**Chapter 5 Arithmetic Progressions**

Maths NCERT solutions class 10 chapter 5 Arthimatic progression

•5.1 Introduction

• 5.2 Arithmetic Progression

• 5.3 nth Term of an AP

• 5.4 Sum of First n Terms of an AP

• 5.5 Summary**Important formula**

*Exercise 5.1*

Maths NCERT solutions class 10 chapter 5 Arthimatic progression

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### Q.1. 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.

It can be observed that

Taxi fare for 1st 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) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1/4 = 3/4th part of air will remain.

Therefore, volumes will be V, 3V/4 , (3V/4)^{2} , (3V/4)^{3}…

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) 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.

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

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be

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

### Q.2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = – 3

(iv) a = -1 d = 1/2

(v) a = – 1.25, d = – 0.25

(i) a = 10, d = 10

Let the series be a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …

a_{1} = a = 10

a_{2} = a_{1} + d = 10 + 10 = 20

a_{3} = a_{2} + d = 20 + 10 = 30

a_{4} = a_{3} + d = 30 + 10 = 40

a_{5} = a_{4} + 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 a_{1}, a_{2}, a_{3}, a_{4} …

a_{1} = a = -2

a_{2} = a_{1} + d = – 2 + 0 = – 2

a_{3} = a_{2} + d = – 2 + 0 = – 2

a_{4} = a_{3} + 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 a_{1}, a_{2}, a_{3}, a_{4} …

a_{1} = a = 4

a_{2} = a_{1} + d = 4 – 3 = 1

a_{3} = a_{2} + d = 1 – 3 = – 2

a_{4} = a_{3} + 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 = 1/2

Let the series be a_{1}, a_{2}, a_{3}, a_{4} …a_{1} = a = -1

a_{2} = a_{1} + d = -1 + 1/2 = -1/2

a_{3} = a_{2} + d = -1/2 + 1/2 = 0

a_{4} = a_{3} + d = 0 + 1/2 = 1/2

Clearly, the series will be-1, -1/2, 0, 1/2

First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25

Let the series be a_{1}, a_{2}, a_{3}, a_{4} …

a_{1} = a = – 1.25

a_{2} = a_{1} + d = – 1.25 – 0.25 = – 1.50

a_{3} = a_{2} + d = – 1.50 – 0.25 = – 1.75

a_{4} = a_{3} + 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.

Q.3. 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) 1/3, 5/3, 9/3, 13/3 ….

(iv) 0.6, 1.7, 2.8, 3.9 …

(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) 1/3, 5/3, 9/3, 13/3 ….

Here, first term, a = 1/3

Common difference, d = Second term – First term

= 5/3 – 1/3 = 4/3

(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

### Q.4. 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) 2, 5/2, 3, 7/2 ….

(iii) -1.2, -3.2, -5.2, -7.2 …

(iv) -10, – 6, – 2, 2 …

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, – 4, – 8, – 12 …

(viii) -1/2, -1/2, -1/2, -1/2 ….

(ix) 1, 3, 9, 27 …

(x) a, 2a, 3a, 4a …

(xi) a, a^{2}, a^{3}, a^{4} …

(xii) √2, √8, √18, √32 …

(xiii) √3, √6, √9, √12 …

(i) 2, 4, 8, 16 …

Here,

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

a_{4} – a_{3} = 16 – 8 = 8

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ….

Here,

a_{2} – a_{1} = 5/2 – 2 = 1/2

a_{3} – a_{2} = 3 – 5/2 = 1/2

a_{4} – a_{3} = 7/2 – 3 = 1/2

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = 1/2 and the given numbers are in A.P.

Three more terms are

a_{5} = 7/2 + 1/2 = 4

a_{6} = 4 + 1/2 = 9/2

a_{7} = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …

Here,

a_{2} – a_{1} = ( -3.2) – ( -1.2) = -2

a_{3} – a_{2} = ( -5.2) – ( -3.2) = -2

a_{4} – a_{3} = ( -7.2) – ( -5.2) = -2

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = -2 and the given numbers are in A.P.

