s, write the first term and the common difference.

(i) 3, 1, − 1, − 3 …

(ii) − 5, − 1, 3, 7 …

(iii) $\f$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}...\f$$

(iv) 0.6, 1.7, 2.8, 3.9

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Answer:

(i) 3, 1, −1, −3 …

Here, first term, a = 3

Common difference, d = Second term − First term

= 1 − 3 = −2

(ii) −5, −1, 3, 7 …

Here, first term, a = −5

Common difference, d = Second term − First term

= (−1) − (−5) = − 1 + 5 = 4

(iii) $\f$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}...\f$$

Here, first term, $\f$a=\frac{1}{3}\f$$

Common difference, d = Second term − First term

$\f$=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\f$$

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, first term, a = 0.6

Common difference, d = Second term − First term

= 1.7 − 0.6

= 1.1

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P99 - Q4

Question:

Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.

(i) 2, 4, 8, 16 …

(ii) $\f$2,\frac{5}{2},3,\frac{7}{2}...\f$$

(iii) − 1.2, − 3.2, − 5.2, − 7.2 …

(iv) − 10, − 6, − 2, 2 …

(v) $\f$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}...\f$$

(vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, − 4, − 8, − 12 …

(viii) $\f$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}...\f$$

(ix) 1, 3, 9, 27 …

(x) a, 2a, 3a, 4a …

(xi) a, a₂, a₃, a₄ …
(xii) $\f$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}...\f$$

(xiii) $\f$\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12}...\f$$

(xiv) 12, 32, 52, 72 …

(xv) 12, 52, 72, 73 …

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Answer:

(i) 2, 4, 8, 16 …

It can be observed that

a₂ − a₁ = 4 − 2 = 2

a₃ − a₂ = 8 − 4 = 4

a₄ − a₃ = 16 − 8 = 8

i.e., a_k+1− a_k is not the same every time. Therefore, the given numbers are not forming an A.P.

(ii) $\f$2,\frac{5}{2},3,\frac{7}{2} ...\f$$

It can be observed that

$\f$a_{2}-a=\frac{5}{2}-2=\frac{1}{2}\f$$

$\f$a_{3}-a_{2}=3-\frac{5}{2}=\frac{1}{2}\f$$

$\f$a_{4}-a_{3}=\frac{7}{2}-3=\frac{1}{2}\f$$

i.e., a_k+1− a_k is same every time.

Therefore, $\f$d=\frac{1}{2}\f$$ and the given numbers are in A.P.

Three more terms are

$\f$a_{5}=\frac{7}{2}+\frac{1}{2}=4\f$$

$\f$a_{6}=4+\frac{1}{2}=\frac{9}{2}\f$$

$\f$a_{7}=\frac{9}{2}+\frac{1}{2}=5\f$$

(iii) −1.2, −3.2, −5.2, −7.2 …

It can be observed that

a₂ − a₁ = (−3.2) − (−1.2) = −2

a₃ − a₂ = (−5.2) − (−3.2) = −2

a₄ − a₃ = (−7.2) − (−5.2) = −2

i.e., a_k+1− a_k is same every time. Therefore, d = −2

The given numbers are in A.P.

Three more terms are

a₅ = − 7.2 − 2 = −9.2

a₆ = − 9.2 − 2 = −11.2

a₇ = − 11.2 − 2 = −13.2

(iv) −10, −6, −2, 2 …

It can be observed that

a₂ − a₁ = (−6) − (−10) = 4

a₃ − a₂ = (−2) − (−6) = 4

a₄ − a₃ = (2) − (−2) = 4

i.e., a_k+1 − a_k is same every time. Therefore, d = 4

The given numbers are in A.P.

Three more terms are

a₅ = 2 + 4 = 6

a₆ = 6 + 4 = 10

a₇ = 10 + 4 = 14

(v) $\f$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...\f$$

It can be observed that

$\f$a_{2}-a_{1}=3+\sqrt{2}-3=\sqrt{2}\f$$
$\f$a_{3}-a_{2}=3+2\sqrt{2}-3-\sqrt{2}=\sqrt{2}\f$$

$\f$a_{4}-a_{3}=3+3\sqrt{2}-3-2\sqrt{2}=\sqrt{2}\f$$

i.e., a_k+1 − a_k is same every time. Therefore, $\f$d=\sqrt{2}\f$$

The given numbers are in A.P.

Three more terms are

$\f$a_{5}=3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}\f$$

$\f$a_{6}=3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\f$$

$\f$a_{7}=3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}\f$$
(vi) 0.2, 0.22, 0.222, 0.2222 ….

It can be observed that

a₂ − a₁ = 0.22 − 0.2 = 0.02

a₃ − a₂ = 0.222 − 0.22 = 0.002

a₄ − a₃ = 0.2222 − 0.222 = 0.0002

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(vii) 0, −4, −8, −12 …

It can be observed that

a₂ − a₁ = (−4) − 0 = −4

a₃ − a₂ = (−8) − (−4) = −4

a₄ − a₃ = (−12) − (−8) = −4

i.e., a_k+1 − a_k is same every time. Therefore, d = −4

The given numbers are in A.P.

Three more terms are

a₅ = − 12 − 4 = −16

a₆ = − 16 − 4 = −20

a₇ = − 20 − 4 = −24

(viii) $\f$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2} ...\f$$

It can be observed that

$\f$a_{2}-a_{2}=\left[-\frac{1}{2} \right]-\left[-\frac{1}{2} \right]=0\f$$

$\f$a_{3}-a_{2}=\left[-\frac{1}{2} \right]-\left[-\frac{1}{2} \right]=0\f$$

$\f$a_{4}-a_{3}=\left[-\frac{1}{2} \right]-\left[-\frac{1}{2} \right]=0\f$$

i.e., a_k+1 − a_k is same every time. Therefore, d = 0

The given numbers are in A.P.

