Question

The $8^{th}$ term of an A.P. is zero. Prove that ${38}^{th}$ term is triple the ${18}^{th}$ term.

Answer 1

Solution:-

$a_8=0\left\{Given\right\}$

$\Rightarrow a+7d=0\{\because a_n=a+(n-1)d\}$

To prove:- $a_{38}=a_{18}$

Proof:-

R.H.S.-

$a_{18}=a+17d$

$\Rightarrow =a+7d+10d\{\because 17d=7d+10d\}$

$\Rightarrow =(a+7d)+10d$

$\Rightarrow =0+10d\{\because a+7d=0\left(Given\right)\}$

$\Rightarrow a_{18}=10d$

L.H.S.-

$a_{38}=a+37d$

$\Rightarrow =a+7d+30d\{\because 37d=7d+30d\}$

$\Rightarrow =(a+7d)+30d$

$\Rightarrow =0+30d\{\because a+7d=0\left(Given\right)\}$

$\Rightarrow a_{38}=30d=3\times 10d=3\times a_{18}\{\because a_{18}=10d\}$

Hence proved that ${38}^{th}$ term is triple the ${18}^{th}$ term.