Three more terms are

a_{5} = – 7.2 – 2 = – 9.2

a_{6} = – 9.2 – 2 = – 11.2

a_{7} = – 11.2 – 2 = – 13.2

(iv) -10, – 6, – 2, 2 …

Here,

a_{2} – a_{1} = (-6) – (-10) = 4

a_{3} – a_{2} = (-2) – (-6) = 4

a_{4} – a_{3} = (2) – (-2) = 4

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = 4 and the given numbers are in A.P.

Three more terms are

a_{5} = 2 + 4 = 6

a_{6} = 6 + 4 = 10

a_{7} = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2

Here,

a_{2} – a_{1} = 3 + √2 – 3 = √2

a_{3} – a_{2} = (3 + 2√2) – (3 + √2) = √2

a_{4} – a_{3} = (3 + 3√2) – (3 + 2√2) = √2

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = √2 and the given numbers are in A.P.

Three more terms are

a_{5} = (3 + √2) + √2 = 3 + 4√2

a_{6} = (3 + 4√2) + √2 = 3 + 5√2

a_{7} = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Here,

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

a_{4} – a_{3} = 0.2222 – 0.222 = 0.0002

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …

Here,

a_{2} – a_{1} = (-4) – 0 = -4

a_{3} – a_{2} = (-8) – (-4) = -4

a_{4} – a_{3} = (-12) – (-8) = -4

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = -4 and the given numbers are in A.P.

Three more terms are

a_{5} = -12 – 4 = -16

a_{6} = -16 – 4 = -20

a_{7} = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Here,

a_{2} – a_{1} = (-1/2) – (-1/2) = 0

a_{3} – a_{2} = (-1/2) – (-1/2) = 0

a_{4} – a_{3} = (-1/2) – (-1/2) = 0

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = 0 and the given numbers are in A.P.

Three more terms are

a_{5} = (-1/2) – 0 = -1/2

a_{6} = (-1/2) – 0 = -1/2

a_{7} = (-1/2) – 0 = -1/2

(ix) 1, 3, 9, 27 …

Here,

a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

a_{4} – a_{3} = 27 – 9 = 18

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …

Here,

a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 4a – 3a = a

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = a and the given numbers are in A.P.

Three more terms are

a_{5} = 4a + a = 5a

a_{6} = 5a + a = 6a

a_{7} = 6a + a = 7a

(xi) a, a^{2}, a^{3}, a^{4} …

Here,

a_{2} – a_{1} = a^{2 }– a = (a – 1)

a_{3} – a_{2} = a^{3 }-a^{2 }= a^{2 }(a – 1)

a_{4} – a_{3} = a^{4} – a^{3 }= a^{3}(a – 1)

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 …

Here,

a_{2} – a_{1} = √8 – √2 = 2√2 – √2 = √2

a_{3} – a_{2} = √18 – √8 = 3√2 – 2√2 = √2

a_{4} – a_{3} = 4√2 – 3√2 = √2

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = √2 and the given numbers are in A.P.

Three more terms are

a_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50

a_{6} = 5√2 +√2 = 6√2 = √72

a_{7} = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Here,

a_{2} – a_{1} = √6 – √3 = √3 × 2 -√3 = √3(√2 – 1)

a_{3} – a_{2} = √9 – √6 = 3 – √6 = √3(√3 – √2)

a_{4} – a_{3} = √12 – √9 = 2√3 – √3 × 3 = √3(2 – √3)

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …

Or, 1, 9, 25, 49 …..

Here,

a_{2} − a_{1} = 9 − 1 = 8

a_{3} − a_{2 }= 25 − 9 = 16

a_{4} − a_{3} = 49 − 25 = 24

⇒ a_{n+1} – a_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xv) 1^{2}, 5^{2}, 7^{2}, 73 …

Or 1, 25, 49, 73 …

Here,

a_{2} − a_{1} = 25 − 1 = 24

a_{3} − a_{2 }= 49 − 25 = 24

a_{4} − a_{3} = 73 − 49 = 24

i.e., a_{k}_{+1 }− a_{k} is same every time.

