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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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