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

Question:

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

Solution:

Let the nth term of the given progressions be tn and Tn, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its nth term is given by
tn = a + (n - 1)
 63 + (n - 1) ⨯ 2
⇒ 61 + 2n

The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its nth term is given by
Tn = A + (n - 1)
 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Nowtn = ​Tn
 
⇒ 61 + 2n​ = 7n - 4​
 65 5n
 n = 13
Hence, the 13th terms of the AP's are the same.

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