⇒ a_{n+1} – a_{n} is same every time.

Therefore, d = 24 and the given numbers are in A.P.

Three more terms are

a_{5} = 73+ 24 = 97

a_{6} = 97 + 24 = 121

a_{7 }= 121 + 24 = 145

Maths NCERT solutions class 10 chapter 5 Arthimatic progression

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*Exercise 5.2*

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### Q.1. 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 | ….. |

(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

d = 18/9 = 2

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

(n – 1) = 22.5/2.5

n – 1 = 9

n = 10

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

Q.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. -3, -1/2, ,2 …. is

(A) 28 (B) 22 (C) – 38 (D)-48½

(i) Given that

A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a_{2} − a_{1 }= 7 − 10 = −3

We know that, a_{n} = a + (n − 1) d

a_{30} = 10 + (30 − 1) (−3)

a_{30} = 10 + (29) (−3)

a_{30} = 10 − 87 = −77

Hence, the correct answer is option C.

(ii) Given that A.P. is -3, -1/2, ,2 …

First term a = – 3

Common difference, d = a_{2} − a_{1} = (-1/2) – (-3)

= (-1/2) + 3 = 5/2

We know that, a_{n} = a + (n − 1) d

a_{11} = 3 + (11 -1)(5/2)

a_{11} = 3 + (10)(5/2)

a_{11} = -3 + 25

a_{11} = 22

Hence, the answer is option B.

### Q.3. In the following APs find the missing term in the boxes.

(i) For this A.P.,

a = 2

a_{3} = 26

We know that, a_{n} = a + (n − 1) d

a_{3} = 2 + (3 – 1) d

26 = 2 + 2d

24 = 2d

d = 12

a_{2} = 2 + (2 – 1) 12

= 14

Therefore, 14 is the missing term.

(ii) For this A.P.,

a_{2} = 13 and

a_{4} = 3

We know that, a_{n} = a + (n − 1) d

a_{2} = a + (2 – 1) d

13 = a + d … (i)

a_{4} = a + (4 – 1) d

3 = a + 3d … (ii)

On subtracting (i) from (ii), we get

– 10 = 2d

d = – 5

From equation (i), we get

13 = a + (-5)

a = 18

a_{3} = 18 + (3 – 1) (-5)

= 18 + 2 (-5) = 18 – 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,

a = 5 and

a_{4} = 19/2

We know that, a_{n} = a + (n − 1) d

a_{4} = a + (4 – 1) d

19/2 = 5 + 3d

19/2 – 5 = 3d3d = 9/2

d = 3/2

a_{2} = a + (2 – 1) d

a_{2} = 5 + 3/2

a_{2} = 13/2

a_{3} = a + (3 – 1) d

a_{3} = 5 + 2×3/2

a_{3} = 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,

a = −4 and

a_{6} = 6

We know that,

a_{n} = a + (n − 1) d

a_{6} = a + (6 − 1) d

6 = − 4 + 5d

10 = 5d

d = 2

a_{2} = a + d = − 4 + 2 = −2

a_{3} = a + 2d = − 4 + 2 (2) = 0

a_{4} = a + 3d = − 4 + 3 (2) = 2

a_{5} = a + 4d = − 4 + 4 (2) = 4

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

(v)

For this A.P.,

a_{2} = 38

a_{6} = −22

We know that

a_{n} = a + (n − 1) d

a_{2} = a + (2 − 1) d

38 = a + d … (i)

a_{6} = a + (6 − 1) d

−22 = a + 5d … (ii)

On subtracting equation (i) from (ii), we get

− 22 − 38 = 4d

−60 = 4d

d = −15

a = a_{2} − d = 38 − (−15) = 53

a_{3} = a + 2d = 53 + 2 (−15) = 23

a_{4} = a + 3d = 53 + 3 (−15) = 8

a_{5} = a + 4d = 53 + 4 (−15) = −7

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

### Q.4. Which term of the A.P. 3, 8, 13, 18, … is 78?

3, 8, 13, 18, …

For this A.P.,

a = 3

d = a_{2} − a_{1} = 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.

