Question

If four numbers are in A.P such that their sum is 20 and sum of their squares is 120, them the numbers are

• A. 1, 4, 7, 10
• B. 3, 5, 7, 9
• C. 2, 4, 6, 8
• D. -1, 3, 7, 11

Answer 1

The correct option is C 2, 4, 6, 8

Let the four numbers in A.P be $a-3d,a-d,a+d,a+3d$. ---- (1)

Given that Sum of the terms $=20.$

$=(a-3d)+(a-d)+(a+d)+(a+3d)=20$

$4a=20$

$a=5$. ---- (2)

Given that sum of squares of the term $=120.$

$=(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$

$=(a^2+9d^2-6ad)+(a^2+d^2-2ad)+(a^2+d^2+2ad)+(a^2+9d^2+6ad)=120$

$=4a^2+20d^2=120$

Substitute $a=5$ from (2) .

$4(5{)}^2+20d^2=120$

$100+20d^2=120$

$20d^2=20$

$d=\pm 1$

Since AP cannot be negative.

Substitute $a=5$ and $d=1$ in (1), we get

$a-3d,a-d,a+d,a+3d=2,4,6,8.$