Question

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.

• A. $a=1,d=2,$ $S_{15}=235$
• B. $a=3,d=4,S_{15}=215$
• C. $a=5,d=3,S_{15}=205$
• D. None of these
  

Answer 1

The correct option is D None of these

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