### Q.5. Find the number of terms in each of the following A.P.

(i) 7, 13, 19, …, 205

(ii) 18,15½, 13,…., -47

(i) For this A.P.,

a = 7

d = a_{2} − a_{1} = 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) For this A.P.,

a = 18

Let there are n terms in this A.P.

a_{n} = 205

a_{n} = a + (n − 1) d

-47 = 18 + (n – 1) (-5/2)

-47 – 18 = (n – 1) (-5/2)

-65 = (n – 1)(-5/2)

(n – 1) = -130/-5

(n – 1) = 26

n = 27

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

### Q.6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …

For this A.P.,

a = 11

d = a_{2} − a_{1} = 8 − 11 = −3

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

We know that,

a_{n} = a + (n − 1) d

-150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

-164 = -3n

n = 164/3

Clearly, n is not an integer.

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

### Q.7. Find the 31^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73.

Given that,

a11 = 38

a16 = 73

We know that,

an = a + (n − 1) d

a11 = a + (11 − 1) d

38 = a + 10d … (i)

Similarly,

a16 = a + (16 − 1) d

73 = a + 15d … (ii)

On subtracting (i) from (ii), we get

35 = 5d

d = 7

From equation (i),

38 = a + 10 × (7)

38 − 70 = a

a = −32

a31 = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178.

### Q.8. 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.

Given that,

a_{3} = 12

a_{50} = 106

We know that,

a_{n} = a + (n − 1) d

a_{3} = a + (3 − 1) d

12 = a + 2d … (i)

Similarly, a_{50 }= a + (50 − 1) d

106 = a + 49d … (ii)

On subtracting (i) from (ii), we get

94 = 47d

d = 2

From equation (i), we get

12 = a + 2 (2)

a = 12 − 4 = 8

a_{29} = a + (29 − 1) d

a_{29} = 8 + (28)2

a_{29} = 8 + 56 = 64

Therefore, 29^{th} term is 64.

### Q.9. 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.

Given that,

a_{3} = 4

a_{9} = −8

We know that,

a_{n} = a + (n − 1) d

a_{3} = a + (3 − 1) d

4 = a + 2d … (i)

a_{9} = a + (9 − 1) d

−8 = a + 8d … (ii)

On subtracting equation (i) from (ii), we get,

−12 = 6d

d = −2

From equation (i), we get,

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.

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

We know that,

For an A.P., a_{n} = a + (n − 1) d

a_{17} = a + (17 − 1) d

a_{17} = a + 16d

Similarly, a_{10} = a + 9d

It is given that

a_{17} − a_{10} = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

### Q.11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^{th} term?

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a_{2} − a_{1} = 15 − 3 = 12

a_{54} = 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.

Or

Let n^{th} term be 132 more than 54^{th} term.

n = 54 + 132/2

= 54 + 11 = 65^{th} term

### Q.12. 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?

Let the first term of these A.P.s be a_{1} and a_{2} respectively and the common difference of these A.P.s be d.

For first A.P.,

a_{100} = a_{1} + (100 − 1) d

= a_{1} + 99d

a_{1000} = a_{1} + (1000 − 1) d

a_{1000} = a_{1} + 999d

For second A.P.,

a_{100} = a_{2} + (100 − 1) d

= a_{2} + 99d

a_{1000} = a_{2} + (1000 − 1) d

= a_{2} + 999d

Given that, difference between

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

Therefore, (a_{1} + 99d) − (a_{2} + 99d) = 100

a_{1} − a_{2} = 100 … (i)

Difference between 1000^{th} terms of these A.P.s

(a_{1} + 999d) − (a_{2} + 999d) = a_{1} − a_{2}

From equation (i),

This difference, a_{1} − a_{2 }= 100

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

### Q.13. How many three digit numbers are divisible by 7?

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 nth 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.

Or

Three digit numbers which are divisible by 7 are 105, 112, 119, …. 994 .

These numbers form an AP with a = 105 and d = 7.