Three more terms are

$\f$a_{5}=-\frac{1}{2}-0=-\frac{1}{2}\f$$

$\f$a_{6}=-\frac{1}{2}-0=-\frac{1}{2}\f$$

$\f$a_{7}=-\frac{1}{2}-0=-\frac{1}{2}\f$$

(ix) 1, 3, 9, 27 …

It can be observed that

a₂ − a₁ = 3 − 1 = 2

a₃ − a₂ = 9 − 3 = 6

a₄ − a₃ = 27 − 9 = 18

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(x) a, 2a, 3a, 4a …

It can be observed that

a₂ − a₁ = 2a − a = a

a₃ − a₂ = 3a − 2a = a

a₄ − a₃ = 4a − 3a = a

i.e., a_k+1 − a_k is same every time. Therefore, d = a

The given numbers are in A.P.

Three more terms are

a₅ = 4a + a = 5a

a₆ = 5a + a = 6a

a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …

It can be observed that

a₂ − a₁ = a²− a = a (a − 1)

a₃ − a₂ = a³ − a² = a² (a − 1)

a₄ − a₃ = a⁴ − a³ = a³ (a − 1)

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xii) $\f$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32} ...\f$$

It can be observed that

$\f$a_{2}-a_{1}=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\f$$

$\f$a_{3}-a_{2}=\sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2}\f$$

$\f$a_{4}-a_{3}=\sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}\f$$

i.e., a_k+1 − a_k is same every time.

Therefore, the given numbers are in A.P.

And, $\f$d=\sqrt{2}\f$$

Three more terms are

$\f$a_{5}=\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}\f$$

$\f$a_{6}=5\sqrt{2}+\sqrt{2}=6\sqrt{2}+\sqrt{72}\f$$

$\f$a_{7}=6\sqrt{2}+\sqrt{2}=7\sqrt{2}=\sqrt{98}\f$$

(xiii) $\f$\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12} ...\f$$

It can be observed that

$\f$a_{2}-a_{1}=\sqrt{6}-\sqrt{3}=\sqrt{3 \asymp 2}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)\f$$

$\f$a_{3}-a_{2}=\sqrt{9}-\sqrt{6}=3-\sqrt{6}=\sqrt{3}(\sqrt{3}-\sqrt{2})\f$$

$\f$a_{4}-a_{3}=\sqrt{12}-\sqrt{9}=2\sqrt{3}-\sqrt{3X3}=\sqrt{3}(2-\sqrt{3})\f$$

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xiv) 1², 3², 5², 7² …

Or, 1, 9, 25, 49 …..

It can be observed that

a₂ − a₁ = 9 − 1 = 8

a₃ − a₂ = 25 − 9 = 16

a₄ − a₃ = 49 − 25 = 24

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xv) 1², 5², 7², 7³ …

Or 1, 25, 49, 73 …

It can be observed that

a₂ − a₁ = 25 − 1 = 24

a₃ − a₂ = 49 − 25 = 24

a₄ − a₃ = 73 − 49 = 24

i.e., a_k+1 − a_k is same every time.

Therefore, the given numbers are in A.P.

And, d = 24

Three more terms are

a₅ = 73+ 24 = 97

a₆ = 97 + 24 = 121

a₇ = 121 + 24 = 145

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P105 - Q1

Question:

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.

	a	d	n	a_n
I	7	3	8	…...
II	− 18	…..	10	0
III	…..	− 3	18	− 5
IV	− 18.9	2.5	…..	3.6
V	3.5	0	105	…..

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Answer:

I. a = 7, d = 3, n = 8, a_n = ?

We know that,

For an A.P. a_n = a + (n − 1) d

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence, a_n = 28

II. Given that

a = −18, n = 10, a_n = 0, d = ?

We know that,

a_n = a + (n − 1) d

0 = − 18 + (10 − 1) d

18 = 9d

$\f$d=\frac{18}{9}=2\f$$

Hence, common difference, d = 2

III. Given that

d = −3, n = 18, a_n = −5

We know that,

a_n = a + (n − 1) d

−5 = a + (18 − 1) (−3)

−5 = a + (17) (−3)

−5 = a − 51

a = 51 − 5 = 46

Hence, a = 46

IV. a = −18.9, d = 2.5, a_n = 3.6, n = ?

We know that,

a_n = a + (n − 1) d

3.6 = − 18.9 + (n − 1) 2.5

3.6 + 18.9 = (n − 1) 2.5

22.5 = (n − 1) 2.5

$\f$(n-1)=\frac{22.5}{2.5}\f$$

$\f$n-1=0\f$$

$\f$n=10\f$$

Hence, n = 10

V. a = 3.5, d = 0, n = 105, a_n = ?

We know that,

a_n = a + (n − 1) d

a_n = 3.5 + (105 − 1) 0

a_n = 3.5 + 104 × 0

a_n = 3.5

Hence, a_n = 3.5

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q2

Question:

Choose the correct choice in the following and justify

I. 30^th term of the A.P: 10, 7, 4, …, is

A. 97 B. 77 C. − 77 D. − 87

II 11^th term of the A.P. is

A. 28 B. 22 C. 3, $\f$-\frac{1}{2},2,...\f$$ is D $\f$-48\frac{1}{2}\f$$

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Answer:

I. Given that

A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a₂ − a₁ = 7 − 10

= −3

We know that, a_n = a + (n − 1) d

a₃₀ = 10 + (30 − 1) (−3)

a₃₀ = 10 + (29) (−3)

a₃₀ = 10 − 87 = −77

Hence, the correct answer is C.

