#### If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.

The given APs are

$9, 7, 5, \ldots ., and 24, 21, 18, \ldots ..\\$

In AP

$9, 7, 5, \ldots ..\\$

$\\ a\textsubscript{1} = a = 9 , a\textsubscript{2}=7\\ d\textsubscript{1} = a\textsubscript{2} - a\textsubscript{1} = 7 - 9 = - 2\\ a\textsubscript{n} = a + (n - 1) d\textsubscript{1 }\\ a\textsubscript{n} = 9 + (n - 1) (-2) \ldots (1)\\ In AP 24, 21, 18 \ldots \\ b\textsubscript{1} =b = 24, b\textsubscript{2 }= 21\\ d\textsubscript{2} = b\textsubscript{2} - b\textsubscript{1} = 21 - 24 = - 3\\ b\textsubscript{n} = b + (n - 1)d\textsubscript{2}\\ b\textsubscript{n}=24+ (n - 1) (-3) \ldots .(2) \\$

Here the number of terms n is equal in both case

According to question

$a\textsubscript{n} = b\textsubscript{n}\\$

$\\9 + (n - 1) (-2) = 24 + (n - 1) (-3)\\ 9 - 2n + 2 = 24 - 3n + 3\\ 9 - 2n + 2 - 24 + 3n - 3 = 0\\ n - 16 = 0\\ n = 16\\ a\textsubscript{16}=a+(16-1)d (using a\textsubscript{n} = a + (n-1)d)\\ = 9 + 15(-2)\\ = 9 - 30 = - 21\\ a\textsubscript{16} = - 21\\ n\textsuperscript{th} term is - 21 and the value of n is 16\\$