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If the sum of $n$ terms of an $A.P.$ is $(pn+qn^2)$ where $p$ and $q$ are constants, then find the common difference $d$.

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

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1 Answer

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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
$S_2-2\times S_1=2a+d-2a=(2p+4q)-(2p+2q)=2q$
$\Rightarrow\:d=2q$
answered Feb 22, 2014 by rvidyagovindarajan_1
 

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