Question

If sum of first n terms of an AP is given by formula 3n square+2n then it's second term is?

Answer 1

Given,
$S_{n}=3n^2+2n$
we know,
$T_n=S_n-S_{n-1}$

we Tn shows nth term of AP , Sn and S_(n-1) shows the sum of nth term and (n-1)th term respectively.

now,
$T_n= 3n^2+2n-3(n-1)^2-2(n-1)\\=3[n^2-(n-1)^2]+2[n -n+1]\\=3[n^2-n^2+2n-1]+2[1]\\=3(2n-1)+2=6n-1$
2nd term of AP = 6 × 2 - 1 = 11