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Question
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
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Solution
Given:
nth term of the A.P. 9, 7, 5... is the same as the nth term of the A.P. 15, 12, 9...
\[\text { Considering } 9, 7, 5...\]
\[a = 9, d = \left( 7 - 9 \right) = - 2\]
\[ n^{th} \text { term } = 9 + (n - 1)( - 2) \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ = 9 - 2n + 2\]
\[ = 11 - 2n . . . (i)\]
\[\text { Considering } 15, 12, 9, ...\]
\[a = 15, d = \left( 12 - 15 \right) = - 3\]
\[ n^{th} \text { term } = 15 + (n - 1)( - 3) \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ = 15 - 3n + 3\]
\[ = 18 - 3n . . . (ii)\]
Equating (i) and (ii), we get:
\[11 - 2n = 18 - 3n\]
\[ \Rightarrow n = 7\]
Thus, 7th terms of both the A.P.s are the same.
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