For what value of n, the nth terms of the arithmetic progressions $63, 65, 67,…$ and $3, 10, 17, …$ are equal?
Given:
nth terms of the arithmetic progressions $63, 65, 67,…$ and $3, 10, 17, …$ are equal.
To do:
We have to find the value of $n$.
 Solution:
$A_1 =63, 65, 67, .....$
Here, $d=65-63=2$
$a_{n_{1}}=a+(n−1)d$
$=63+(n−1)2$
$=63+2n−2$
$=2n+61$
$A_2=3, 10, 17, ......$
Here, $d=10-3=7$
$a_{n_{2}}=a+(n−1)d$
$=3+(n−1)7$
$=3+7n−7$
$=7n-4$
$a_{n_{1}}=a_{n_{2}}$ (Given)
$\Rightarrow 2n+61=7n-4$
$\Rightarrow 7n−2n=61+4$
$\Rightarrow 5n=65$
$\Rightarrow n=\frac{65}{5}$
$\Rightarrow n=13$
The value of $n$ is $13$.
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