Question

Write the common difference of an A.P. the sum of whose first n terms is $\frac{P}{2}{n}^2+Qn$

Answer 1

We have,

$S_n=\frac{P}{2}{n}^2+Qn$

$\therefore S_2=\frac{P}{2}(2{)}^2+2Q$

$\Rightarrow S_2=2P+2Q$

$S_3=\frac{P}{2}(3{)}^2+3Q$

$\Rightarrow S_3=\frac{9p}{2}+3Q$

and, $S_4=\frac{P}{2}(4{)}^2+4Q$

$\Rightarrow S_4=8P+4Q$

Now,

$T_4=S_4-S_3=8P+4Q-\frac{9p}{2}-3Q$

$\Rightarrow T_4=\frac{7P}{2}+Q$

and,

$T_3=S_3-S_2=\frac{9p}{2}+3Q-2P-2Q$

$=\frac{5P}{2}+Q$

$\therefore T_4-T_3=\frac{7P}{2}+Q-\frac{5P}{2}-Q$

$=\frac{7P}{2}-\frac{5P}{2}$

$=\frac{2P}{2}$

$=P$

$\therefore T_4-T_3=P$

$\therefore$ Common difference = P.