# Question-1

**Find the sum of the following A.P. 1, 3, 5, 7, ............,199.****Solution:**

Given, a = 1, d = 2 , a

_{n}= l = 199,

a + (n â€“ 1) d = 199

1 + (n â€“ 1) 2 = 199

â‡’ 1 + 2n â€“ 2 = 199

â‡’ 2n = 200

âˆ´ n =

n = 100.

S

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

= 50(1 + 199)

= 50(200)

= 10000

# Question-2

**Find the A.P. whose 10**^{th}term is 5 and 18^{th}term is 77.**Solution:**

given, 10

^{th}term of an A.P= 5

Ãž a + (10 â€“ 1) d = 5

â‡’ a + 9d = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

and 18

^{th}term = 77

Ãža + (18 â€“ 1) d = 77

â‡’ a + 17d = 77 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

(ii) â€“ (i), 8d = 72

âˆ´ d = 9

Substituting the value of d = 9 in (i),

a + 81 = 5

a = 5 â€“ 81 = - 76

âˆ´ The A.P. is â€“ 76, - 67, â€¦â€¦â€¦â€¦.

# Question-3

**In a certain A.P the 24**^{th}term is twice the 10^{th}term. Prove that the 72^{nd}term is twice the 34^{th}term.**Solution:**

Given, a

_{24}

^{ }= 2a

_{10 }

a

_{24}

^{ }= a + 23d and a

_{10}

^{ }= a + 9d

To prove: a

_{72}

^{ }= 2 a

_{34}

a

_{72}

^{ }= a + 71d

a

_{34}

^{ }= a + 33d

a

_{24}

^{ }= 2a

_{10 }(Given)

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a â€“ 5d = 0

a = 5dâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

a

_{72}= 2 a

_{34}

a + 71d = 2(a + 33d)

a + 71d = 2a + 66d

a â€“ 5d = 0

a = 5dâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

from, (1) and (2) a

_{72}= 2 a

_{34}

Hence proved.

# Question-4

**a, b and c are in A.P. Prove that b + c, c + a and a + b are in A.P.****Solution:**

Given, a, b and c are in A.P.

âˆ´ b â€“ a = c â€“ b

To prove :

**b + c, c + a and a + b are in A.P.**

c + a â€“ (b + c) = a + b â€“ (c + a)

â‡’ c + a â€“ b â€“ c = a + b â€“ c â€“ a

a â€“ b = b â€“ c

â‡’ b â€“ a = c â€“ b

âˆ´ a, b, c are in A.P.

âˆ´ b + c, c + a and a + b are in A.P.

# Question-5

**If 9**

^{th}term of an A.P. is zero, prove that its 29^{th}term is double the 19^{th}term.**Solution:**

9

^{th}term = 0

a

_{1}+ 8d = 0

a

_{29 }= a

_{1}+ 28d = a

_{1}+ 8d + 20d = 0 + 20d = 20d

a

_{19}= a

_{1}+ 18d = a

_{1}+ 8d + 10d = 0 + 10d = 10d

a

_{29 }= 2a

_{19}.

# Question-6

**Determine the A.P whose third term is 16 and the difference of 5**^{th}from 7^{th}term is 12.**Solution:**

Let the A.P. be a, a + d, a + 2d, â€¦â€¦..

â‡’ The third term = a

_{3}= a + 2d = 16 â€¦â€¦â€¦â€¦â€¦â€¦(i)

and seventh term = a

_{7}= a + 6d

Given that a

_{7}â€“ a

_{5 }= 12

â‡’ (a + 6d) â€“ (a + 4d) = 12

â‡’ a + 6d â€“ a â€“ 4d = 12

â‡’ 2d = 12

â‡’ d = 6

Substituting the value of d = 6 in (i),

a + 12 = 16

a = 4

âˆ´ The first term of the A.P. is 4 and the common difference is 6.

âˆ´ The A.P. is 4, 10, 16, 22, 28, 34, â€¦

âˆ´ The fifth term = a

_{5}= a + 4d.

