Question

If the sum of four consecutive terms of an AP is 20 and the sum of their squares is 120, then the numbers are .

• A. $2,4,6,8$
• B. $1,2,3,4$
• C. $4,5,6,7$

Answer 1

The correct option is A $2,4,6,8$

Let the terms be $a-3d,a-d,a+d$ and $a+3d.$
Then,
$(a-3d)+(a-d)+(a+d)+(a+3d)=20$
$\Rightarrow 4a=20$
$\Rightarrow a=5.$
Also, given that $(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120.$
i.e., $a^2-6ad+9d^2+a^2-2ad+d^2+a^2+2ad+d^2+a^2+6ad+9d^2=120$
$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow a^2+5d^2=30$
$\Rightarrow 5^2+5d^2=30$
$\Rightarrow 5d^2=5$
$\Rightarrow d=\pm 1$
Taking $d=1,$ we get $a-3d,a-d,a+d,a+3d=5-3,5-1,5+1,5+3=2,4,6,8.$