# Exercise 5.1

**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.**

**Answer:**

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.**

**Answer:**

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.**

**Answer:**

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.**

**Answer:**

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

#### Therefore, after every year, our money will be

#### It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)

#### (Difference between consecutive terms is not equal)

#### Therefore, it is not an Arithmetic Progression.

**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

**Answer:**

(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.

**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 …

**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) 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

**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 …

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …

(xv) 1^{2}, 5^{2}, 7^{2}, 7^{3 }…

**Answer:**

(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} – an 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)

⇒ an+1 – an 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) 12, 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

# Exercise 5.2

**1. Find the missing variable from a, d, n and an, where a is the first term, d is the common difference and an is the nth term of AP.**

#### (i) a = 7, d = 3, n = 8

#### (ii) a = –18, n = 10, a_{n}=0

#### (iii) d = –3, n = 18, a_{n}=-5

#### (iv) a = –18.9, d = 2.5, a_{n}=-3.6

#### (v) a = 3.5, d = 0, n = 105

**Answer:**

#### (i) a = 7, d = 3, n = 8

#### We know that,

#### = 7 + (8 − 1) 3

#### = 7 + (7) 3

#### = 7 + 21 = 28

#### (ii) a = –18, n= 10,

#### We need to find d here.

#### Using formula

#### Putting values of a,n and ,

#### 0 = –18 + (10 – 1) d

#### ⇒ 0 = −18 + 9d

#### ⇒ 18 = 9d ⇒ d = 2

#### (iii) d = –3, n = 18,

#### We need to find a here.

#### Using formula

#### Putting values of d, ,

#### –5 = a + (18 – 1) (–3)

#### ⇒ −5 = a + (17) (−3)

#### ⇒ −5 = a – 51 ⇒ a = 46

#### (iv) a = –18.9, d = 2.5,

#### We need to find n here.

#### Using formula

#### Putting values of d, a,and ,

#### 3.6 = –18.9 + (n – 1) (2.5)

#### ⇒ 3.6 = −18.9 + 2.5n − 2.5

#### ⇒ 2.5n = 25 ⇒ n = 10

#### (v) a = 3.5, d = 0, n = 105

#### We need to find here.

#### Using formula

#### Putting values of d, n and a,

#### an = 3.5 + (105 − 1) (0)

#### ⇒

**2. Choose the correct choice in the following and justify:**

#### (i) 30^{th} term of the AP: 10, 7, 4… is

#### (A) 97

#### (B) 77

#### (C) –77

#### (D) –87

#### (ii) 11^{th} term of the AP: −3, −12, 2… is

#### (A) 28

#### (B) 22

#### (C) –38

#### (D)

**Answer:**

#### (i) 10, 7, 4…

#### First term = a = 10, Common difference = d = 7 – 10 = 4 – 7 = –3

#### And n = 30{Because, we need to find 30^{th} term}

#### ⇒ = 10 + (30 − 1) (−3) = 10 – 87 = −77

#### Therefore, the answer is (C).

#### (ii) −3, −½, 2…

#### First term = a = –3, Common difference = d = − − (−3) =

#### And n = 11 (Because, we need to find 11th term)

#### = −3 + (11 – 1) = −3 + 25 = 22

#### Therefore 11^{th} term is 22 which means answer is (B).

**3. In the following AP’s find the missing terms:**

#### (i) 2, __ , 26

#### (ii) __, 13, __, 3

#### (iii) 5, __, __,

#### (iv) –4. __, __, __, __, 6

#### (v) __, 38, __, __, __, –22

**Answer:**

#### (i) For this A.P.,

a = 2

a_{3} = 26

We know that, an = 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, an = 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, an = 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.

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

**Answer:**

3, 8, 13, 18, …

For this A.P.,

a = 3

d = a_{2} − a1 = 8 − 3 = 5

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

an = 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.

**5. Find the number of terms in each of the following APs:**

#### (i) 7, 13, 19…., 205

#### (ii) 18, , 13…, −47

**Answer:**

#### (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) 18, , 13 …, −47

#### First term = a =18, Common difference = d =

#### Using formula , to find nth term of arithmetic progression,

#### −47 = 18 + (n − 1)

#### =

#### Therefore, there are 27 terms in the given arithmetic progression.

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

**Answer:**

For this A.P.,

a = 11

d = a_{2} − a_{1} = 8 − 11 = −3

Let −150 be the nth term of this A.P.

