If the $ n $th terms of the two APs: $ 9,7,5, \ldots $ and $ 24,21,18, \ldots $ are the same, find the value of $ n $. Also find that 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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Updated on: 10-Oct-2022

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