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from dps, indirapuram , asked a question
Subject: Math , asked on 21/1/14

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

PLEASE SEND THE SOLUTION SOON. IT IS VERY URGENT.

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63, 65, 67, …

a = 63

d = a 2a 1 = 65 − 63 = 2

n th term of this A.P. = a n = a + (n − 1) d

a n = 63 + (n − 1) 2 = 63 + 2n − 2

a n = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a 2a 1 = 10 − 3 = 7

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

a n = 3 + 7n − 7

a n = 7n − 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 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

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

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n th term is given by  a+(n-1)d

thus for the first a.p. we have nth term =63 +(n-1)(2)= 63 +2n -2 = 61+2n

and for the second a.p. nth term is =3 +(n-1)7= 7n -4

now they equal when 7n -4= 61+2n

5n=65

n=13 ans

Hi Farha, thanku for the solution.

May I know in which school ur studing.

a = 63 , a = 3

d = 2 , d= 7

acc. to an formula

63 + ( n - 1 ) 2 = 3 + ( n - 1) 7

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

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

60 = 5n - 5

60 + 5 = 5n

65 = 5n

65/ 5 = n

n = 13

if right then give me thumbs up