Question

If the sum of first $n$ terms of an $A.P.$ is given by $Sn=6n+7n^2$, then find the $10th$ term of $A.P.$

Answer 1

$S_n=6n+{7n}^2$

If $n=1,$

$S_1=6+7=13$

$S_2=6\left(2\right)+7\left(4\right)=40$

$a_1=13=a$

$a_1+a_2=40$

$a+a+d=40$

$2a+d=40$

$2\times 13+d=40d=14.$

$a_{10}=a+9d=13+126=139$