II. Given that, A.P. $\f$-3,-\frac{1}{2},2,...\f$$

First term a = −3

Common difference, d = a₂ − a₁

$\f$=-\frac{1}{2}-(-3)\f$$

$\f$=-\frac{1}{2}+3=\frac{5}{2}\f$$

We know that,

$\f$a_{n}=a+(n-1)d\f$$

$\f$a_{11}=-3+(11-1)(\frac{5}{2})\f$$

$\f$a_{11}=-3+(10)(\frac{5}{2})\f$$

$\f$a_{11}=-3+25\f$$

$\f$a_{11}=22\f$$

Hence, the answer is B.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q4

Question:

Which term of the A.P. 3, 8, 13, 18, … is 78?

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Answer:

3, 8, 13, 18, …

For this A.P.,

a = 3

d = a₂ − a₁ = 8 − 3 = 5

Let n^th term of this A.P. be 78.

a_n = a + (n − 1) d

78 = 3 + (n − 1) 5

75 = (n − 1) 5

(n − 1) = 15

n = 16

Hence, 16^th term of this A.P. is 78.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q5

Question:

Find the number of terms in each of the following A.P.

I. 7, 13, 19, …, 205

II.18, $\f$15\frac{1}{2}\f$$ ,13,...,-47

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Answer:

I. 7, 13, 19, …, 205

For this A.P.,

a = 7

d = a₂ − a₁ = 13 − 7 = 6

Let there are n terms in this A.P.

a_n = 205

We know that

a_n = a + (n − 1) d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

II.18, $\f$15\frac{1}{2}\f$$ ,13,...,-47

For this A.P,

$\f$a=18\f$$

$\f$d=a_{2}-a_{1}=15\frac{1}{2}-18\f$$

$\f$d=\frac{31-36}{2}=-\frac{5}{2}\f$$

Let there are n terms in this A.P.

Therefore, a_n = −47 and we know that,

$\f$a_{n}=a+(n-1)d\f$$

$\f$-47=18+(n-1)(-\frac{5}{2})\f$$

$\f$-47-18=(n-1)(-\frac{5}{2})\f$$

$\f$-65=(n-1)(-\frac{5}{2})\f$$

$\f$(n-1)=\frac{-130}{-5}\f$$

$\f$(n-1)=26\f$$

$\f$n=27\f$$

Therefore, this given A.P. has 27 terms in it.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q6

Question:

Check whether − 150 is a term of the A.P. 11, 8, 5, 2, …

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Answer:

For this A.P.,

a = 11

d = a₂ − a₁ = 8 − 11 = −3

Let −150 be the n^th term of this A.P.

We know that,

$\f$a_{n}=a+(n-1)d\f$$

$\f$-150=11+(n-1)(-3)\f$$

$\f$-150=11-3n+3\f$$

$\f$n=\frac{164}{3} \f$$

Clearly, n is not an integer.

Therefore, −150 is not a term of this A.P.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q7

Question:

Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73

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Answer:

Given that,

a₁₁ = 38

a₁₆ = 73

We know that,

a_n = a + (n − 1) d

a₁₁ = a + (11 − 1) d

38 = a + 10d (1)

Similarly,

a₁₆ = a + (16 − 1) d

73 = a + 15d (2)

On subtracting (1) from (2), we obtain

35 = 5d

d = 7

From equation (1),

38 = a + 10 × (7)

38 − 70 = a

a = −32

a₃₁ = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31^st term is 178.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q8

Question:

An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term

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Answer:

Given that,

a₃= 12

a₅₀ = 106

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

12 = a + 2d (I)

Similarly, a₅₀ = a + (50 − 1) d

106 = a + 49d (II)

On subtracting (I) from (II), we obtain

94 = 47d

d = 2

From equation (I), we obtain

12 = a + 2 (2)

a = 12 − 4 = 8

a₂₉ = a + (29 − 1) d

a₂₉ = 8 + (28)2

a₂₉ = 8 + 56 = 64

Therefore, 29^th term is 64.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q9

Question:

If the 3^rd and the 9^th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

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Answer:

Given that,

a₃ = 4

a₉ = −8

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

4 = a + 2d (I)

a₉ = a + (9 − 1) d

−8 = a + 8d (II)

On subtracting equation (I) from (II), we obtain

−12 = 6d

d = −2

From equation (I), we obtain

4 = a + 2 (−2)

4 = a − 4

a = 8

Let n^th term of this A.P. be zero.

a_n = a + (n − 1) d

0 = 8 + (n − 1) (−2)

0 = 8 − 2n + 2

2n = 10

n = 5

Hence, 5^th term of this A.P. is 0.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q10

Question:

If 17^th term of an A.P. exceeds its 10^th term by 7. Find the common difference.

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Answer:

We know that,

For an A.P., a_n = a + (n − 1) d

a₁₇ = a + (17 − 1) d

a₁₇ = a + 16d

Similarly, a₁₀ = a + 9d

It is given that

a₁₇ − a₁₀ = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q11

Question:

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^th term?

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Answer:

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a₂ − a₁ = 15 − 3 = 12

a₅₄ = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

Let n^th term be 771.

a_n = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65^th term was 132 more than 54^th term.

Alternatively,
Let n^th term be 132 more than 54^th term.
$\f$n=54+\frac{132}{12}\f$$

$\f$=54+11=65^{th} term\f$$

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q12

Question:

Two APs have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^th terms?

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Answer:

Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.

