If the $n$ th term of the A.P. $9, 7, 5, . . . . .$ is same as the $n$ th term of the A.P $15, 12, 9, . . . .$ . Find $n$ .

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Solution

Given first A.P is
$9, 7, 5, . . . . .$
first term of A.P is $a_{1} = 9$
second term of A.P is $a_{2} = 7$
common difference $d = a_{2} - a_{1} = 7 - 9 = - 2$
hence the nth term of this A.P is given by
$a_{n} = a_{1} + (n - 1) d$
$a_{n} = 9 + (n - 1) (- 2)$
$a_{n} = 9 - 2 n + 2$
$a_{n} = 11 - 2 n \dots . . . e q (1)$

given second A.P is
$15, 12, 9, . . . . .$
first term of A.P is $a_{1} = 15$
second term of A.P is $a_{2} = 12$
common difference $d = a_{2} - a_{1} = 12 - 15 = - 3$
hence the nth term of this A.P is given by
$a_{n} = a_{1} + (n - 1) d$
$a_{n} = 15 + (n - 1) (- 3)$
$a_{n} = 15 - 3 n + 3$
$a_{n} = 18 - 3 n \dots . . . e q (2)$

if nth term of both A.P are same equate eq(1) and eq(2)
$⟹ 11 - 2 n = 18 - 3 n$
$⟹ 3 n - 2 n = 18 - 11$
$⟹ n = 7$

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