the sum of 4 no in an ap is 20.and the sum of their squares is 120.find the nos?

Answer 1

Let the numbers be (a-d),a,(a+d),(a+2d)
Sum of all 4 numbers = 4a+2d = 20
2a+d= 10
d=10-2a.....(1)

sum of their squares =120
=(a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2
= 2a^2 +2d^2 + a^2 + a^2 + 4ad+ 4d^2
=4a^2 + 4ad + 6d^2
=120

2a^2 + 2ad + 3d^2=60
Substitute value of d from equation (1)

2a^2 + 2a(10-2a) + 3(10-2a)^2=60
a^2 + a(10-2a) + 6(5-a)^2 = 30

5a^2-50a+120=0
a^2-10a+24 = 0

(a-6)(a-4) = 0
a = 6 or a = 4

_____ ______ ______ _____
For a= 6; d= 10-2*6 =(-2)

Series is 8,6,4,2
_____ ______ ______ _____
For a= 4; d= 10-2*4 = 2

Series is 2,4,6,8

_______ _______ _______ _________
Check:
Sum = 2+4+6+8 =20
Sum of squares = 4+16+36+64= 120

SOLVED!!

Answer 2

2, 4, 6, 8

2 + 4 + 6 + 8 = 20
4 + 16 + 36 + 64 = 120

Answer 3

LET THE NO.S BE (a-3d),(a-d),(a+d),(a+3d)
AS THEIR SUM IS 20
(a-3d)+(a-d)+(a+d)+(a+3d)=20
4a=20
a=5
Also, sum of the squares of four no.s =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
AS a=5
4(5)^2+20(d)^2=120
100+20d^2=120
20d^2=20
d^2=1
d=+1 or d=-1
NO.S ARE
If d=1 ,then,(a-3d)=5-3=2
(a-d)=5-1=4
(a+d)=5+1=6
(a+3d)=5+3=8
therefore no.s are 2,4,6,8.
BUT IF d=-1,then,(a-3d)=5-3(-1)=8
(a-d)=5-(-1)=6
(a+d)=5+(-1)=4
(a+3d)=5+3(-1)=2
therefore no.s are 8,6,4,2.

Answer 4

let the nos be a-3d,a-d,a+d,a+3d
sum=4a=20
so a=5
sum of the squares=(a-3d)^2+(a+3d)^2+(a-d)^2+(a+d)^2
2(a^2+9d^2)+2(a^2+d^2)
=4a^2+20d^2=120
substituting a
4*25+20d^2=120
20d^2=20
d=1
so the A.P. is 2,4,6,8
check 2+4+6+8=20
4+16+36+64=120