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# 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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