Let number of three-digit numbers divisible by 7 be n, a_{n} = 994

⇒ a + (n – 1) d = 994

⇒ 105 + (n – 1) × 7 = 994

⇒7(n – 1) = 889

⇒ n – 1 = 127

⇒ n = 128

### Q.14. How many multiples of 4 lie between 10 and 250?

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.

a = 12

d = 4

a_{n} = 248

a_{n} = a + (n – 1) d

248 = 12 + (n – 1) × 4

236/4 = n – 1

59 = n – 1

n = 60

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

Or

Multiples of 4 lies between 10 and 250 are 12, 16, 20, …., 248.

These numbers form an AP with a = 12 and d = 4.

Let number of three-digit numbers divisible by 4 be n, a_{n} = 248

⇒ a + (n – 1) d = 248

⇒ 12 + (n – 1) × 4 = 248

⇒4(n – 1) = 248

⇒ n – 1 = 59

⇒ n = 60

### Q.15. For what value of n, are the n^{th} terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

63, 65, 67, …

a = 63

d = a_{2} − a_{1} = 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 … (i)

3, 10, 17, …

a = 3

d = a_{2} − a_{1} = 10 − 3 = 7

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

a_{n}= 3 + 7n − 7

a_{n} = 7n − 4 … (ii)

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.

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

a_{3} = 16

a + (3 − 1) d = 16

a + 2d = 16 … (i)

a_{7} − a_{5} = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (i), we get,

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

### Q.17. Find the 20^{th} term from the last term of the A.P. 3, 8, 13, …, 253.

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

a_{20} = a + (20 − 1) d

a_{20} = 253 + (19) (−5)

a_{20} = 253 − 95

a = 158

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

### Q.18. 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.

We know that,

a_{n} = a + (n − 1) d

a_{4} = a + (4 − 1) d

a_{4} = a + 3d

Similarly,

a_{8} = a + 7d

a_{6} = a + 5d

a_{10} = a + 9d

Given that, a_{4} + a_{8} = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 … (i)

a_{6} + a_{10} = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 … (ii)

On subtracting equation (i) from (ii), we get,

2d = 22 − 12

2d = 10

d = 5

From equation (i), we get

a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a_{2} = a + d = − 13 + 5 = −8

a_{3} = a_{2} + d = − 8 + 5 = −3

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

### Q.19. 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?

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 11th year, his salary will be Rs 7000.

### Q.20. 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.

Given that,

a = 5

d = 1.75

a_{n }= 20.75

n = ?

a_{n} = a + (n − 1) d

20.75 = 5 + (n – 1) × 1.75

15.75 = (n – 1) × 1.75

(n – 1) = 15.75/1.75 = 1575/175

= 63/7 = 9

n – 1 = 9

n = 10

Hence, n is 10.

Maths NCERT solutions class 10 chapter 5 Arthimatic progression

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*Exercise 5.3*

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### Q.1. 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) 1/15, 1/12, 1/10, …… , to 11 terms

(i) 2, 7, 12 ,…, to 10 terms

For this A.P.,

a = 2

d = a_{2} − a_{1} = 7 − 2 = 5

n = 10

We know that,

S_{n} = n/2 [2a + (n – 1) d]

S_{10} = 10/2 [2(2) + (10 – 1) × 5]

= 5[4 + (9) × (5)]

= 5 × 49 = 245

(ii) −37, −33, −29 ,…, to 12 terms

For this A.P.,

a = −37

d = a_{2} − a_{1} = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

S_{n} = n/2 [2a + (n – 1) d]

S_{12} = 12/2 [2(-37) + (12 – 1) × 4]

= 6[-74 + 11 × 4]

= 6[-74 + 44]

= 6(-30) = -180

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

a = 0.6

d = a_{2} − a_{1} = 1.7 − 0.6 = 1.1

n = 100

We know that,

S_{n} = n/2 [2a + (n – 1) d]

S_{12} = 50/2 [1.2 + (99) × 1.1]

= 50[1.2 + 108.9]

= 50[110.1]

