Question

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

Answer 1

We have

$a_4=9$

$a+3d=9---\left(i\right)$

$a_6+a_{15}=40$

$2a+6d=40---\left(ii\right)$

On solving equation $I)$ and $\left(II\right)$ we get

$a+3d=9$

$2a+6d=40$

$----------$

$13d=22$

$\because d=\frac{22}{13}$

Now,

On putting the value of $d$ in equation $\left(ii\right)$ we get

$2a+6\times \frac{22}{13}=18$

$a+6\times \frac{11}{13}=9$

$a=9-\frac{66}{13}$

$a=\frac{57}{13}$

$A.P=\frac{51}{13},\frac{51+22}{13},\frac{51+44}{13},\frac{51+66}{13}$

$=\frac{51}{13},\frac{73}{13},\frac{95}{13},\frac{117}{13}=9$

Which is the required answer.