Question

the sum of first n terms of an A.P is n/2*(3n+13)

Answer 1

S1=a1
 => a1=8
a2=S2-S1
    =19-8
    = 11

d = 11-8
   = 3

a30= 8+29(3)
       = 95

Answer 2

Given sn =  n/2 * (3n + 13).

Take n = 1.

s1 = 1/2 * (3(1) + 13)

     = 1/2(3 + 13)

     = 1/2(16)

     = 8.

So the first term of AP = 8.

Sum of 1st term will be the first term.

So first term a = 8.


Take n = 2.

s2 = 2/2 * (3(2) + 13)

     = 1 * (6 + 13)

    = 19.


Sum of first 2 terms is 19.

First term + second term = 19

8 + a2 = 19

a2 = 19 - 8

     = 11.



Hence in AP,

First term a = 8.

Common difference d = 11 - 8

                                     = 3.


Now,

We know that sum of n terms of an AP an = a + (n - 1) * d

30th term a30 = 8 + (30 - 1) * 3

                         = 8 + 29 * 3

                         =  8 + 87

                        = 95.



Hope this helps!