The $ n $th terms of two APs $ 9,7,5, \ldots $ and $ 24,21,18, \ldots $ are equal. Find the value of $ n $ and also that equal term.
Given:
The \( n \)th terms of two APs \( 9,7,5, \ldots \) and \( 24,21,18, \ldots \) are equal.
To do:
We have to find the value of \( n \) and also that equal term.
Solution:
The first A.P is \( 9,7,5, \ldots \)
$a_1 = 9, d_1 = 7-9 = - 2$
The second A.P is \( 24,21,18, \ldots \)
$b_1 = 24, d_2 = 21-24 =-3$
$n^{th}$ term of both the A.P.s are equal.
So, $a_n=b_n$
We know that,
$a_n=a+ (n-1)d$
This implies,
$a_n= 9 + (n-1)(-2)$
$b_n= 24+(n-1)(-3)$
$9 + (n-1) (-2) = 24 + (n-1) (-3)$
$9 - 2n + 2 = 24 - 3 n + 3$
$11 - 2n =27 - 3n$
$3n - 2n = 27 - 11$
$n = 16$
Therefore,
$a_{16}=9+(16-1)(-2)$
$=9+15(-2)$
$=9-30$
$=-21$
The value of $n$ is $16$ and the equal term is $a_{16}=b_{16}=-21$.
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