For first A.P.,

a₁₀₀ = a₁ + (100 − 1) d

= a₁ + 99d

a₁₀₀₀ = a₁ + (1000 − 1) d

a₁₀₀₀ = a₁ + 999d

For second A.P.,

a₁₀₀ = a₂ + (100 − 1) d

= a₂ + 99d

a₁₀₀₀ = a₂ + (1000 − 1) d

= a₂ + 999d

Given that, difference between

100^th term of these A.P.s = 100

Therefore, (a1 + 99d) − (a₂ + 99d) = 100

a₁− a₂ = 100 (1)

Difference between 1000th terms of these A.P.s

(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂

From equation (1),

This difference, a₁ − a₂= 100

Hence, the difference between 1000th terms of these A.P. will be 100.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q13

Question:

How many three digit numbers are divisible by 7

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Answer:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

The series is as follows.

105, 112, 119, …, 994

Let 994 be the n^th term of this A.P.

a = 105

d = 7

a_n = 994

n = ?

a_n = a + (n − 1) d

994 = 105 + (n − 1) 7

889 = (n − 1) 7

(n − 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q14

Question:

How many multiples of 4 lie between 10 and 250?

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Answer:

First multiple of 4 that is greater than 10 is 12. Next will be 16.

Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows.

12, 16, 20, 24, …, 248

Let 248 be the n^th term of this A.P.

$\f$a=12\f$$

$\f$d=4\f$$

$\f$a_{n}=248\f$$

$\f$a_{n}=a+(n-1)d\f$$

$\f$248=12+(n-1)4\f$$

$\f$\frac{236}{4}=n-1\f$$

$\f$59=n-1\f$$

$\f$n=60\f$$

Therefore, there are 60 multiples of 4 between 10 and 250.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q15

Question:

For what value of n, are the n^th terms of two APs 63, 65, 67, and 3, 10, 17, … equal

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Answer:

63, 65, 67, …

a = 63

d = a₂ − a1 = 65 − 63 = 2

n^th term of this A.P. = a_n = a + (n − 1) d

a_n= 63 + (n − 1) 2 = 63 + 2n − 2

a_n = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a₂ − a₁ = 10 − 3 = 7

n^th term of this A.P. = 3 + (n − 1) 7

a_n = 3 + 7n − 7

a_n = 7n − 4 (2)

It is given that, n^th term of these A.P.s are equal to each other.

Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13^th terms of both these A.P.s are equal to each other.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P106 - Q16

Question:

Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12.

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Answer:

=a₃ = 16

a + (3 − 1) d = 16

a + 2d = 16 (1)

a₇− a₅ = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P107 - Q17

Question:

Find the 20^th term from the last term of the A.P. 3, 8, 13, …, 253

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Answer:

Given A.P. is

3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 5

For this A.P.,

a = 253

d = 248 − 253 = −5

n = 20

a₂₀ = a + (20 − 1) d

a₂₀ = 253 + (19) (−5)

a₂₀ = 253 − 95

a = 158

Therefore, 20^th term from the last term is 158.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P107 - Q18

Question:

The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.

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Answer:

We know that,

a_n = a + (n − 1) d

a₄ = a + (4 − 1) d

a₄ = a + 3d

Similarly,

a₈ = a + 7d

a₆ = a + 5d

a₁₀ = a + 9d

Given that, a₄ + a₈ = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 (1)

a₆ + a₁₀ = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 (2)

On subtracting equation (1) from (2), we obtain

2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain

a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a₂ = a + d = − 13 + 5 = −8

a₃= a₂ + d = − 8 + 5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P107 - Q19

Question:

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

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Answer:

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are

5000, 5200, 5400, …

Here, a = 5000

d = 200

Let after n^th year, his salary be Rs 7000.

Therefore, a_n = a + (n − 1) d

7000 = 5000 + (n − 1) 200

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11^th year, his salary will be Rs 7000.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P107 - Q20

Question:

Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^th week, her week, her weekly savings become Rs 20.75, find n.

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Answer:

Given that,

a = 5

d = 1.75

a_n = 20.75

n = ?

a_n = a + (n − 1) d

$\f$20.75=5+(n-1)1.75\f$$

$\f$15.75=(n-1)1.75\f$$

$\f$(n-1)=\frac{15.75}{1.75}=\frac{1575}{175}\f$$

$\f$\frac{63}{7}=9\f$$

$\f$n-1=9\f$$

$\f$n=10\f$$

Hence, n is 10.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P112 - Q1

Question:

Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms.

(ii) − 37, − 33, − 29 ,…, to 12 terms

(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms

(iv) $\f$\frac{1}{15},\frac{1}{12},\frac{1}{10}\f$$ ,………, to 11 terms

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Answer:

(i)2, 7, 12 ,…, to 10 terms

For this A.P.,

a = 2

d = a₂ − a₁ = 7 − 2 = 5

n = 10

We know that,

$\f$s_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$s_{10}=\frac{10}{2}\left[2(2)+(10-1)5 \right]\f$$

$\f$=5\left[4+(9) X (5) \right]\f$$

$\f$5 X 49 = 245\f$$

(ii)−37, −33, −29 ,…, to 12 terms

For this A.P.,

a = −37

d = a₂ − a₁ = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

$\f$s_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$s_{12}=\frac{12}{2}\left[2(-37)+(12-1)4 \right]\f$$

$\f$=6\left[-74+11 X 4 \right]\f$$

$\f$=6\left[-74+44 \right]\f$$

$\f$=6(-30)=-180\f$$

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

a = 0.6

d = a₂− a₁ = 1.7 − 0.6 = 1.1

n = 100

We know that,

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{100}=\frac{100}{2}\left[2(0.6)+(100-1)1.1 \right]\f$$

$\f$=50\left[1.2+108.9 \right]\f$$

$\f$50\left[110.1 \right]\f$$

$\f$=5505\f$$

(iv) $\f$\frac{1}{15},\frac{1}{12},\frac{1}{10}\f$$ ,………, to 11 terms

For this A.P.

