Question:
If 7 times the $7^{\text {th }}$ term of an AP is equal to 11 times its $11^{\text {th }}$ term, show that the $18^{\text {th }}$ term of the AP is zero.
Solution:
Show that: $18^{\text {th }}$ term of the AP is zero.
Given: $7 a_{7}=11 a_{11}$
(Where $a_{7}$ is Seventh term, $a_{11}$ is Eleventh term, $a_{n}$ is nth term and $d$ is common difference of given AP)
Formula Used: $a_{n}=a+(n-1) d$
$7(a+6 d)=11(a+10 d)$
$7 a+42 d=11 a+110 d \rightarrow 68 d=(-4 a)$
$a+17 d=0 \ldots .$ equation (i)
Now $a_{18}=a+(18-1) d$
So a + 17d = 0 [by using equation (i)]
HENCE PROVED
[NOTE: If $n$ times the $n^{\text {th }}$ term of $A P$ is equal to $m$ times the $m^{\text {th }}$ term of same AP then its $(m+n)^{\text {th }}$ term is equal to zero]