$\begin{array}{1 1}p \\ q\\ 2p \\2q\end{array} $

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- Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
- $t_n=a+(n-1)d$

Given: Sum of $n$ terms of an A.P.=$S_n=pn+qn^2$

$\therefore\:$$S_1=a=1^{st}\:term=p+q$......(i)

$S_2=$Sum of first two terms.$=t_1+t_2=a+(a+d)=2p+4q$

$\Rightarrow\:2a+d=2p+4q$.....(ii)

Subtracting (ii)-$2\times $(i) we get

$S_2-2\times S_1=2a+d-2a=(2p+4q)-(2p+2q)=2q$

$\Rightarrow\:d=2q$

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