Question

If in an arithmetic progression pth terms is $\frac{1}{q}$ and qth term is $\frac{1}{p}$, then prove that
${\left(pq\right)}^{th}$ term of arithmetic progression is $1$.

Answer 1

Formula,

$t_n=a+(n-1)d$

Given,

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

$a+(q-1)d=\frac{1}{p}$........(2)

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

$\frac{p-q}{pq}=d(p-q)$

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

substituting the value of d in (1) we get,

$a=\frac{1}{pq}$

$t_{pq}=a+(pq-1)d$

$=\frac{1}{pq}+1-\frac{1}{pq}$

$=1$

Hence proved.