We know that,

an = a + (n − 1) d

-150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

-164 = -3_{n}

n = 164/3

Clearly, n is not an integer.

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

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

^{st}term of an A.P. whose 11

^{th}term is 38 and the 16

^{th}term is 73.

**Answer:**

Given that,

a_{11} = 38

a_{16 }= 73

We know that,

a_{n} = a + (n − 1) d

a_{11} = a + (11 − 1) d

38 = a + 10d … (i) Similarly,

a_{16} = 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

a_{31} = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31^{st} term is 178.

**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.**

**Answer:**

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.

**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.**

**Answer:**

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 nth term of this A.P. be zero.

a_{n} = a + (n − 1) d

0 = 8 + (n − 1) (−2)

0 = 8 − 2n + 2

2_{n} = 10

n = 5

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

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

**Answer:**

We know that,

For an A.P., an = 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.

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

**Answer:**

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.

an = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

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

**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?**

**Answer:**

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

= a2 + 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.

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

**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 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.

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

**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 nth 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.

**15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?**

**Answer:**

63, 65, 67, …

a = 63

d = a_{2} − a1 = 65 − 63 = 2

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

a_{n}= 63 + (n − 1) 2 = 63 + 2_{n} − 2

a_{n} = 61 + 2_{n} … (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 + 7_{n }− 7

a_{n} = 7_{n} − 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 + 2_{n} = 7_{n} − 4

61 + 4 = 5_{n}

5_{n} = 65

n = 13

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

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

^{th}term exceeds the 5

^{th}term by 12.

**Answer:**

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, …

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

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

**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

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.

**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.

^{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.

**Answer:**

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.

**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?**

**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 nth 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.

**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 nth week, her week, her weekly savings become Rs 20.75, find n.**

**Answer:**

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.

# Exercise 5.3

**1. Find the sum of the following AP’s.**

#### (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

**Answer:**

#### (i) 2, 7, 12… to 10 terms

#### Here First term = a = 2, Common difference = d = 7 – 2 = 5 and n = 10

#### Applying formula,to find sum of n terms of AP,

#### (ii) –37, –33, –29… to 12 terms

#### Here First term = a = –37, Common difference = d = –33 – (–37) = 4

#### And n = 12

#### Applying formula,to find sum of n terms of AP,

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

#### Here First term = a = 0.6, Common difference = d = 1.7 – 0.6 = 1.1

#### And n = 100

#### Applying formula,to find sum of n terms of AP,

#### (iv) to 11 terms

#### Here First tern =a = 1/15Common difference =

#### Applying formula,to find sum of n terms of AP,

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

(i) 7 + 21/2+ 14 + ……………… +84

(ii)+ 14 + ………… + 84

(ii) 34 + 32 + 30 + ……….. + 10

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

**Answer:**

#### (i)

#### Here First term = a = 7, Common difference = d =

#### And Last term = l = 84

#### We do not know how many terms are there in the given AP.

#### So, we need to find n first.

#### Using formula , to find n^{th} term of arithmetic progression,

#### [7 + (n − 1) (3.5)] = 84

#### ⇒ 7 + (3.5) n − 3.5 = 84

#### ⇒ 3.5n = 84 + 3.5 – 7

#### ⇒ 3.5n = 80.5

#### ⇒ n = 23

#### Therefore, there are 23 terms in the given AP.

#### It means n = 23.

#### Applying formula, to find sum of n terms of AP,

#### (ii) 34 + 32 + 30 + … + 10

#### Here First term = a = 34, Common difference = d = 32 – 34 = –2

#### And Last term = l = 10

#### We do not know how many terms are there in the given AP.

#### So, we need to find n first.

#### Using formula , to find nth term of arithmetic progression,

#### [34 + (n − 1) (−2)] = 10

#### ⇒ 34 – 2n + 2 = 10

#### ⇒ −2n = −26⇒ n = 13

#### Therefore, there are 13 terms in the given AP.

#### It means n = 13.

#### Applying formula, to find sum of n terms of AP,

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

#### Here First term = a = –5, Common difference = d = –8 – (–5) = –8 + 5 = –3

#### And Last term = l = −230

#### We do not know how many terms are there in the given AP.

#### So, we need to find n first.

#### Using formula , to find nth term of arithmetic progression,

#### [−5 + (n − 1) (−3)] = −230

#### ⇒ −5 − 3n + 3 = −230

#### ⇒ −3n = −228 ⇒ n = 76

#### Therefore, there are 76 terms in the given AP.

#### It means n = 76.

#### Applying formula, to find sum of n terms of AP,

**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 an.

(vii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.

