Question

If $p^{th}$ term of an AP is $a$ and its $q^{th}$ term is $b$, then what is the sum of the first $(p+q)$ terms of the AP?

Answer 1

Let A and D be the first term and the common difference of the A.P, then
$T_p=a=A+(p-1)D$ ...... (i)

$T_q=b=A+(q-1)D$ ...... (ii)

By (i) - (ii), we have $T_p-T_q=a-b=(p-q)D$.......(iii)

$\Rightarrow D=\frac{a-b}{p-q}$

Using this in (i) gives

$A=a-(p-1)\left(\frac{a-b}{p-q}\right)$

Since, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Now, the sum of the first $(p+q)$ terms is $S_{p+q}=\left(\frac{p+q}{2}\right)(2A+(p+q-1\left)D\right)$

$=\left(\frac{p+q}{2}\right)\left(2\right(a-(p-1)\left(\frac{a-b}{p-q}\right))+(p+q-1\left)\left(\frac{a-b}{p-q}\right)\right)$

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

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

$\Rightarrow S_{p+q}=\frac{p+q}{2}[a+(a-\frac{a-b}{p-q}(p-q))+\frac{a-b}{p-q}]$......(iv)

From equation(iii), $a-b=(p-q)D$, where $D=\frac{a-b}{p-q}$

$\Rightarrow a-(p-q)\frac{a-b}{p-q}=b$

Substitute above equation in equation(iv), we get

$S_{p+q}=\left(\frac{p+q}{2}\right)(a+b+\frac{a-b}{p-q})$