Answer :


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 second series tn2

tn1 = a1 + (n-1)d1

tn2 = a2 + (n-1)d2


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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