Question:

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Solution:

Let a be the first term and d be the common difference of the AP. Then,

$a_{4}=9$

$\Rightarrow a+(4-1) d=9 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+3 d=9 \quad \ldots .(1)$

Now,

$a_{6}+a_{13}=40$                   (Given)

$\Rightarrow(a+5 d)+(a+12 d)=40$

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

From (1) and (2), we get

$2(9-3 d)+17 d=40$

$\Rightarrow 18-6 d+17 d=40$

$\Rightarrow 11 d=40-18=22$

$\Rightarrow d=2$

Putting d = 2 in (1), we get

$a+3 \times 2=9$

$\Rightarrow a=9-6=3$

Hence, the AP is 3, 5, 7, 9, 11, ... .