Question

The pth term of an A.P. is a and qth term is b. Prove that sum of its $(p+q)$ terms is
$\frac{p+q}{2}[a+b+\frac{a-b}{p-q}]$.

Answer 1

Let the first term be A and common difference D.
$S_{p+1}=\frac{p+q}{2}[2A+(p+q-1\left)D\right]$ $\left(1\right)$
It is given that $T_p=a$ and $T_q=b$
$\therefore \left\{\begin{array}{c}A+(p-1)D=a\\ A+(q-1)D=b\end{array}\right\}$
Substracting, $(p-q)D=a-b$
$\therefore D=\frac{a-b}{p-q}$ $\left(2\right)$
Adding, $2A+(p+q-2)D=a+b$
or $2A+(p+q-1)D=a+b+D$
$=a+b+\frac{a-b}{p-q}$, by $\left(2\right)$ ..$\left(3\right)$
Hence from $\left(1\right)$ and $\left(3\right)$,
$S_{p+q}=\frac{p+q}{2}[a+b+\frac{a-b}{p-q}]$.