Question

If the $m^{th}$ term of an AP is $20$ and $n^{th}$ term is $10$, then show that sum of its first $\frac{m+n}{2}[30+\frac{10}{m-n}]$

Answer 1

Let a & d be first term & common

difference of AP

$m^{th}$ term is 20

a+(m-1)d=20 --(1)

$n^{th}$ term is 10

a+(n-1)d=10-----(2)

sub (2) & (1)

a+(m-1) d - a (n-1) d = 20-10

=( m - 1 - n+1) d=10

= (m-n) d = 10

$\therefore d=\frac{10}{m-n}$

substituting

$a+(m-1)\frac{10}{m-n}=20$

$=a=20-\frac{(m-1)10}{m-n}$

$=\frac{20m-20n-10m+10}{m-n}=\frac{10m+10-20n}{m-n}$

sum of (m+n) terms

$\frac{(m+n)}{2}[2a+(m+n-1\left)d\right]$

$=\frac{(m+n)}{2}[\frac{2(10m+10-20n)}{m+n}+(m+n-1\left)\frac{10}{m-n}\right]$

$=\frac{m+n}{2}[\frac{20m+20-40n}{m+n}+\frac{10m+10n-1}{m-n}]$

$=\left(\frac{m+n}{2}\right)\left[\frac{20m^2+20m-40mn-20mn-20n+40n^2+10m^2+10mn-m+10mn+10n^2-n}{(m+n)-(m-n)}\right]$

= $\left(\frac{m+n}{2}\right)[30+\frac{10}{m-n}]$