Browse Questions

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

$\begin{array}{1 1}0 \\ p+q \\ q \\p\end{array}$

Toolbox:
• Sum of first $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d] Given that the sum of first p terms=sum of first q terms of an A.P. \Rightarrow\:S_p=S_q \Rightarrow\:\large\frac{p}{2}$$[2a+(p-1)d]=\large\frac{q}{2}$$[2a+(q-1)d] \Rightarrow\:p[2a+(p-1)d]-q[2a+(q-1)d]=0 \Rightarrow\:2a(p-q)+d(p^2-p-q^2+q)=0 \Rightarrow\:2a(p-q)+d[(p^2-q^2)-(p-q)]=0 p^2-q^2=(p-q)(p+q) \Rightarrow\:(p-q)[2a+d(p+q-1)]=0..........(A) But we know that sum of first p+q terms= S_{p+q}=\large\frac{p+q}{2}$$[2a+(p+q-1)d]$
$\therefore\:$from (A) $S{p+q}=0$