Question

If the $p^{th}$ term of an A.P. is $q$ and $q^{th}$ term is $p$, prove that its $n^{th}$ term is $(p+q-n)$.

Answer 1

Let $a$ be the first term and $d$ the common difference of the given A.P.

Then,
$p^{th}$ term $=q⟹a+(p-1)d=q$ ...(i)
$q^{th}$ term $=p⟹a+(q-1)d=p$ ...(ii)
Subtracting equation (ii) from equation (i), we get
$(p-q)d=(q-p)⟹d=-1$
Putting $d=-1$ in equation (i), we get
$a+(p-1)\times (-1)=q⟹a=(p+q-1)$

$\therefore n^{th}$ term $=a+(n-1)d=(p-q-1)+(n-1)\times (-1)=(p+q-n)$