Question

Find four numbers in AP, whose sum is $20$ and the sum of whose squares is $120.$

• A. Numbers are $3,6,9,12$
  
• B. Numbers are $1,3,5,7$
  
• C. Numbers are $2,4,6,8$
• D. Data insufficient
  

Answer 1

The correct option is C Numbers are $2,4,6,8$

Let the numbers be$(a-3d),(a-d),(a+d),(a+3d)$.
Then, Sum $=20\Rightarrow (a-3d)+(a-d)+(a+d)+(a+3d)=20$
$\Rightarrow 4a=20$

$\Rightarrow a=5$ ...(1)
Now, sum of their squares $=120$
$\Rightarrow (a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$
$\Rightarrow (a^2-6ad+9d^2)+(a^2-2ad+d^2)+(a^2+2ad+d^2)+(a^2+6ad+9d^2)=120$
$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow a^2+5d^2=30$
$\Rightarrow 25+5d^2=30$ ($\because a=5$)
$\Rightarrow 5d^2=5$
$\Rightarrow d=\pm 1$
If, $d=1$, then the numbers are $2,4,6,8$ and If $d=-1$, then the numbers are $8,6,4,2$.