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The sum of first p terms and the first term is given. We have to find next q terms. So, the total terms become p+q.\par

Hence, sum of all terms minus the sum of first p terms will give the sum of next q terms. However, sum of first p terms is zero, so the sum of next q terms will be same as sum of all terms. \par

Sum of n terms is given byS_{n}= \left( \frac{n}{2} \right) \left( 2a+ \left( n-1 \right) d \right)

required sum=S_{p+q}-S_{p}=\frac{p+q}{2} \left( 2a+ \left( p+q-1 \right) d \right) \ldots \ldots \left( i \right)

Replacing the value of d in equation (i)\par

 \[ required sum=\frac{p+q}{2} \left( 2a+ \left( p+q-1 \right)  \left( -\frac{2a}{p-1} \right)  \] \par

 \[ =\frac{p+q}{2} \left( 2a-\frac{2ap+2aq-2a}{p-1} \right)  \] \par

 \[ =\frac{p+q}{2} \left( 2a \right)  \left( 1-\frac{p-1}{p-1}-\frac{q}{p-1} \right) ~ \] \par

 \[ \text{~ On simplifying, } \] \par

 \[ =a \left( p+q \right)  \left( -\frac{q}{p-1} \right) =-\frac{a \left( p+q \right) q}{p-1} \] \par

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