# Question-7

**The sum of the first six terms of an A.P is zero and the fourth term is 2. Find the sum of its first 30 terms.****Solution:**

Let the sum of first 30 terms be S

_{30}, first term be a, fourth term be a

_{4}and the sum of first six terms be S

_{6}.

Given that S

_{6}= 0 and fourth term a

_{4}= 2

â‡’ a + 3d = 2 â€¦â€¦â€¦â€¦..(i)

S

_{6}= 0

= 0

â‡’ 2a + 5d = 0 â€¦â€¦â€¦â€¦..(ii)

(i) Ã— 2,

2a + 6d = 4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

(iii) â€“ (ii),

âˆ´ d = 4

Substituting the value of d = 4 in (i),

a + 3 Ã— (4) = 2

â‡’a = 2 - 12 = - 10

âˆ´ a

_{30}= a + 29d

= - 10 + 29 Ã— (4)

= â€“ 10 + 116

= 106

âˆ´ Sum to first 30 terms = S

_{30}= (a + l)

=(- 10 + 106)

= 15 Ã— 96

= 1140.

# Question-8

**An A.P consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32**^{nd}term.**Solution:**

Given, n = 60, a

_{1}= 7,

and a

_{60}= 125

â‡’ a

_{1}+ 59d = 125

7 + 59d = 125

59d = 118

d = 118/59 = 2

a

_{32}= a

_{1}+ 31d = 7 + 31(2) = 7 + 62

âˆ´ a

_{32}= 69.

# Question-9

**Find the sum of the series 51 + 50 + 49 + ..... + 21**

**.**

**Solution:**

51 + 50 + 49 + ..... + 21

a = 51, d = - 1, a

_{n}= 21

âˆ´ a + (n â€“ 1) d = a

_{n}

51 + (n â€“ 1) (- 1) = 21

(n â€“ 1) (- 1) = 21 â€“ 51

n â€“ 1 = 30

âˆ´ n = 31

âˆ´ Sum of the series = (51 + 21)

= Ã— 72

= 1116

âˆ´ The sum of the series 51 + 50 + 49 + ..... + 21 = 1116.

# Question-10

**Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.****Solution:**

Let the three numbers be a â€“ d, a, a+ d.

a â€“ d + a + a + d = 27

3a = 27

a = 9

(a â€“ d)(a)(a + d) = 648

a(a

^{2}â€“ d

^{2}) = 648

9(9

^{2}â€“ d

^{2}) = 648

9

^{2}â€“ d

^{2}= 72

d

^{2}= 81 - 72

d

^{2}= 9

d = 3

The numbers are 6, 9, 12.

# Question-11

**How many terms of A.P -10, -7, -4, -1, ........ must be added to get the sum -104?****Solution:**

-10, -7, -4, -1, ........

a = - 10, d = 3

S

_{n}= n{2a + (n â€“ 1) d}

- 104 = n {2(-10) + (n â€“ 1) 3}

= n (- 20 + 3n â€“ 3)

-208 = n (3n â€“ 23)

3n

^{2}â€“ 23n + 208 = 0

3n

^{2}â€“ 39n + 16n + 208 = 0

3n (n â€“ 13) + 16 (n â€“ 13) = 0

(n â€“ 13) (3n + 16) = 0

âˆ´ n = 13

âˆ´ 13 terms must be added to get the sum of the A.P â€“ 104.

# Question-12

**If the sum of p terms of an A.P is 3p**^{2}+ 4p, find its n^{th}term.**Solution:**

S

_{p}= 3p

^{2}+ 4p

t

_{n}= S

_{n}- S

_{n-1 }= (3n

^{2}+ 4n) â€“ [3(n - 1)

^{2}+ 4(n - 1)]

= (3n

^{2}+ 4n) â€“ [3(n

^{2}â€“ 2n + 1) + 4(n - 1)]

= (3n

^{2}+ 4n) â€“ [3n

^{2}â€“ 6n + 3 + 4n - 4]

= (3n

^{2}+ 4n) â€“ [3n

^{2}â€“ 2n - 1]

= 3n

^{2}+ 4n â€“ 3n

^{2}+ 2n + 1

= 6n + 1

Therefore the n

^{th}term is 6n + 1.