Question

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

Answer 1

Let the value of An be X(for both the sequence)
Then, X = 63 + (n-1)(2)
=63+2n-2=X
=61+2n=X — (eq.(1))

Again,
X=3 + (n-1)(7)
61+2n=3+(n-1)×7
61-3=7n-7 -2n
58=5n-7
58+7=5n
65/5=n
13=n

Therefore, your answer is 13.

Answer 2

a+(n-1)d = a+(n-1)d

1st AP = 2nd AP

63+(n-1)2 = 3+(n-1)7
63+2n-2 = 3+7n-7
61+2n = 7n-4
5n = 65
n = 13
The 13th term of the AP'S are equal.

VERIFICATION:-

1st AP :- a+12d
= 63 + 12(2)
= 63 + 24
= 87

2nd AP :- a+12d
= 3 + 12(7)
= 3 + 84
= 87

Hence verified!!

Hope this helps!!
cheers!! (:

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