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

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## CERTIFIED ANSWER

63, 65, 67, …

*a* = 63

*d* = *a* _{2} − *a* _{1} = 65 − 63 = 2

*n* ^{th} term of this A.P. = *a* _{ n } = *a* + (*n* − 1) *d*

*a* _{ n }= 63 + (*n* − 1) 2 = 63 + 2*n* − 2

*a* _{ n } = 61 + 2*n* (1)

3, 10, 17, …

*a* = 3

*d* = *a* _{2} − *a* _{1} = 10 − 3 = 7

*n* ^{th} term of this A.P. = 3 + (*n* − 1) 7

*a* _{ n } _{ }= 3 + 7*n* − 7

*a* _{ n } = 7*n* − 4 (2)

It is given that, *n* ^{th} term of these A.P.s are equal to each other.

Equating both these equations, we obtain

61 + 2*n* = 7*n* − 4

61 + 4 = 5*n*

5*n* = 65

*n* = 13

Therefore, 13^{th} terms of both these A.P.s are equal to each other.

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