Question

If the $p^{th}$ term of an $A.P.$ is $1/q$ and $q^{th}$ term is $1/p$ and sum of $pq$ terms is $25pq$, then find the value of $pq$.

Answer 1

Given that, $p^{th}$ term is $\frac{1}{q}$

$\Rightarrow a_p=a+(p-1)d=\frac{1}{q}$

$\Rightarrow aq+q(p-1)d=1$

$\Rightarrow aq+(pq-q)d=1$ ...$\left(1\right)$

Given that $q^{th}$ term is $\frac{1}{p}$

$\Rightarrow a_q=a+(q-1)d=\frac{1}{p}$

$\Rightarrow ap+p(q-1)d=1$

$\Rightarrow ap+(pq-p)d=1$ ...$\left(2\right)$

From $\left(1\right),\left(2\right)$ we can say that

$\Rightarrow aq+(pq-q)d=ap+(pq-p)d$

$\Rightarrow aq-ap=d(pq-p-pq+q)$

$\Rightarrow a(q-p)=d(q-p)$

$\Rightarrow a=d$

So, equation $\left(1\right)$ becomes

$\Rightarrow dq+(pq-q)d=1$

$\Rightarrow dq+pqd-dq=1$

$\Rightarrow pqd=1$

$\Rightarrow d=\frac{1}{pq}$

$\Rightarrow Hencea=\frac{1}{pq}$

Consider $S_{pq}=\frac{pq}{2}[2a+(pq-1\left)d\right]$

$\Rightarrow S_{pq}=\frac{pq}{2}\left[2\right(\frac{1}{pq})+(pq-1\left)\right(\frac{1}{pq}\left)\right]$

$\Rightarrow S_{pq}=\frac{pq}{2}[\frac{2}{pq}+1-\frac{1}{pq}]$

$\Rightarrow S_{pq}=\frac{pq}{2}[\frac{1}{pq}+1]$

$\Rightarrow S_{pq}=\frac{1}{2}+\frac{pq}{2}$

$\Rightarrow S_{pq}=\frac{1}{2}(1+pq)$

But given that $S_{pq}=25pq$

Therefore, $\frac{1}{2}(1+pq)=25pq$

$\Rightarrow (1+pq)=50pq$

$\Rightarrow 1=49pq$

$\Rightarrow pq=\frac{1}{49}$