$\f$a=\frac{1}{15}\f$$

n = 11

$\f$d=a_{2}-a_{1}=\frac{1}{12}-\frac{1}{15}\f$$

$\f$=\frac{5-4}{60}=\frac{1}{60}\f$$

We know that,

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{11}=\frac{11}{2}\left[2\left(\frac{1}{15} \right)+(11-1)\frac{1}{60} \right]\f$$

$\f$=\frac{11}{2}\left[\frac{2}{15}+\frac{10}{60} \right]\f$$

$\f$=\left(\frac{11}{2} \right)\left(\frac{9}{0} \right)=\frac{33}{20}\f$$

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P112 - Q2

Question:

Find the sums given below

(i) 7 + $\f$10\frac{1}{2}\f$$ + 14 + ………… + 84

(ii) 34 + 32 + 30 + ……….. + 10

(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

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Answer:

(i) 7 + $\f$10\frac{1}{2}\f$$ + 14 + ………… + 84

For this A.P.,

a = 7

l = 84

$\f$d=a_{2}-a_{1}=10\frac{1}{2}-7=\frac{21}{2}-7=\frac{7}{2}\f$$

Let 84 be the n^th term of this A.P.

l = a + (n − 1)d

$\f$84=7+(n-1)\frac{7}{2}\f$$

$\f$77=(n-1)\frac{7}{2}\f$$

22 = n − 1

n = 23

We know that,

$\f$S_{n}=\frac{n}{2}\left[7+84 \right]\f$$

$\f$S_{n}=\frac{23}{2}\left[7+84 \right]\f$$

$\f$=\frac{23X91}{2}=\frac{2093}{2}\f$$

$\f$1046\frac{1}{2}\f$$

(ii)34 + 32 + 30 + ……….. + 10

For this A.P.,

a = 34

d = a₂ − a₁ = 32 − 34 = −2

l = 10

Let 10 be the n^th term of this A.P.

l = a + (n − 1) d

10 = 34 + (n − 1) (−2)

−24 = (n − 1) (−2)

12 = n − 1

n = 13

$\f$S_{n}=\frac{n}{2}(a+l)\f$$

$\f$=\frac{13}{2}(34+10)\f$$

$\f$=\frac{13X44}{2}=13X22\f$$

$\f$=286\f$$

(iii)(−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,

a = −5

l = −230

d = a₂ − a₁ = (−8) − (−5)

= − 8 + 5 = −3

Let −230 be the n^th term of this A.P.

l = a + (n − 1)d

−230 = − 5 + (n − 1) (−3)

−225 = (n − 1) (−3)

(n − 1) = 75

n = 76

And, $\f$S_{n}=\frac{n}{2}\left(a+l \right)\f$$

$\f$=\frac{76}{2}\left[\left(-5 \right)+\left(-230 \right) \right]\f$$

$\f$=38(-235)\f$$

$\f$=-8930\f$$

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P112 - Q3

Question:

In an AP

(i) Given a = 5, d = 3, a_n = 50, find n and Sn.

(ii) Given a = 7, a₁₃ = 35, find d and S13.

(iii) Given a₁₂ = 37, d = 3, find a and S12.

(iv) Given a₃ = 15, S₁₀ = 125, find d and a10.

(v) Given d = 5, S₉ = 75, find a and a9.

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.

(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.

(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.

(ix) Given a = 3, n = 8, S = 192, find d.

(x)Given l = 28, S = 144 and there are total 9 terms. Find a.

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Answer:

(i) Given that, a = 5, d = 3, a_n = 50

As a_n = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

$\f$S_{n}=\frac{n}{2}\left[a+a_{n} \right]\f$$

$\f$S_{16}=\frac{16}{2}\left[5+50 \right]\f$$

$\f$8X55\f$$

$\f$=440\f$$

(ii) Given that, a = 7, a₁₃ = 35

As a_n = a + (n − 1) d,

∴ a₁₃ = a + (13 − 1) d

35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

$\f$d=\frac{7}{3}\f$$

$\f$S_{n}=\frac{n}{2}\left[a+a_{n} \right]\f$$

$\f$S_{13}=\frac{n}{2}\left[a+a_{13} \right]\f$$

$\f$\frac{13X42}{2}=13X21\f$$

$\f$=273\f$$

(iii)Given that, a₁₂ = 37, d = 3

As a_n = a + (n − 1)d,

a₁₂ = a + (12 − 1)3

37 = a + 33

a = 4

$\f$S_{n}=\frac{n}{2}\left[a+a_{n} \right]\f$$

$\f$S_{n}=\frac{12}{2}\left[4+37 \right]\f$$

$\f$S_{n}=6(41)\f$$

$\f$S_{n}=246\f$$

(iv) Given that, a₃ = 15, S₁₀ = 125

As a_n = a + (n − 1)d,

a₃ = a + (3 − 1)d

15 = a + 2d (i)

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{10}=\frac{10}{2}\left[2a+(10-1)d \right]\f$$

$\f$125=5(2a+9d)\f$$

$\f$25=2a+9d (ii)\f$$

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a₁₀ = a + (10 − 1)d

a₁₀ = 17 + (9) (−1)

a₁₀ = 17 − 9 = 8

(v)Given that, d = 5, S₉ = 75

As, $\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{9}=\frac{9}{2}\left[2a+(9-1)5 \right]\f$$

$\f$75=\frac{9}{2}\left(2a+40 \right)\f$$

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

$\f$a=\frac{-35}{3}\f$$

a_n = a + (n − 1)d

a₉ = a + (9 − 1) (5)

$\f$=\frac{-35}{3}+8(5)\f$$

$\f$=\frac{-35}{3}+40\f$$

$\f$=\frac{-35+120}{3}+\frac{85}{3}\f$$

(vi) Given that, a = 2, d = 8, S_n = 90

As, $\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$90=\frac{n}{2}\left[4+(n-1)8 \right]\f$$

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n2 − 2n

4n² − 2n − 90 = 0

4n² − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or $\f$n=-\frac{18}{4}=\frac{-9}{2}\f$$

However, n can neither be negative nor fractional.