= 5505

(iv) 1/15, 1/12, 1/10, …… , to 11 terms

For this A.P.,

### Q.2. Find the sums given below

(i) 7 +10½ + 14 + …………. +84

(ii)+ 14 + ………… + 84

(ii) 34 + 32 + 30 + ……….. + 10

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

(i) For this A.P.,

a = 7

l = 84

d = a_{2} − a_{1} = 10½ – 7 = 21/2 – 7 = 7/2

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

l = a (n – 1)d

84 = 7 + (n – 1) × 7/2

77 = (n – 1) × 7/2

22 = n − 1

n = 23

We know that,

S_{n} = n/2 (a + l)

S_{n} = 23/2 (7 + 84)

= (23×91/2) = 2093/2

= 1046½

(ii) 34 + 32 + 30 + ……….. + 10

For this A.P.,

a = 34

d = a_{2} − a_{1} = 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

S_{n} = n/2 (a + l)

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286

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

For this A.P.,

a = −5

l = −230

d = a_{2} − a_{1} = (−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,

S_{n} = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930

### Q.3. In an AP

(i) Given a = 5, d = 3, a_{n} = 50, find n and S_{n}.

(ii) Given a = 7, a_{13} = 35, find d and S_{13}.

(iii) Given a_{12} = 37, d = 3, find a and S_{12}.

(iv) Given a_{3} = 15, S_{10} = 125, find d and a_{10}.

(v) Given d = 5, S_{9} = 75, find a and a_{9}.

(vi) Given a = 2, d = 8, S_{n} = 90, find n and a_{n}.

(vii) Given a = 8, a_{n} = 62, S_{n} = 210, fi

(i) Given that, a = 5, d = 3, a_{n} = 50

As a_{n} = a + (n − 1)d,

⇒ 50 = 5 + (n – 1) × 3

⇒ 3(n – 1) = 45

⇒ n – 1 = 15

⇒ n = 16

Now, S_{n} = n/2 (a + a_{n})

S_{n} = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a_{13} = 35

As a_{n} = a + (n − 1)d,

⇒ 35 = 7 + (13 – 1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, S_{n} = n/2 (a + a_{n})

S_{13} = 13/2 (7 + 35) = 273

(iii)Given that, a_{12} = 37, d = 3

As a_{n} = a + (n − 1)d,

⇒ a_{12} = a + (12 − 1)3

⇒ 37 = a + 33

⇒ a = 4

S_{n} = n/2 (a + a_{n})

S_{n} = 12/2 (4 + 37)

= 246

(iv) Given that, a_{3} = 15, S_{10} = 125

As a_{n} = a + (n − 1)d,

a_{3} = a + (3 − 1)d

15 = a + 2d … (i)

S_{n} = n/2 [2a + (n – 1)d]

S_{10} = 10/2 [2a + (10 – 1)d]

125 = 5(2a + 9d)

25 = 2a + 9d … (ii)

On multiplying equation (i) by (ii), we get

30 = 2a + 4d … (iii)

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

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a_{10} = a + (10 − 1)d

a_{10} = 17 + (9) (−1)

a_{10} = 17 − 9 = 8

(v) Given that, d = 5, S_{9} = 75

As S_{n} = n/2 [2a + (n – 1)d]

S_{9} = 9/2 [2a + (9 – 1)5]

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

a = -35/3

a_{n} = a + (n − 1)d

a_{9} = a + (9 − 1) (5)

= -35/3 + 8(5)

= -35/3 + 40

= (35+120/3) = 85/3

(vi) Given that, a = 2, d = 8, S_{n} = 90

As S_{n} = n/2 [2a + (n – 1)d]

90 = n/2 [2a + (n – 1)d]

⇒ 180 = n(4 + 8n – 8) = n(8n – 4) = 8n^{2} – 4n

⇒ 8n^{2} – 4n – 180 = 0

⇒ 2n^{2} – n – 45 = 0

⇒ 2n^{2} – 10n + 9n – 45 = 0

⇒ 2n(n -5) + 9(n – 5) = 0

⇒ (2n – 9)(2n + 9) = 0

So, n = 5 (as it is positive integer)

∴ a_{5 }= 8 + 5 × 4 = 34

(vii) Given that, a = 8, a_{n} = 62, S_{n} = 210

As S_{n} = n/2 (a + a_{n})

