The sum of the first $9$ terms of an $A.P$ is $81$ and the sum of it's first $20$ terms is $400.$ Find the first term, the common difference and the sum upto $15th$ term.

Sum of the term $S_n=\frac{n}{2}\left[2\times a+\left(n-1\right)d\right]$
Given sum of $9$ terms $=81$
$81=\frac{9}{2}\left[2a+\left(9-1\right)d\right]...(1)$
Sum of $20$ terms $=400$
$400=\frac{20}{2}\left[2a+\left(20-1\right)d\right]...(2)$
$\Rightarrow 81=\frac{9}{2}\left[2a+8d\right]$
$\Rightarrow 400=\frac{9}{2}\left[2a+19d\right]$
$\Rightarrow 162=18a+72d...(3)$
$\Rightarrow 400=20a+190d....(4)$
Multiply eq$(3)$ by $20$ and and eq$(4)$ by $18$ and subtract both the equation
$\Rightarrow 360a+3420d=7200$
$\Rightarrow 360a+1440d=3240$
$1980d=3690$
$\Rightarrow d=2$
Now substitute the $d=2$ in eq$(4)$
$\Rightarrow 400=20a+190\times 2$
$\Rightarrow 400=20a+380$
$\Rightarrow 20a=20$
$\Rightarrow a=1$
Sum of the 15th Term
$S_{15}=\frac{15}{2}\left[2\times 1+\left(15-1\right)2\right]$
$S_{15}=\frac{15}{2}\left[2+\left(14\right)2\right]$
$S_{15}=\frac{15}{2}\left[2+28\right]$
$S_{15}=\frac{15}{2}\left[30\right]$
$S_{15}=15\times 15=225$
Hence $a=1,d=2,S_{15}=225$