Therefore, n = 5

a_n = a + (n − 1)d

a₅ = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210

$\f$S_{n}=\frac{n}{2}\left[a+a_{n} \right]\f$$

$\f$210=\frac{n}{2}\left[8+62 \right]\f$$

$\f$210=\frac{n}{2}\left[70 \right]\f$$

n = 6

a_n = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

$\f$d=\frac{54}{5}\f$$

(viii) Given that, a_n = 4, d = 2, S_n = −14

a_n = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

$\f$S_{n}=\frac{n}{2}\left[a+a_{n} \right]\f$$

$\f$-14 =\frac{n}{2}\left[ a+ 4 \right]\f$$

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n² + 10n

2n² − 10n − 28 = 0

n² − 5n −14 = 0

n² − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

(ix)Given that, a = 3, n = 8, S = 192

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$192=\frac{8}{2}\left[2X3+\left(8-1)d \right) \right]\f$$

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given that, l = 28, S = 144 and there are total of 9 terms.

$\f$S_{n}=\frac{n}{2}(a+l)\f$$

$\f$144=\frac{9}{2}(a+28)\f$$

(16) × (2) = a + 28

32 = a + 28

a = 4

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q4

Question:

How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

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Answer:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a₂− a₁ = 17 − 9 = 8

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$636=\frac{n}{2}\left[2a+(n-1)8 \right]\f$$

$\f$636=\frac{n}{2}\left[18+(n-1)8 \right]\f$$

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n² + 5n − 636 = 0

4n² + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

$\f$n=\frac{-53}{4}\f$$ or n = 12

n cannot be $\f$-\frac{53}{4}\f$$ . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q5

Question:

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

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Answer:

Given that,

a = 5

l = 45

S_n = 400

$\f$S_{n}=\frac{n}{2}(a+l)\f$$

$\f$400=\frac{n}{2}(5+45)\f$$

$\f$400=\frac{n}{2}(50)\f$$

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d
$\f$d=\frac{40}{15}=\frac{8}{3}\f$$

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q6

Question:

The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

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Answer:

Given that,

a = 17

l = 350

d = 9

Let there be n terms in the A.P.

l = a + (n − 1) d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38

$\f$s_{n}=\frac{n}{2}(a+l)\f$$

$\f$\Rightarrow S_{n}=\frac{38}{2}\left(17+350 \right)=19(367)=6973\f$$

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q7

Question:

Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.

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Answer:

d = 7

a₂₂ = 149

S₂₂ = ?

a_n = a + (n − 1)d

a₂₂ = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2

$\f$S_{n}=\frac{n}{2}(a+a_{n})\f$$

$\f$=\frac{22}{2}(2+149)\f$$

$\f$=11(151)=1661\f$$

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q8

Question:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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Answer:

Given that,

a₂ = 14

a₃ = 18

d = a₃ − a₂ = 18 − 14 = 4

a₂ = a + d

14 = a + 4

a = 10

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{51}=\frac{51}{2}\left[2(10)+(51-1)4 \right]\f$$

$\f$=\frac{51}{2}\left[20+(50)(4) \right]\f$$

$\f$\frac{51(220)}{2}=51(110)\f$$

= 5610

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q9

Question:

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

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Answer:

Given that,

S₇ = 49

S₁₇ = 289

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{7}=\frac{7}{2}\left[2a+(7-1)d \right]\f$$

$\f$49=\frac{7}{2}(2a+6d)\f$$

7 = (a + 3d)

a + 3d = 7 (i)

Similarly, $\f$S_{7}=\frac{17}{2}\left[2a+(17-1)d \right]\f$$

$\f$289=\frac{17}{2}\left[2a+16d \right]\f$$

17 = (a + 8d)

a + 8d = 17 (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i),

a + 3(2) = 7

a + 6 = 7

a = 1

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$=\frac{n}{2}\left[2(1)+(n-1)2 \right]\f$$

$\f$=\frac{n}{2}(2+2n-2)\f$$

$\f$=\frac{n}{2}(2n)\f$$
= n²

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q10

Question:

Show that a₁, a₂ … , a_n , … form an AP where a_n is defined as below

(i) a_n = 3 + 4n

(ii) a_n = 9 − 5_n

Also find the sum of the first 15 terms in each case.

