If the nth term of the A.P. $9, 7, 5, …$ is same as the nth term of the A.P. $15, 12, 9, …$ find $n$.

Given:

nth term of the A.P. $9, 7, 5, …$ is same as the nth term of the A.P. $15, 12, 9, …$

To do:

We have to find the value of $n$.

Solution:

We know that,

$a_{n}=a+(n-1)d$
Therefore,

In the A.P. $9, 7, 5, …$,

$a_1=a=9, a_2=7, a_3=5$ and $d=a_2-a_1=7-9=-2$

$a_n=9+(n-1)(-2)$

$a_n=9+(-2)n-1(-2)$

$a_n=9-2n+2$

$a_n=11-2n$

In the A.P. $15, 12, 9, …$,

$a_1=a=15, a_2=12, a_3=9$ and $d=a_2-a_1=12-15=-3$

$a_n=15+(n-1)(-3)$

$a_n=15+(-3)n-1(-3)$

$a_n=15-3n+3$

$a_n=18-3n$

This implies,

$11-2n=18-3n$

$3n-2n=18-11$

$n=7$

The value of $n$ is $7$.

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Updated on: 10-Oct-2022

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