Question

If the nth term of an A.P. is 3n+2, find the sum of first 'n terms of that AP.​

Answer 1

Sn = (3/2)n² + (7/2)n

Explanation:

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

a = 3(1)+2 = 5

d = second term - 5 = 3(2)+2 - 5 = 3

Sn = (n/2)[2(5) + (n-1)(3)]

= (n/2)[10 + 3n - 3]

= (n/2)(7 + 3n)

Sn = (3/2)n² + (7/2)n

Answer 2

`S_(n)` =
`(n)/(2)` (3n + 7)

Explanation:

We require to find the first term a₁ and the common difference d

The n th term is given by 3n + 2, thus

a₁ = 3(1) + 2 = 3 + 2 = 5

a₂ = 3(2) + 2 = 6 + 2 = 8

d = 8 - 5 = 3

`S_(n)` =
`(n)/(2)` [ 2a₁ + (n - 1)d ], substitute values

`S_(n)` =
`(n)/(2)` [ (2 × 5) + 3(n - 1) ] =
`(n)/(2)` (10 + 3n - 3) =
`(n)/(2)` (3n + 7)