Question

If A.M. between $p^{th}$ and $q^{th}$ terms of an A.P. be equal to the A.M. between $r^{th}$ and $s^{th}$ terms of the A.P., then prove that $p+q=r+s$

Answer 1

A.M between $p^{th}$ term and $q^{th}$ term $=$ A.M between $r^{th}$ term and $s^{th}$ term of the A.P.

Let 'a' be first term of A.P and 'd' be common difference.

$p^{th}$ term $=a+(p-1)d$

$q^{th}$ term $=a+(q-1)d$

$r^{th}$ term $=a+(r-1)d$

$s^{th}$ term $=a+(s-1)d$

$\frac{a+(p-1)d+a+(q-1)d}{\text{/}2}=\frac{a+(r-1)d+a+(s-1)d}{\text{/}2}$

$\text{/}2a+(p+q)d-\text{/}2d=\text{/}2a+(r+s)d-\text{/}2d$

$p+q=r+s$.