Question

for what value of n, the nth terms of the AP 63,65,67,............and 3,10,17,......are equal ?

Answer 1

$an = a + (n - 1)d$
$an = 63 + (n - 1)2$
$am = 3 + (m - 1)7$
am=an
63+(n-1)2 = 3+(n-1)7
63+2n-2 = 3+7n-7
61+2n=7n-4
61+4=5n
5n=65
n= 65/5
n=13

Answer 2

In 1st AP:

Given AP is 63,65,67.....

First-term a = 63.

Common difference d = 65 - 63 = 2.

We know that sum of n terms an = a + (n - 1) * d

                                                       = 63 + (n - 1) * 2

                                                       = 63 + 2n - 2

                                                       = 61 + 2n  ------- (1)


In 2nd AP : 

Given AP is 3,10,17.

First-term a = 3.

Common difference d = 10 - 3 = 7.

We know that sum of n terms an = a + (n-1) d
     
                                                        = 3 + (n - 1) * 7

                                                        = 3 + 7n - 7

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


Given that nth term of these AP are equal.

From (1) and (2), we get

61 + 2n = 7n - 4

61 + 4 = 7n - 2n

61 + 4 = 5n

65 = 5n

n = 13.

Therefore 13th term of these AP's are equal.

Hope this helps!