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

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Solution

In the first AP,
$a = 63 d = 2 T_{n} = 63 + (n - 1) \times 2 = 2 n + 61$

In the second AP,
$a = 3 d = 7 T_{n} = 3 + (n - 1) \times 7 = 7 n - 4$

Now, $7 n - 4 = 2 n + 61 \Rightarrow 5 n = 65 \Rightarrow n = 13$

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