Question

vue of 'n' of which N th terms of two A.P.'s 63,65,67.........and 3,10,17....... are equal

Answer 1

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Given that

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

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

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

$61+2n = -4+7n$

$61 +2n+4-7n = 0$

$65-5n = 0$

$5n = 65$

$x = 65/5$

$n = 13$

Answer 2

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For 1st A.P.,

First term ( a ) = 63

Common difference ( d ) = 65 - 63 = 2.

Nth term = a + ( N - 1 ) d

n = 63 + ( N - 1 ) 2

n = 63 + 2 N - 2

n = 61 + 2 N                       --------------------------- equation 1

For 2nd A.P.,

First term ( a ) = 3

Common difference ( d ) = 10 - 3 = 7.

Nth term = a + ( N - 1 ) d

n = 3 + ( N - 1 ) 7

n = 3 + 7 N - 7

n = 7 N - 4                   -------------------------------  equation 2


From equation 1 and equation 2 , we get,

7 N - 4 = 61 + 2 N

7 N - 2 N = 61 + 4

5 N = 65

N = 65 / 5

N = 13.

By substituting the value of 'N' in equation 1,

n = 61 + 2 N

n = 61 + 2 x 13

n = 61 + 26

n = 87.


So the 13th term of both A.P. will be equal that's equal to 87.


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