(viii) Given an = 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.

**Answer:**

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

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression,

#### ⇒ 50 = 5 + (n − 1) (3)

#### ⇒ 50 = 5 + 3n − 3

#### ⇒ 48 = 3n⇒ n = 16

#### Applying formula,to find sum of n terms of AP,

#### Therefore, n = 16 and S_{n }= 440

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

#### Using formula a_{n} = a + (n-1)d, to find n^{th} term of arithmetic progression,

#### a_{n} = a + (n-1)d

#### a_{13}= 7 + (13 − 1) (d)

#### ⇒ 35 = 7 + 12d

#### ⇒ 28 = 12d⇒ d = 28/12 = 7/3

#### Applying formula,to find sum of n terms of AP,

#### Therefore, d = 7/3 and S_{13} = 273

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

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression,

#### a_{n} = a + (n-1)d

#### a_{12 }= a + (12 − 1) 3

#### ⇒ 37 = a + 33 ⇒ a = 4

#### Applying formula,to find sum of n terms of AP,

#### Therefore, a = 4 and S12 = 246

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

#### Using formula a_{n} = a + (n-1)d, to find n^{th} term of arithmetic progression,

#### a_{n} = a + (n-1)d

#### a_{3} = a + (3 − 1) (d)

#### ⇒ 15 = a + 2d

#### ⇒ a = 15 − 2d… (1)

#### Applying formula,to find sum of n terms of AP,

#### ⇒ 125 = 5 (2a + 9d) = 10a + 45d

#### Putting (1) in the above equation,

#### 125 = 5 [2 (15 − 2d) + 9d] = 5 (30 − 4d + 9d)

#### ⇒ 125 = 150 + 25d

#### ⇒ 125 – 150 = 25d

#### ⇒ −25 = 25d⇒ d = −1

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression,

#### a_{n} = a + (n-1)d

#### a_{10}= a + (10 − 1) d

#### Putting value of d and equation (1) in the above equation,

#### a_{10}= 15 − 2d + 9d = 15 + 7d

#### = 15 + 7 (−1) = 15 – 7 = 8

#### Therefore, d = −1 and a_{10}= 8

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

#### Applying formula,to find sum of n terms of AP,

#### ⇒ 150 = 18a + 360

#### ⇒ −210 = 18a

#### ⇒ a =

#### Using formula a_{n} = a + (n-1)d, to find n^{th} term of arithmetic progression,

#### a_{9} = + (9 − 1) (5)

#### a_{9} =

#### Therefore, a = and a9 =

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

#### Applying formula,to find sum of n terms of AP,

#### ⇒ 2n (n − 5) + 9 (n − 5) = 0

#### ⇒ (n − 5) (2n + 9) = 0

#### ⇒ n = 5,−9/2

#### We discard negative value of n because here n cannot be in negative or fraction.

#### The value of n must be a positive integer.

#### Therefore, n = 5

#### Using formula a_{n} = a + (n-1)d ,to find nth term of arithmetic progression,

#### a_{5 }= 2 + (5 − 1) (8) = 2 + 32 = 34

#### Therefore, n = 5 and a5 = 34

#### (vii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression,

#### 62 = 8 + (n − 1) (d) = 8 + nd – d

#### ⇒ 62 = 8 + nd − d

#### ⇒ nd – d = 54

#### ⇒ nd = 54 + d… (1)

#### Applying formula,to find sum of n terms of AP,

#### ⇒ n = 6

#### Putting value of n in equation (1),

#### 6d = 54 + d ⇒ d =

#### Therefore, n = 6 and d =

#### (viii) Given Given a_{n} = 4, d = 2, S_{n} = − 14, find n and a

#### Using formula a_{n} = a + (n-1)d , to find n^{th} term of arithmetic progression,

#### 4 = a + (n − 1) (2) = a + 2n − 2

#### ⇒ 4 = a + 2n – 2

#### ⇒ 6 = a + 2n

#### ⇒ a = 6 − 2n… (1)

#### Applying formula,to find sum of n terms of AP,

#### ⇒ (n + 2) (n − 7) = 0

#### ⇒ n = −2, 7

#### Here, we cannot have negative value of n.

#### Therefore, we discard negative value of n which means n = 7.

#### Putting value of n in equation (1), we get

#### a = 6 − 2n = 6 – 2 (7) = 6 – 14 = −8

#### Therefore, n = 7 and a = −8

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

#### Using formula, a_{n} = a + (n-1)d to find sum of n terms of AP, we get

#### 192 = [6 + (8 − 1) d] = 4 (6 + 7d)

#### ⇒ 192 = 24 + 28d

#### ⇒ 168 = 28d ⇒ d = 6

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

#### Applying formula, , to find sum of n terms, we get

#### 144 = [a + 28]

#### ⇒ 288 = 9 [a + 28]

#### ⇒ 32 = a + 28⇒ a = 4

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

