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# If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

10.   If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

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Let first term of AP = a and common difference = d.

Then,

$S_p=\frac{p}{2}[2a+(p-1)d]$

$S_q=\frac{q}{2}[2a+(q-1)d]$

Given : $S_p=S_q$

$\Rightarrow \frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$

$\Rightarrow p[2a+(p-1)d]=q[2a+(q-1)d]$

$\Rightarrow 2ap+p^2d-pd=2aq+q^2d-qd$

$\Rightarrow 2ap+p^2d-pd-2aq-q^2d+qd=0$

$\Rightarrow 2a(p-q)+d(p^2-p-q^2+q)=0$

$\Rightarrow 2a(p-q)+d((p-q)(p+q)-(p-q))=0$

$\Rightarrow 2a(p-q)+d[(p-q)(p+q-1)]=0$

$\Rightarrow (p-q)[2a+d(p+q-1)]=0$

$\Rightarrow 2a+d(p+q-1)=0$

$\Rightarrow d(p+q-1)=-2a$

$\Rightarrow d=\frac{-2a}{p+q-1}$

Now, $S_(_p_+_q_)=$ $\frac{p+q}{2}[2a+(p+q-1)d]$

$=\frac{p+q}{2}[2a+(p+q-1)\frac{-2a}{p+q-1}]$

$=\frac{p+q}{2}[2a+(-2a)]$

$=\frac{p+q}{2}[0]=0$

Thus, sum of p+q terms of AP is 0.

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