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Answer:

(i) a_n = 3 + 4n

a₁ = 3 + 4(1) = 7

a₂= 3 + 4(2) = 3 + 8 = 11

a₃ = 3 + 4(3) = 3 + 12 = 15

a₄ = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂ − a₁ = 11 − 7 = 4

a₃ − a₂ = 15 − 11 = 4

a₄ − a₃ = 19 − 15 = 4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{15}=\frac{15}{2}\left[2(7)+(15-1)4 \right]\f$$

$\f$=\frac{15}{2}\left[(14)+56 \right]\f$$

$\f$=\frac{15}{2}(70)\f$$

= 15 × 35

= 525

(ii) a_n = 9 − 5n

a₁ = 9 − 5 × 1 = 9 − 5 = 4

a₂ = 9 − 5 × 2 = 9 − 10 = −1

a₃ = 9 − 5 × 3 = 9 − 15 = −6

a₄ = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a₂ − a₁ = − 1 − 4 = −5

a₃ − a₂ = − 6 − (−1) = −5

a₄ − a₃ = − 11 − (−6) = −5

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{15}=\frac{15}{2}\left[2(4)+(15-1)(-5) \right]\f$$

$\f$=\frac{15}{2}\left[8+14(-5) \right]\f$$

$\f$\frac{15}{2}(8-70)\f$$

$\f$\frac{15}{2}(-62)=15(-31)\f$$

= −465

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q11

Question:

If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

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Answer:

Given that,

S_n = 4n − n²

First term, a = S₁ = 4(1) − (1)2 = 4 − 1 = 3

Sum of first two terms = S₂

= 4(2) − (2)2 = 8 − 4 = 4

Second term, a₂ = S₂ − S₁ = 4 − 3 = 1

d = a₂ − a = 1 − 3 = −2

a_n = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1

a₁₀ = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10^th, and nth terms are −1, −15, and 5 − 2n respectively.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q118

Question:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? $\f$\left(Take \pi \right)=\frac{22}{7}\f$$

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Answer:

Semi-perimeter of circle = πr

I₁ = π(0.5) $\f$=\frac{\pi }{2}cm\f$$

I₂ = π(1) = π cm

I₃= π(1.5) = $\f$=\frac{3\pi }{2}cm\f$$

Therefore, I₁, I₂, I₃ ,i.e. the lengths of the semi-circles are in an A.P.,

$\f$\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi ,..........\f$$

$\f$a=\frac{\pi }{2}\f$$

$\f$d=\pi -\frac{\pi }{2}=\frac{\pi }{2}\f$$

S₁₃ =?

We know that the sum of n terms of an a A.P. is given by

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$\frac{13}{2}\left[2\left(\frac{\pi }{2} \right)+(13-1)\left(\frac{\pi }{2} \right) \right]\f$$

$\f$=\frac{13}{2}\left[\pi +\frac{12\pi }{2} \right]\f$$

$\f$=\left(\frac{13}{2} \right)(7x)\f$$

$\f$=\frac{91\pi }{2}\f$$

$\f$=\frac{91X22}{2X7}=13X11\f$$

= 143

Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q12

Question:

Find the sum of first 40 positive integers divisible by 6.

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Answer:

The positive integers that are divisible by 6 are

6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6

S₄₀ =?

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{40}=\frac{40}{2}\left[2(6)+(40-1)6 \right]\f$$

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q13

Question:

Find the sum of first 15 multiples of 8.

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Answer:

The multiples of 8 are

8, 16, 24, 32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

S₁₅ =?

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right] \f$$

$\f$=\frac{15}{2}\left[2(8)+(15-1)8 \right] \f$$

$\f$=\frac{15}{2}\left[16+14(8) \right]\f$$

$\f$=\frac{15}{2}\left[16+112 \right]\f$$

$\f$\frac{15(128)}{2}=15X64\f$$

= 960

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q14

Question:

Find the sum of the odd numbers between 0 and 50.

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Answer:

The odd numbers between 0 and 50 are

1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P.

a = 1

d = 2

l = 49

l = a + (n − 1) d

49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

$\f$S_{n}=\frac{n}{2}(a+l)\f$$

$\f$S_{n}=\frac{25}{2}(1+49)\f$$

$\f$=\frac{25(50)}{2}=(25)(25)\f$$

= 625

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q15

Question:

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

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Answer:

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

a = 200

d = 50

Penalty that has to be paid if he has delayed the work by 30 days = S₃₀

$\f$=\frac{30}{2}\left[2(200)+(30-1)50 \right]\f$$

= 15 [400 + 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q16

Question:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

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Answer:

Let the cost of 1^st prize be P.

Cost of 2^nd prize = P − 20

And cost of 3^rd prize = P − 40

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.

a = P

d = −20

Given that, S₇ = 700

$\f$\frac{7}{2}\left[2a+(7-1)d \right]=700\f$$

$\f$\frac{\left[2a+(6)(-20) \right]}{2}=100\f$$

a + 3(−20) = 100

a − 60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P113 - Q17

Question:

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

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Answer:

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term, a = 1

Common difference, d = 2 − 1 = 1

$\f$Sn=\frac{n}{2}\left[2a+(n-1)d \right] \f$$

$\f$S_{13}=\frac{12}{2}\left[2(1)+(12-1)(1) \right] \f$$

= 6 (2 + 11)

= 6 (13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3 × 78 = 234

Therefore, 234 trees will be planted by the students.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P114 - Q19

Question:

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

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Answer:

It can be observed that the numbers of logs in rows are in an A.P.

20, 19, 18…

For this A.P.,

a = 20

d = a₂ − a₁ = 19 − 20 = −1

Let a total of 200 logs be placed in n rows.

S_n = 200

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$200=\frac{n}{2}\left[2(20)+(n-1)(-1) \right]\f$$

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n²

n² − 41n + 400 = 0

n² − 16n − 25n + 400 = 0

n (n − 16) −25 (n − 16) = 0

(n − 16) (n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0

n = 16 or n = 25

a_n = a + (n − 1)d

a₁₆ = 20 + (16 − 1) (−1)

a₁₆ = 20 − 15

a₁₆ = 5

Similarly,

a₂₅ = 20 + (25 − 1) (−1)

a₂₅ = 20 − 24

= −4

Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P114 - Q20

Question:

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

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Answer:

The distances of potatoes are as follows.

