Question

the sum of two numbers of an A.P. is 20 and sum of their squares is 120. Find the number?

Answer 1

Say the four numbers be $a,(a+d),(a+2d),(a+3d)<span>$
sum =$4a+6d=20</span><span>$      
$2a +3d=10</span><span>$
Sum of squares =
a^2+(a^2 +2ad +d^2)+(a^2 +4ad + 4d^2)+(a^2+6ad +9d^2)= 4a^2+12ad+14 d^2=120 plug $2a=10-3d</span><span>$
$(10 - 3d)^2+6d (10-3d)+14d^2=120<span>$ 
$100 -60d +9d^2 +60d -18d^2+14d^2 = 120<span>$
$5d^2 =20</span><span>$
$d=2<span>$   $a=2</span><span>$     
The numbers are $2,4,6,8</span><span>$

Answer 2

=A+P=20
=A=20-p........................i
 =(A+P)^2=120
sunstiturw value of equatino i to formula.
(20-P+P)^2=120 
done