210 = n/2 (8 + 62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8 + 5d

⇒ 5d = 62 – 8 = 54

⇒ d = 54/5 = 10.8

(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)

S_{n} = n/2 (a + a_{n})

-14 = n/2 (a + 4)

−28 = n (a + 4)

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

−28 = n (− 2n + 10)

−28 = − 2n^{2} + 10n

2n^{2} − 10n − 28 = 0

n^{2} − 5n −14 = 0

n^{2} − 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 get

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

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

As S_{n} = n/2 [2a + (n – 1)d]

192 = 8/2 [2 × 3 + (8 – 1)d]

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.

S_{n} = n/2 (a + l)

144 = 9/2 (a + 28)

(16) × (2) = a + 28

32 = a + 28

a = 4

### Q.4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be n terms of this A.P.

For this A.P., a = 9

d = a_{2} − a_{1} = 17 − 9 = 8

As S_{n} = n/2 [2a + (n – 1)d]

636 = n/2 [2 × a + (8 – 1) × 8]

636 = n/2 [18 + (n- 1) × 8]

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n^{2} + 5n − 636 = 0

4n^{2} + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

n = (-53/4) or n = 12

n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

### Q.5. 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.

Given that,

a = 5

l = 45

S_{n} = 400

S_{n} = n/2 (a + l)

400 = n/2 (5 + 45)

400 = n/2 (50)

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

d = 40/15 = 8/3

### Q.6. 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?

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

S_{n} = n/2 (a + l)

S_{38} = 13/2 (17 + 350)

= 19 × 367

= 6973

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

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

d = 7

a_{22} = 149

S_{22} = ?

a_{n} = a + (n − 1)d

a_{22} = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2

S_{n} = n/2 (a + a_{n})

= 22/2 (2 + 149)

= 11 × 151

= 1661

### Q.8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Given that,

a_{2} = 14

a_{3} = 18

d = a_{3} − a_{2} = 18 − 14 = 4

a_{2} = a + d

14 = a + 4

a = 10

S_{n} = n/2 [2a + (n – 1)d]

S_{51} = 51/2 [2 × 10 + (51 – 1) × 4]

= 51/2 [2 + (20) × 4]

= 51×220/2

= 51 × 110

= 5610

### Q.9. 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.

Given that,

S_{7} = 49

S_{17} = 289

S_{7}

= 7/2 [2a + (n – 1)d]

S_{7} = 7/2 [2a + (7 – 1)d]

49 = 7/2 [2a + 16d]

7 = (a + 3d)

a + 3d = 7 … (i)

Similarly,

S_{17} = 17/2 [2a + (17 – 1)d]

289 = 17/2 (2a + 16d)

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

S_{n} = n/2 [2a + (n – 1)d]

= n/2 [2(1) + (n – 1) × 2]

= n/2 (2 + 2n – 2)

= n/2 (2n)

= n^{2}

### Q. 10. Show that a_{1}, a_{2 }… , a_{n} , … form an AP where a_{n} is defined as below

(i) a_{n} = 3 + 4n

(ii) a_{n} = 9 − 5n

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

(i) a_{n} = 3 + 4n

a_{1} = 3 + 4(1) = 7

a_{2} = 3 + 4(2) = 3 + 8 = 11

a_{3} = 3 + 4(3) = 3 + 12 = 15

a_{4} = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a_{2} − a_{1} = 11 − 7 = 4

a_{3} − a_{2} = 15 − 11 = 4

a_{4} − a_{3} = 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.

S_{n} = n/2 [2a + (n – 1)d]

S_{15 }= 15/2 [2(7) + (15 – 1) × 4]

= 15/2 [(14) + 56]

= 15/2 (70)

= 15 × 35

= 525

(ii) a_{n} = 9 − 5n

a_{1} = 9 − 5 × 1 = 9 − 5 = 4

a_{2} = 9 − 5 × 2 = 9 − 10 = −1

a_{3} = 9 − 5 × 3 = 9 − 15 = −6

a_{4} = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a_{2} − a_{1} = − 1 − 4 = −5

a_{3} − a_{2} = − 6 − (−1) = −5

a_{4} − a_{3} = − 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.

