Question

Question 12
If 7 times the $7^{th}$ term of an AP is equal to 11 times its ${11}^{th}$ term, then its ${18}^{th}$ term will be

A) 7
B) 11
C) 18
D) 0

Answer 1

According to the question,
$7a_7=11a_{11}$
$\Rightarrow 7[a+(7-1\left)d\right]=11[a+(11-1\left)d\right][\because a_n=a+(n-1\left)d\right]$
$\Rightarrow 7(a+6d)=11(a+10d)$
$\Rightarrow 7a+42d=11a+110d$
$\Rightarrow 4a+68d=0$
$\Rightarrow 2(2a+34d)=0[\because 2\ne 0]$
$\Rightarrow 2a+34d=0$
$\Rightarrow a+17d=0\cdots \left(i\right)$
$\therefore$ 18th term of the AP,
$a_{18}=a+(18-1)d$
$=a+17d=0[fromeq.(i\left)\right]$