5, 8, 11, 14…

It can be observed that these distances are in A.P.

a = 5

d = 8 − 5 = 3

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$S_{10}=\frac{10}{2}\left[2(5)+(10-1)3 \right]\f$$

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run = 2 × 185

= 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,……….

a = 10

d = 16 − 10 = 6

S₁₀ =?

$\f$S_{10}=\frac{10}{2}\left[2(10)+(10-1)6 \right]\f$$

= 5[20 + 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P115 - Q1

Question:

Which term of the A.P. 121, 117, 113 … is its first negative term?

[Hint: Find n for a_n < 0]

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Answer:

Given A.P. is 121, 117, 113 …

a = 121

d = 117 − 121 = −4

a_n = a + (n − 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to find the first negative term of this A.P.

$\f$Therefore,a_{n}\ <0\f$$

$\f$125-4n<0\f$$

$\f$125<4n\f$$

$\f$n>\frac{125}{4}\f$$

$\f$n>31.25\f$$

Therefore, 32^nd term will be the first negative term of this A.P.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P115 - Q2

Question:

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

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Answer:

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

a₃ = a + 2d

Similarly, a₇ = a + 6d

Given that, a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a3) × (a7) = 8

(a + 2d) × (a + 6d) = 8

From equation (i),

$\f$(3-4d+2d)X(3-4d+6d)=8 \f$$

$\f$ (3-2d)X(3+2d)=8\f$$

$\f$9-4d^{2}=8\f$$

$\f$4d^{2}=9-8=1\f$$

$\f$d^{2}=\frac{1}{4}\f$$

$\f$d=\pm \frac{1}{2}\f$$

$\f$d=\frac{1}{2} or -\frac{1}{2}\f$$

From equation (i),

$\f$\left(when d is \frac{1}{2} \right)\f$$

a=3-4d

$\f$a=3-4(\frac{1}{2})\f$$

=3-2=1

$\f$(When d is -\frac{1}{2})\f$$

$\f$a=3-4\left(-\frac{1}{2} \right)\f$$

$\f$a=3-2=5\f$$

$\f$S_{n}=\frac{n}{2}\left[2a+(n-1)d \right]\f$$

$\f$(When a is 1 and d is \frac{1}{2})\f$$

$\f$S_{16}=\frac{16}{2}\left[2(1)+(16-1)\left(\left(\frac{1}{2} \right) \right) \right]\f$$

$\f$=8\left[2+\frac{15}{2} \right]\f$$

$\f$=4(19)=76\f$$

$\f$(When a is 5 and d is -\frac{1}{2})\f$$

$\f$S_{16}=\frac{16}{2}\left[2(25)+(16-1)(-\frac{1}{2}) \right]\f$$

$\f$=8\left[10+(15)(-\frac{1}{2}) \right]\f$$

$\f$=8\left(\frac{5}{2} \right)\f$$

=20

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P115 - Q3

Question:

A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are m apart, what is the length of the wood required for the rungs?

[Hint: number of rungs $\f$\frac{250}{25}\f$$ ]

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Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs are $\f$2\frac{1}{2}\f$$ m apart.

∴ Total number of rungs $\f$=\frac{2\frac{1}{2}X100}{25}+1=\frac{250}{25}+1=11\f$$

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, a = 45

Last term, l = 25

n = 11

$\f$S_{n}=\frac{n}{2}(a+l)\f$$

$\f$S_{10}=\frac{11}{2}(45+25)=\frac{11}{2}(70)=385cm\f$$

Therefore, the length of the wood required for the rungs is 385 cm.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P115 - Q4

Question:

The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

[Hint S_x − 1 = S₄₉ − Sx]

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Answer:

The number of houses was

1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of x^th house was like this.

We know that,

Sum of n terms in an A.P. $\f$=\frac{n}{2}\left[ 2a+(n-1)d\right]\f$$

Sum of number of houses preceding xth house = S_x − 1

$\f$=\frac{(x-1)}{2}\left[2a+(x-1-1)d \right]\f$$

$\f$=\frac{x-1}{2}\left[2(1)+(x-2)(1) \right]\f$$

$\f$=\frac{(x)(x-1)}{2}\f$$

Sum of number of houses following x^th house = S₄₉− S_x

$\f$=\frac{49}{2}\left[2(1)+(49-1)(1) \right]-\frac{x}{2}\left[2(1)+(x-1)(1) \right]\f$$

$\f$=\frac{49}{2}(2+49-1)-\frac{x}{2}(2+x-1)\f$$

$\f$=\left(\frac{49}{2} \right)(50)-\frac{x}{2}(x+1)\f$$

$\f$=25(49)-\frac{x(x+1)}{2}\f$$

It is given that these sums are equal to each other.

$\f$\frac{x(x-1)}{2}=25(49)-x(\frac{x+1}{2})\f$$

$\f$\frac{x^{2}}{2}-\frac{x}{2}=1225-\frac{x^{2}}{2}-\frac{x}{2}\f$$

$\f$x^{2}=1225\f$$

$\f$x=\pm 35\f$$

However, the house numbers are positive integers.

The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

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CBSE 10th Maths Chap 5 - Arithmetic Progressions - P115 - Q5

Question:

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of $\f$\frac{1}{4}\f$$ m and a tread of $\f$\frac{1}{2}\f$$ m (See figure) calculate the total volume of concrete required to build the terrace.

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Answer:

From the figure, it can be observed that

1^st step is $\f$\frac{1}{2}\f$$ m wide,

2^nd step is 1 m wide,

3^rd step is $\f$\frac{3}{2}\f$$ m wide.

Therefore, the width of each step is increasing by $\f$\frac{1}{2}\f$$ m each time whereas their height $\f$\frac{1}{4}\f$$ m and length 50 m remains the same.