Question

Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.

Answer 1

Given T₃ = 16

T₇ = T₅ + 12

We know that,

T_n = a + (n – 1)d

T₃ = a + (3 – 1)d

⇒ a + 2d = 16   ...(1)

T₇ = T₅ + 12

a + (7 – 1)d = a + (5 – 1)d + 12

a + 6d = a + 4d + 12

6d – 4d = 12

2d = 12

d = 6

From equation (1)

a + 2(b) = 16

a + 12 = 16

a = 16 – 12

a = 4

∴ AP is 4, 10, 16, 22 .....

We know `S_n = n/2 [2a + (n - 1)d]`

`S_29 = 29/2 [2 xx 4 + (29 - 1) xx 6]`

= `29/2 [8 + 168]`

= `29/2 xx 176`

= 29 × 88

= 2552