For what value of *n*, are the n^{th} terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

#### Solution 1

Consider the A.P 63, 65, 67, …

a = 63

d = a_{2} − a_{1} = 65 − 63 = 2

n^{th} term of this A.P. = a_{n} = a + (n − 1) d

a_{n}= 63 + (n − 1) 2 = 63 + 2n − 2

a_{n} = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a_{2} − a_{1} = 10 − 3 = 7

n^{th} term of this A.P. = 3 + (n − 1) 7

a_{n}_{ }= 3 + 7n − 7

a_{n} = 7n − 4 (2)

It is given that, n^{th} term of these A.P.s are equal to each other.

Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13^{th} terms of both these A.P.s are equal to each other.

#### Solution 2

Consider the A.P 63, 65, 67,...

Here, First term (a) = 63 and common difference (d) = 2

a_{n} = a + (n − 1)d

a_{n} = 63 + (n − 1)2

a_{n} = 61 + 2n .....(1)

Consider another A.P 3, 10, 17,...

Here, First term (a') = 3 and common difference (d') = 7

`a_n^'`= a' + (n − 1)d'

`a_n^'` = 3 + (n − 1)7

`a_n^'` = 7n − 4 .....(2)

According to question,

61 + 2n = 7n − 4

⇒ 65 = 5n

⇒ n = 13