Write the common difference of an A.P. the sum of whose first n t…

Question

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

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Answer

We have,
$\text{S}_\text{n}=\frac{\text{p}}{2}\text{n}^2+\text{Qn}$
$\therefore\text{S}_2=\frac{\text{p}}{2}(2)^2+\text{Qx}^2$
$\Rightarrow\text{S}_2=2\text{p}+2\text{Q}$
$\text{S}_3=\frac{\text{p}}{2}(3)^2+3\text{Q}$
$\Rightarrow\text{S}_3=\frac{\text{qp}}{2}+3\text{Q}$
and,
$\text{S}_4=\frac{\text{P}}{2}(4)^2+4\text{Q}$
$\Rightarrow\text{S}_4=8\text{P}+4\text{Q}$
Now,
$\text{T}_4=\text{S}_4-\text{S}_3=8\text{p}+4\text{Q}-\frac{\text{pq}}{2}-3\text{Q}$
$\Rightarrow\text{T}_4=\frac{7\text{P}}{2}+\text{Q}$
and,
$\text{T}_3=\text{S}_3-\text{S}_2=\frac{\text{pq}}{2}+3\text{Q}-2\text{P}-2\text{Q}$
$=\frac{5\text{P}}{2}+\text{Q}$
$\therefore\text{T}_4-\text{T}_3=\frac{7\text{p}}{2}+\text{Q}-\frac{5\text{p}}{2}-\text{Q}$
$=\frac{7\text{P}}{2}-\frac{5\text{P}}{2}$
$=\frac{2\text{P}}{2}$
$=\text{P}$
$\therefore\text{T}_4-\text{T}_3=\text{P}$
$\therefore$ Common difference = P.

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