Find an A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.

Given: An A.P., its fourth term, $a_{4} =9$
Sum of its sixth term and thirteenth term, $a_{6} +a_{13} =40$

To do: To find the A.P.

Solution: 
Let the first term of the given A.P. is$\ a\ $and its common difference is $d.$

As known $n^{th}$ term of an A.P., $a_{n} =a+( n-1) d$

$a_{4} =a+( 4-1) d $ 

$( here\ a_{4} =9\ and\ n=4) $ 
 
$\Rightarrow a+3d=9$                          $\ \ \ \ \dotsc \dotsc \dotsc ..( 1)$

and similiarly$\ a_{6} =a+( 6-1) d=a+5d$

and $a_{13} =a+( 13-1) d=a+12d\ \ $

as given $a_{6} +a_{13} =a+5d+a+12d=40$

$\Rightarrow 2a+17d=40$                $\dotsc \dotsc \dotsc \dotsc ( 2)$ 

On sollving equations $( 1)$and $( 2)$ 
$d=2$ and $a=3$

$\therefore$ The A.P. is 3, 5, 7,  9, ................
 

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Updated on: 10-Oct-2022

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