Question

The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.

Answer 1

Here a = 20 and S₇ = 2100

We know that,

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

`=> S_7 = 7/2 [2 xx 20 + (7 - 1)d]`

`=> 2100 = 7/2 (40 + 6d)`

`=> 4200 = 7(40 + 6d)`

`=> 600 = 40 + 6d`

`=> d = 560/6`

To find: t₃₁ = ?

t_n = a + (n – 1)d

`=> t_31 = 20 + (31 - 1) 560/6`

= `20 + 30 xx 560/6`

= 20 + 5 × 560

= 2820

Therefore, the 31^st term of the given A.P. is 2820.