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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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