Question

If $7$ times the$7^{th}$ term of an A.P $\ge$ $11$ times its ${11}^{th}$ term , then find its ${18}^{th}$ term.

Answer 1

$7$th term in AP$=a+(7-1)d$

$=a+6d$

$11$th term in AP$=a+(11-1)d$

$=a+10d$

Since $7$ times the $7$th term of A.P$\ge 11$ times its ${11}^{th}$ time

$(a+6d)7=11(a+10d)$

$7a+42d=11a+110d$

$7a-11a=110d-42d$

$-4a=68d$

$a=\frac{68d}{-4}$

$a=-17d$ ………$\left(1\right)$

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

$=a+17d$

Sub the value of "a"

$=-17d+17d$

$=0$

$\therefore$ The ${18}^{th}$ term is $0$.