If $7$ times of the $7^{th}$ term of an A.P. is equal to $11$ times its $11^{th}$ term, then find its $18^{th}$ term.

Given: $7$ times of the $7^{th}$ term of an A.P. is equal to $11$ times its $11^{th}$ term.

To do: 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, $a_11=a+( 11-1)d$

$Rightarrow a_11=a+10d\ .........\ ( ii)$

As 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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Updated on: 10-Oct-2022

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