Question

The sum of the first n terms of an AP is ( 3n^2 + 6n ). Find the nth term and the 15th term of this AP.

Answer 1

Hope you like my process
=======================
Formula to be used
=-=-=-=-=-=-=-=-=-=-=-
$= > sum = \bf\frac{n}{2} (2a + (n - 1)d) \\ \\ = > n {}^{th} \: \: term \: = \bf a + (n - 1)d \\ \\ where \: \: \bf \: a = \it \: first \: \: term \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf d = \it \: difference$

___________________________
Given
-----------
$= > sum \: = (3 {n}^{2} + 6n) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{n}{2} (6n + 12) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{n}{2} (18 + (n - 1)6) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \frac{n}{2} (2(9) + (n - 1)6)$

thus

comparing we get
----------------------------
=> a = 9

=> d = 6
____________________

Hence .

$= > {n}^{th } \: \: term \: = \bf \: a + (n - 1)d \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 + (n - 1)6 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 + 6n - 6 = \bf \underline{3 + 6n}$

=>15th term = 3 +6(15)= 3+90 =93

===========================
Hope this is ur required answer

Proud to help u.

Answer 2

Given, Sum of first n terms of an AP is Sn = 3n^2 + 6n.

When n = 1:

⇒ S1 = 3(1)^2 + 6(1)

        = 3 + 6

        = 9.

When n = 2:

⇒ S2 = 3(2)^2 + 6(2)

         = 12 + 12

         = 24.

First term, a1 = 9.

Second term, a2 = 24 - 9

                            = 15.

Common difference d = 15 - 9 = 6.

nth term of the AP:

⇒ an = a + (n - 1) * d

         = 9 + (n - 1) * 6

         = 9 + 6n - 6

         = 6n + 3.

15th term of the AP:

⇒ a15 = a + (15 - 1) * d

          = 9 + (15 - 1) * 6

          = 9 + 84

          = 93.

Therefore, the nth term of the AP is 6n + 3 and 15th term of the AP is 93.

Hope this helps!