Question

The $8^{th}$ terms of an AP is zero. Prove that its ${38}^{th}$ term is triple its ${18}^{th}$ term.

Answer 1

$n^{th}$ term $t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

${38}^{th}$ term$=a+(38-1)d$

${18}^{th}$ term$=a+(18-1)d$

$8^{th}$ term$=a+7d$

$a+7d=0$

$a=-7d$

${38}^{th}$ term$=-7d+37d=30d$

${18}^{th}$ term$=-7d+17d=10d$

${38}^{th}$ term$=3\left({18}^{th}term\right)$