Question

the sum of first 6 terms of ap is 96 and first 10 terms is 240. Find tge sum of nth term in ap

Answer 1

I am not pretty sure but i hope that's the right answer. till 'a' it is definitely right

Answer 2

$S_{n} = n[2n + 4]$

Step-by-step explanation:

Let the A.P. be a, a + d, a + 2d, a + 3d, ........

Now, the sum of first n terms of the A.P. is given by

$S_{n} = \frac{n}{2}[a + (n - 1)d]$

Therefore, the sum of first 6 terms of the A.P. will be given by

$S_{6} = \frac{6}{2}[a + (6 - 1)d] = 3a + 15d = 96$ {Given}

⇒ a + 5d = 32 ......... (1)

Again, the sum of first 10 terms of the A.P. will be given by

$S_{10} = \frac{10}{2}[a + (10 - 1)d] = 5a + 45d = 240$ {Given}

⇒ a + 9d = 48 ......... (2)

Now, solving equations (1) and (2) we get,

4d = 16

⇒ d = 4

Now, from equation (1) we get,

a = 32 - 5d = 12

Therefore, the sum of first n terms of the A.P. will be

$S_{n} = \frac{n}{2}[12 + (n - 1) 4] = n[2n + 4]$ (Answer) {Since a = 12 and d = 4}