Let there be an A.P. with first term ‘a’, common difference, d. Ifdenotes its n^th term andthe sum of first n terms, find.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i)

Given, a_n= 50

a + (n – 1) d = 50

5 + (n – 1) 3 = 50

(n – 1)3 = 45

(n -1) = 15

n = 16

S_n= [a + l]

= 8 [5 + 50] = 8 [55] = 440

(ii)

a_n= 4, d = 2, S_n= -14

a + (n – 1) 2 = 4

a + 2n = 6, and

[2a + (n – 1)2] = -14

n [ 2a + 2n – 2] = -14, or

[a + a_n] = -14

[a + 4] = -14

n [6 -2n +4] = -28

n [10 -2n] = -28

2n² – 10n – 28 = 0

2(n² – 5n – 14) = 0

(n + 2) (n – 7) = 0

n = -2, n = 7

Therefore, n = -2 is not a natural number. So, n = 7

(iii)

Given, a = 3, n = 8, S_n= 192

S_n = {2a + (n – 1) d]

192 * 2 = 8 [6 + (8 – 1)d]

= 6 + 7d

48 = 6 + 7d

7d = 42

d = 6

(iv)

a_n= 28, S_n = 144, n = 9

S_n= [a + l]

144 = [a + 28]

= a = 28

a + 28 = 32

a = 4

(v)

Given, a = 8, a_n= 62 and S_n= 210

S_n= [a + l]

210 = [8 + 62]

210 * 2 = n (70)

n = = 6

a + (n – 1) d = 62

8 + (6 – 1) d = 62

5d = 54

d = 10.8

(vi)

Given, a = 2, d = 8 and S_n = 90

90 = [4 + (n – 1)d] {Therefore, S_n= [2a + (n – 1)d]}

180 = n [4 + 8n – 8]

8n² – 4n – 180 = 0

4 (2n² – n – 45) = 0

2n² – n – 45 = 0

(2n + 1) (n – 5) = 0

Therefore, n = is not a natural number

n = 5

a_n= a + (n – 1)d

= 2 + 4(8) = 32