Question

If $7$ times $7^{th}$ term of an $AP$ is equal to $11$ times the ${11}^{th}$ term, find its ${18}^{th}$ term.

Answer 1

Given $7a_7={}^{11}{a}_{11}$

we know that $T_n=a+(n-1)d$

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

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

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

$4a=-68d$

$4a+68d=0$

$4(a+17d)=0$

$a+17d=0$

$a+(18-1)d=0$

$\therefore a_{18}=0$