Question

If the $pth$term of an A.P. is $x$ and $qth$ term is $y,$ show that the sum of $(p+q)$ terms is $\frac{p+q}{2}[x+y+\left(\frac{x-y}{p-q}\right)]$

Answer 1

Given: $a_p=x$ and $a_q=y$
We know that,
$a_n=a+(n-1)d$
$a_p=a+(p-1)d$
$\Rightarrow x=a+(p-1)d...\left(i\right)$
now,
$a_q=a+(q-1)d$
$\Rightarrow y=a+(q-1)d...\left(ii\right)$
From equation. $\left(i\right)$ and $\left(ii\right),$ we get
$x-(p-1)d=y-(q-1)d$
$\Rightarrow x-y=(p-1)d-(q-1)d$
$\Rightarrow x-y=d[p-1-q+1]$
$\Rightarrow x-y=d[p-q]$
$\Rightarrow d=\frac{x-y}{p-q}...\left(iii\right)$
Adding Eq $\left(i\right)$ and $\left(ii\right),$ we get
$x+y=2a+(p-1)+(q-1)d$
$\Rightarrow x+y=2a+d[p+q-1-1]$
$\Rightarrow x+y=2a+d(p+q-1)-d$
$\Rightarrow x+y+d=2a+(p+q-1)d...\left(iv\right)$
We know that,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow S_{p+q}=\frac{p+q}{2}[2a+(p+q-1\left)d\right]$
$\Rightarrow S_{p+q}=\frac{p+q}{2}[x+y+d]$ [using$\left(iv\right)$]
$\Rightarrow S_{p+q}=\frac{p+q}{2}[x+y+\frac{x-y}{p-q}$ [using$\left(iii\right)$]
Hence Proved