For what value of n, the nth term of two APs: 63, 65, 61,… 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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Updated on: 10-Oct-2022

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