Q11 of 47 Page 1

For what value of n are the nth terms of two A.P.’s 63, 65, 67, ... and 3, 10, 17, ... equal?

Given:

1st A.P. series: 63, 65, 67 …


2nd A.P. series: 3, 10, 17


Therefore, the nth term of the first series tn1 must be equal to nth term of the first series tn2


tn1 = a1 + (n-1)d1


tn2 = a2 + (n-1)d2


where


a1 & a2 are the first terms of the 1st and 2nd A.P. Series respectively


d1 & d2 are Common differences of the 1st and 2nd A.P. Series respectively


According to the problem:


tn1 = tn2


⇒ a1 + (n-1)d1 = a2 + (n-1)d2


Putting the values, we get


63 + (n-1) × 2 = 3 + (n-1) × 7


⇒ 60 = 5(n-1)


⇒ 12 = n-1


Answer: The Value of n = 13


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