If 7 times the 7th term of an AP is equal to 11 times its 11th term,

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]