Question

If (m+1) th term of an A.P. is twice the (n+1) th term, then show that (3m+1) th term is twice the (m+n+1) th term.

Answer 1

Given $(m+1{)}^{th}$ term of an AP is twice of $(n+1{)}^{th}$ term, Let a be first term & d be common difference $a_{m+1}=2a_{n+1}$

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

$a+md=2a+2nd$

$a+md=2a+2nd$

$d(m-2n)=a$

$d=\frac{a}{m-2n}$

Now $(3m+1{)}^{th}$ term of AP is

$a_{3m+1}=a+(3m+1-1)d=a+3md$

putting values of d we get

$a_{3m+1}=a+3m\left(\frac{a}{m-2n}\right)=\frac{am-2am+3am}{m-2n}$

$a_{3m+1}=\frac{4am-2an}{m-2n}$ ……..$\left(1\right)$

Now $(m+n+1{)}^{th}$ term of AP is

$a_{m+n+1}=a+(m+n+1-1)d$

$=a+(m+n)d=a+\frac{(m+n)a}{m-2n}$

$a_{m+n+1}=\frac{am-2an+am+an}{m-2n}$

$a_{m+n+1}=\frac{2am-an}{m-2n}$ ………..$\left(2\right)$

on comparing equation $\left(1\right)$ & $\left(2\right)$ we get

$[a_{3m+1}=2_{am+n+1}]$.