**Answer:**

#### First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636

#### Applying formula,to find sum of n terms of AP, we get

#### 636 =[18 + (n − 1) (8)]

#### ⇒ 1272 = n (18 + 8n − 8)

#### ⇒

#### We discard negative value of n here because n cannot be in negative, n can only be a positive integer.

#### Therefore, n = 12

#### Therefore, 12 terms of the given sequence make sum equal to 636.

**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.**

**Answer.**

#### First term = a = 5, Last term = l = 45,

#### Applying formula, to find sum of n terms of AP, we get

#### ⇒ n = 16

#### Applying formula,to find sum of n terms of AP and putting value of n, we get

#### We get,

#### 45 = 5 + (16 – 1)d

#### 45 = 5 + (15)d

#### 45 – 5 = 15d

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

**Answer:**

#### First term = a = 17, Last term = l = 350 and Common difference = d = 9

#### Using formula , to find nth term of arithmetic progression, we get

#### 350 = 17 + (n − 1) (9)

#### ⇒ 350 = 17 + 9n − 9

#### ⇒ 342 = 9n ⇒ n = 38

#### Applying formula,to find sum of n terms of AP and putting value of n, we get

#### =19 (34 + 333) = 6973

#### Therefore, there are 38 terms and their sum is equal to 6973.

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

**Answer:**

#### It is given that 22nd term is equal to 149

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression, we get

#### a_{n} = a + (n-1)d

#### 149 = a + (22 − 1) (7)

#### ⇒ 149 = a + 147⇒ a = 2

#### Applying formula,to find sum of n terms of AP and putting value of a, we get

#### = 11 (4 + 147)

#### ⇒ 11(151)= 1661

#### Therefore, sum of first 22 terms of AP is equal to 1661.

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

**Answer:**

#### It is given that second and third term of AP are 14 and 18 respectively.

#### Using formula a_{n} = a + (n-1)d, to find nth term of arithmetic progression, we get

#### 14 = a + (2 − 1) d

#### ⇒ 14 = a + d … (1)

#### And, 18 = a + (3 − 1) d

#### ⇒ 18 = a + 2d … (2)

#### These are equations consisting of two variables.

#### Using equation (1), we get, a = 14 − d

#### Putting value of a in equation (2), we get

#### 18 = 14 – d + 2d

#### ⇒ d = 4

#### Therefore, common difference d = 4

#### Putting value of d in equation (1), we get

#### 18 = a + 2 (4)

#### ⇒ a = 10

#### Applying formula,to find sum of n terms of AP, we get

#### Therefore, sum of first 51 terms of an AP is equal to 5610.

**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.**

**Answer:**

#### It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.

#### Applying formula,to find sum of n terms of AP, we get

#### ⇒ 98 = 7 (2a + 6d)

#### ⇒ 7 = a + 3d ⇒ a = 7 − 3d … (1)

#### And, s_{17}= 289

#### ⇒ 578 = 17 (2a + 16d)

#### ⇒ 34 = 2a + 16d

#### ⇒ 17 = a + 8d

#### Putting equation (1) in the above equation, we get

#### 17 = 7 − 3d + 8d

#### ⇒ 10 = 5d ⇒ d = 2

#### Putting value of d in equation (1), we get

#### a = 7 − 3d = 7 – 3 (2) = 7 – 6 = 1

#### Again applying formula,to find sum of n terms of AP, we get

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

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

(ii) a_{n} = 9 − 5_{n}

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

_{1}, a

_{2}… , a

_{n}, … form an AP where an is defined as below

**Answer:**

#### (i)

#### is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

#### Therefore, sum of first 15 terms of AP is equal to 525.

#### (ii)

#### is same every time. Therefore, this is an AP with common difference as -5 and first term as 4.

#### Therefore, sum of first 15 terms of AP is equal to –465.

**11. If the sum of the first n terms of an AP is 4**_{n} − 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 nth terms.

_{n}− 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 nth terms.