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 *n*^{th }terms of both the sequences are equal. We need to find *n*

So let us first find the *n*^{th} 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 *n*^{th} term,

`a_n = 63 + (n-1)2`

= 63 + 2n - 2

= 61 + 2n ....................(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 *n*^{th} term,

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

= 3 + 7n - 7

= - 4 + 7 n ...................(2)

Now, we are given that the *n*^{th} terms for both the A.P. sequences are equal, we equate (1) and (2),

61 + 2n = - 4 + 7n

2n - 7n = -4 - 61

- 5n = - 65

`n =(-65)/(-5)`

n = 13

Therefore,** n = 13 **