Question

The sum of 'n' terms of an ap is 3n^2+4n. find the nth term and show that the sequence is an ap

Answer 1

$S_n=3n^2+4n\\\\we\:know,\\a_n=S_n-S_{n-1}\\\\=3n^2+4n-3(n-1)^2-4(n-1)\\\\=3[n^2-(n-1)^2]+4(n-n+1)\\\\=3[n^2-n^2+2n-1]+4\\\\=3(2n-1)+4\\\\=6n+1\\\\\\a_n=6n+1$

put n = 1 ,
$a_1=6\times\:1+1=7$
put n = 2 ,
$a_2=6\times\:2+1=13$
put n = 3,
$a_3=6\times3+1=19$
here, we see ,
$a_3-a_2=a_2-a_1=6$

e.g., common difference of series is constant so, an is in AP