Question

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

Answer 1

13

Step-by-step explanation:

d for 1st AP is 65 - 63 = 2

d for 2nd AP is 10 - 3 = 7

Let those be equal for n = x, where

xth term of 1st AP is 63 + (x - 1)2 = 2x + 61

xth term of 2nd AP is 3 + (x - 1)7 = 7x - 4

As both are equal,

=> 2x + 61 = 7x - 4

=> 61 + 4 = 7x - 2x

=> 65 = 5x

=> 13 = x

Hence, for n = 13, nth term of both the APs is equal.

Answer 2

Ans

Step-by-step explanation

In Ist AP a= 63, d=2

a in IInd AP=3, d=7

for equal AP condition

63+[n-1]2=3+[n-1]7

63+2n-2=3+7n-7

61+2n=7n-4

61+4=7n-2n

65=5n

n=13