Question

If 7 times the 7 th terms of an AP is equal to II time its 11 th terms , then find 18 th term.

Answer 1

Given:- $7a_7=11a_{11}$

To find:- $a_{18}=?$

As we know that $n^{th}$ term in an A.P. is given as-

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

Whereas, $a$ and $d$ are the first term and common difference of the A.P.

Therefore,

$a_7=a+6d$

$a_{11}=a+10d$

Now,

$7a_7=11a_{11}\left(\text{Given}\right)$

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

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

$11a+110d-7a-42d=0$

$4a+68d=0$

$4(a+17d)=0$

$a+17d=0$

$a_{18}=0(\because a_{18}=a+17d)$

Thus the ${18}^{th}$ term will be $0$.

Hence the correct answer is $0$.