S_{n} = n/2 [2a + (n – 1)d]

S_{15 }= 15/2 [2(4) + (15 – 1) (-5)]

= 15/2 [8 + 14(-5)]

= 15/2 (8 – 70)

= 15/2 (-62)

= 15(-31)

= -465

### Q.11. If the sum of the first n terms of an AP is 4n − n^{2}, what is the first term (that is S_{1})? 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.

Given that,

S_{n} = 4n − n^{2}

First term, a = S_{1} = 4(1) − (1)^{2} = 4 − 1 = 3

Sum of first two terms = S_{2}

= 4(2) − (2)^{2} = 8 − 4 = 4

Second term, a_{2} = S_{2} − S_{1} = 4 − 3 = 1

d = a_{2} − a = 1 − 3 = −2

a_{n} = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a_{3} = 5 − 2(3) = 5 − 6 = −1

a_{10} = 5 − 2(10) = 5 − 20 = −15

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

### Q.12. Find the sum of first 40 positive integers divisible by 6.

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

S_{40} = ?

S_{n} = n/2 [2a + (n – 1)d]

S_{40 }= 40/2 [2(6) + (40 – 1) 6]

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

= 20(12 + 234)

= 20 × 246

= 4920

### Q.13. Find the sum of first 15 multiples of 8.

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

S_{15} = ?

S_{n} = n/2 [2a + (n – 1)d]

S_{15} = 15/2 [2(8) + (15 – 1)8]

= 15/2[6 + (14) (8)]

= 15/2[16 + 112]

= 15(128)/2

= 15 × 64

= 960

### Q.14. Find the sum of the odd numbers between 0 and 50.

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

S_{n} = n/2 (a + l)

S_{25} = 25/2 (1 + 49)

= 25(50)/2

=(25)(25)

= 625

### Q.15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas 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.

AP- 200,250,300,350…

a = 200

d = 50

S_{30} = ?

S_{n }= n/2 [2a+(n-1)d]

S_{30 }=30/2 [2×200+29×50]

S_{30} = 15 [400+1450]

S_{30} = 15×1850

S_{30} = 27,750

### Q.16. 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.

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, S_{7} = 700

7/2 [2a + (7 – 1)d] = 700

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.

### Q.17. 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?

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

S_{n} = n/2 [2a + (n – 1)d]

S_{12} = 12/2 [2(1) + (12 – 1)(1)]

= 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.

### Q.18. 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? (Take π = 22/7)

perimeter of semi-circle = πr_{P1} = π(0.5) = π/2 cm_{P2} = π(1) = π cm_{P3} = π(1.5) = 3π/2 cm_{P1}, P_{2}, P_{3} are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ….

P1 = π/2 cm_{P2} = π cm

d = P2- P1 = π – π/2 = π/2

First term = P1 = a = π/2 cm

S_{n} = n/2 [2a + (n – 1)d]

Therefor, Sum of the length of 13 consecutive circles

S_{13} = 13/2 [2(π/2) + (13 – 1)π/2]

= 13/2 [π + 6π]

=13/2 (7π)

= 13/2 × 7 × 22/7

= 143 cm

### Q.19. 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?

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 = a_{2} − a_{1} = 19 − 20 = −1

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

S_{n} = 200

S_{n} = n/2 [2a + (n – 1)d]

S_{12} = 12/2 [2(20) + (n – 1)(-1)]

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n^{2}

n^{2} − 41n + 400 = 0

n^{2} − 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

a_{16} = 20 + (16 − 1) (−1)

a_{16} = 20 − 15

a_{16} = 5

Similarly,

a_{25} = 20 + (25 − 1) (−1)

a_{25} = 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.

### Q.20. 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.

Q.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

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

S_{10} =?

S_{10} = 12/2 [2(20) + (n – 1)(-1)]

= 5[20 + 54]

= 5 (74)

= 370

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

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Maths NCERT solutions class 10 chapter 5 Arthimatic progression

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