Question

Find four numbers in an $AP$ whose sum is $20$ and sum of whose square is $120$.

• A. $2,4,6,8$
• B. $2,-4,6,8$
• C. $2,4,7,8$
• D. $0,4,6,8$

Answer 1

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

Let the four numbers be $a-3d,a-d,a+d$ and $a+3d$

$\therefore a-3d+a-d+a+d+a+3d$

$\therefore 4a=20$

$\therefore a=\frac{20}{4}=5$

$\therefore$ four numbers are :-

$5-3d,5-d,5+d,5+3d$

$\therefore (5-3d{)}^2+(5-d{)}^2+(5+d{)}^2+(5+3d{)}^2=120$

$\therefore 25-30d+9d^2+25+30d+9d^2+25+d^2-10d+25+10d+d^2=120$

$\therefore 100+20d^2=120$

$\therefore d^2=\frac{20}{20}$

$\therefore d=\pm 1$

taking $d=1$ , we get four number as:-

$5-3=2$

$5-1=4$

$5+1=6$

$5+3=8$

$\therefore$ four numbers are $2,4,6$ and $8.$