Question

The $8th$ term of an $A.P$ is $0$ prove that its ${38}^{th}$ term is triple its ${18}^{th}$ term.

Answer 1

We know $a_n=a+(n-1)d$

$a+7d=0$ [ Given ]

$\therefore$ $a=-7d$ -----( 1 )

For the ${38}^{th}$ term,

$a_{38}=a+(38-1)d$

$a_{38}=a+37d$

$a_{38}=-7d+37d$ [ From ( 1 ) ]

$\therefore$ $a_{38}=30d$ ----- ( 2 )

For ${18}^{th}$ term,

$a_{18}=a+(18-1)d$

$a_{18}=a+17d$

$a_{18}=-7d+17d$

$\therefore$ $a_{18}=10d$ ----- ( 3 )

From ( 2 ) and ( 3 ),

$a_{38}=3a_{18}$ ---- Hence proved