(ii) Given,

First term, a = -18

Common difference, d = ?

Number of terms, n = 10

Nth term, a_n = 0

we know Nth term for an A.P.,

a_n = a+(n−1)d

0 = − 18 +(10−1)d

18 = 9d

d = 18/9 = 2

Hence, common difference, d = 2

(iii) Given,

First term, a = ?

Common difference, d = -3

Number of terms, n = 18

Nth term, a_n = -5

we know Nth term for an A.P.,

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) Given,

First term, a = -18.9

Common difference, d = 2.5

Number of terms, n = ?

Nth term, a_n = 3.6

we know Nth term for an A.P.,

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) Given,

First term, a = 3.5

Common difference, d = 0

Number of terms, n = 105

Nth term, a_n = ?

we know Nth term for an A.P.,

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

Question 2

Answer

(i) Given

A.P. = 10, 7, 4, …

Therefore, we

First term, a = 10

Common difference, d = a₂ − a₁ = 7−10 = −3

As we know nth term of an A.P.,

a_n = a +(n−1)d

a₃₀ = 10+(30−1)(−3)

a₃₀ = 10+(29)(−3)

a₃₀ = 10−87 = −77

Hence, the correct answer is option C.

(ii) Given

A.P. = -3, -1/2, ,2 …

Therefore,

First term a = – 3

Common difference, d = a₂ − a₁ = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

As we know nth term of an A.P.,

a_n = a+(n−1)d

a₁₁ = 3+(11-1)(5/2)

a₁₁ = 3+(10)(5/2)

a₁₁ = -3+25

a₁₁ = 22

Hence, the answer is option B.

Question 3

Answer

(i) For the given A.P., 2,2 , 26

The first and third term are;

a = 2

a₃ = 26

As we know, nth term of an A.P.,

a_n = a+(n −1)d

a₃ = 2+(3-1)d

26 = 2+2d

24 = 2d

d = 12

a₂ = 2+(2-1)12

= 14

So, 14 is the missing term.

(ii) For the given A.P., , 13, ,3

a₂ = 13 and

a₄ = 3

As we know nth term of an A.P.,

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

– 10 = 2d

d = – 5

From equation (i) putting the value of d,we get

13 = a+(-5)

a = 18

a₃ = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

So the missing terms are 18 and 8 respectively.

<(iii) For the given A.P.,

a = 5 and

a₄ = 19/2

As we know,nth term of an A.P.,

a_n = a+(n−1)d

Therefore, putting the values here,

a₄ = a+(4-1)d

19/2 = 5+3d

(19/2) – 5 = 3d

3d = 9/2

d = 3/2

a₂ = a+(2-1)d

a₂ = 5+3/2

a₂ = 13/2

a₃ = a+(3-1)d

a₃ = 5+2×3/2

a₃ = 8

So the missing terms are 13/2 and 8 respectively.

(iv) For the given A.P.,

a = −4 and

a₆ = 6

As we know, for an A.P.,

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

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

(v) For the given A.P.,

a₂ = 38

a₆ = −22

As we know nth term of an A.P.,

a_n = a+(n −1)d

a₂ = a+(2−1)d

38 = a+d ……………………. (i)

a₆ = a+(6−1)d

−22 = a+5d …………………. (ii)

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

− 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

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

Question 4

Answer

Given the A.P. series as3, 8, 13, 18, …

First term, a = 3

Common difference, d = a₂ − a₁ = 8 − 3 = 5

n^th term of given A.P. be 78. Now ,we know,

a_n = a+(n−1)d

So,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

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

Question 5

(i) 7, 13, 19, …, 205

(ii) 18, 15

Answer

(i) Given, 7, 13, 19, …, 205 is the A.P

Therefore

First term, a = 7

Common difference, d = a₂ − a₁ = 13 − 7 = 6

n terms in this A.P.

a_n = 205

As we know, for nth term A.P.,

a_n = a + (n − 1) d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

So, this given series has 34 terms in it.

(ii) 18, 15

First term, a = 7

d = (31-36)/2 = -5/2

Let there are n terms in this A.P.

a_n = 205

As we know, for an A.P.,

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

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

Question 6

Answer

For the given series, A.P. 11, 8, 5, 2..

First term, a = 11

Common difference, d = a₂−a₁ = 8−11 = −3

Let n^th term of this A.P. is −150

As we know, for n^th term of A.P.,

a_n = a+(n−1)d

-150 = 11+(n -1)(-3)

-150 = 11-3n +3

-164 = -3n

n = 164/3

n should be positive integer but value n is fraction.

So, – 150 is not a term of this A.P.

Question 7

Answer

Given

a₁₁ = 38

a₁₆ = 73

We know that,

a_n = a+(n−1)d

a₁₁ = a+(11−1)d

38 = a+10d ............. (i)

Again,

a₁₆ = a +(16−1)d

73 = a+15d ................(ii)

On subtracting equation (i) from (ii),

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, 31^st term is 178.

