For what value of n, the nth terms of the arithmetic progressions
Question:

For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, … and 3, 10, 17, … are equal?

Solution:

Here, we are given two A.P. sequences. We need to find the value of n for which the nth terms of both the sequences are equal. We need to find n

So let us first find the nth term for both the A.P.

First A.P. is 63, 65, 67 …

Here,

First term (a) = 63

Common difference of the A.P. $(d)=65-63=2$

Now, as we know,

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

So, for nth term,

$a_{n}=63+(n-1) 2$

$=63+2 n-2$

 

$=61+2 n$…….(1)

Second A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P $(d)=10-3=7$

Now, as we know,

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

So, for nth term,

$a_{n}=3+(n-1) 7$

$=3+7 n-7$

 

$=-4+7 n$…………(2)

Now, we are given that the nth terms for both the A.P. sequences are equal, we equate (1) and (2),

$61+2 n=-4+7 n$

2 n-7 n=-4-61

$-5 n=-65$

$n=\frac{-65}{-5}$

$n=13$

Therefore, $n=13$

 

 

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