Question

In a certain A.P. ${32}^{th}$ term is twice the ${12}^{th}$ term. Prove that ${70}^{th}$ term is twice the ${31}^{th}$ terms.

Answer 1

General terms of an A.P is $t_n=a+(n-1)d$

Given,

$a+(32-1)d=2[a+(12-1\left)d\right]$

$a+31d=2a+22d$

$a=9d$

$T_{31}=a+(31-1)d$

$=9d+30d$

$=39d$

$T_{70}=a+(70-1)d$

$=9d+69d$

$=78d$

$=2\times 39d$

$=2T_{31}$

Hence proved.