Question

the sum of the first n terms of an ap is 3n²+6n.find the nth term of this ap

Answer 1

I hope it'll help you and I think it's appropriate

Answer 2

Hi

Here is ur answer

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

Given sn = 3n² + 6n

Therefore, n/2[2a+(n-1)d] = 3n² + 6n

n/2[2a+(n-1)d] = n (3n + 6)

2a + nd - d = 2n ( 3n + 6 ) / n

2a + nd - d = 2 (3n + 6)

2a + nd - d = 6n + 12

2a + (n-1)d = 18 + 6n - 6

 2a + (n-1)d = 18 + 6(n-1)

Compare both sides...

2a = 18

a = 9

Therefore, d = 6

Then,

an = a + ( n - 1 ) d

an = 9 + ( n - 1 ) 6

    = 9 + 6n - 6

    = 3 + 6n

Hope it helps U