Choose the correct answer from the given four options:
If 7 times the $ 7^{\text {th }} $ term of an AP is equal to 11 times its $ 11^{\text {th }} $ term, then its 18 th term will be
(A) 7
(B) 11
(C) 18
(D) 0
Given:
$7$ times of the $7^{th}$ term of an A.P. is equal to $11$ times its $11^{th}$ term.
To do:
We have to find its $18^{th}$ term.
Solution:
Let $a$ be the first term and $d$ be the common difference of the A.P.
Therefore, $a_7=a+( 7-1)d$
$\Rightarrow a_7=a+6d\ ........\ ( i)$
Similarly,
$a_{11}=a+( 11-1)d$
$\Rightarrow a_{11}=a+10d\ .........\ ( ii)$
Given,
$7.a_7=11.a_{11}$
$\Rightarrow 7( a+6d)=11( a+10d)$
$\Rightarrow 7a+42=11a+110d$
$\Rightarrow 11a-7a+110d-42=0$
$\Rightarrow 4a+68d=0$
$\Rightarrow 4( a+17d)=0$
$\Rightarrow a+17d=0\ ........\ ( iii)$
$\therefore a_{18}=a+( 18-1)d$
$\Rightarrow a_{18}=a+17d$
$\Rightarrow a_{18}=0$ [From $( iii)$]
Thus, $18^{th}$ term of the A.P. is $0$.
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