Question 8

Answer

a₃ = 12

a₅₀ = 106

We know that,

a_n = a+(n−1)d

a₃ = a+(3−1)d

12 = a+2d .....................(i)

In the same way,

a₅₀ = a+(50−1)d

106 = a+49d...............(ii)

On subtracting equation (i) from (ii)

94 = 47d

d = 2

From equation (i)

12 = a+2(2)

a = 12−4 = 8

a₂₉ = a+(29−1) d

a₂₉ = 8+(28)2

a₂₉ = 8+56 = 64

So 29^th term is 64.

Question 9

Answer

Given

a₃ = 4

a₉ = −8

We know

a_n = a+(n−1)d

Therefore,

a₃ = a+(3−1)d

4 = a+2d ................(i)

a₉ = a+(9−1)d

−8 = a+8d ............... (ii)

On subtracting equation (i) from (ii)

−12 = 6d

d = −2

From equation (i)

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.

Question 10

Answer

We know that, for nth term of A.P series;

a_n = a+(n−1)d

a₁₇ = a+(17−1)d

a₁₇ = a +16d

a₁₀ = a+9d

According to question,

a₁₇ − a₁₀ = 7

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

the common difference d=1.

Question 11

Answer

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a₂ − a₁ = 15 − 3 = 12

We know

a_n = a+(n−1)d

So

a₅₄ = a+(54−1)d

⇒3+(53)(12)

⇒3+636 = 639

a₅₄ = 639

Let a_n be 132 more than its 54th term
a_n= 54th + 132
a_n= 639 +132=771.

a_n = 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.

Question 12

Answer

the first term of two APs be a₁ and a₂ respectively

And the common difference of these APs be d.

For the first A.P

a_n = a+(n−1)d

a₁₀₀ = a₁+(100−1)d

= a₁ + 99d

a₁₀₀₀ = a₁+(1000−1)d

a₁₀₀₀ = a₁+999d

For second A.P

a_n = a+(n−1)d

a₁₀₀ = a₂+(100−1)d

= a₂+99d

a₁₀₀₀ = a₂+(1000−1)d

= a₂+999d

here difference between 100^th term of the two APs = 100

So (a₁+99d) − (a₂+99d) = 100

a₁−a₂ = 100……………………………………………………………….. (i)

Difference between 1000^th terms of the two APs

(a₁+999d) − (a₂+999d) = a₁−a₂

From equation (i)

This difference, a₁−a₂ = 100

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

Question 13

Answer

We know that first number 105 and last number 994 is 3-digit number which is divisible by 7

Now the series is as follows.

105, 112, 119, …, 994

Let 994 be the nth term of this A.P.

first term, a = 105

common difference, 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

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

Question 14

Answer

The first multiple of 4 that is greater than 10 is 12.

Next multiple will be 16.

Therefore, the series formed as;

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, now;

12, 16, 20, 24, …, 248

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

first term, a = 12

common difference, d = 4

a_n = 248

As we know,

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.

Question 15

Answer

Given two APs as; 63, 65, 67,… and 3, 10, 17,….

Taking AP

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 ………………………………………. (i)

3, 10, 17, …

a = 3

d = a₂ − a₁ = 10 − 3 = 7

We know

n^th term of this A.P. = 3+(n−1)7

a_n = 3+7n−7

a_n = 7n−4 ……………………………………………………….. (ii)

Given, n^th term of these A.P.s are equal to each other.

Equating both these equations, we get,

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.

Question 16

Answer

Given,

a₃ = 16

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

a₇ − a₅ = 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

From equation (i),

a+2(6) = 16

a+12 = 16

a = 4

Therefore, A.P. will be 4, 10, 16, 22, …

Question 17

Answer

Given A.P. is 3, 8, 13, …, 253

d= 5.

Therefore, we can write the given AP in reverse order as;

253, 248, 243, …, 13, 8, 5

Now for the new AP,

first term, a = 253

and common difference, d = 248 − 253 = −5

n = 20

Therefore, using nth term formula, we get,

a₂₀ = a+(20−1)d

a₂₀ = 253+(19)(−5)

a₂₀ = 253−95

a = 158

Therefore, 20^th term from the last term of the AP 3, 8, 13, …, 253is 158.

Question 18

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 …………………………………………………… (i)

a₆+a₁₀ = 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

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.

Question 19

Answer

annual salary received by Subha Rao in the year 1995,1996,1997 is

a = 5000

d = 200

n^th year, his salary be Rs 7000.

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.

Question 20

Answer

Given that, a=5 d = 1.75. a_n =20.75 n=?

a = 5

d = 1.75

a_n = 20.75

n = ?

As we know

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.