Question

Four number are in AP. If their sum is 20 and sum of their square is 120. Find the number

Answer 1

Hey

Here is your answer,

(a-3d) + (a-d) + (a+d) + (a+3d) = 20
4a = 20
a = 5

(a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 120
(5-3d)² + (5-d)² + (5+d)² + (5+3d)² = 120
25 - 30d + 9d² + 25 - 10d + d² + 25 + 10d + d² + 25 + 30d + 9d² = 120
100 + 20d² = 120
20d² = 20
d² = 1
d = 1

So the 4 numbers are:
a - 3d = 5 - 3 = 2
a - d = 5 - 1 = 4
a + d = 5 + 1 = 6
a + 3d = 5 + 3 = 8

Numbers are 2, 4, 6, 8

Hope it helps you!

Answer 2

Hey mate!

Here's your answer!!
__________________

Let four numbers in A.P. be,

a-3d, a-d, a+d, a+3d

Now, their sum=(a-3d)+(a-d)+(a+d)+(a+3d) = 20

=> 4a=20
=> a=5

Also sum of squares = 120

i.e. (a-3d)² + (a-d)²+(a+d)²+(a+3d)²=120


=>a²-6ad+9d²-2ad +d²+a²+2ad+d²+a +6ad+9d² = 120

=>4a² = 20d² = 120

=>4(25)=20d²=120

=>20d²+20 =>d²=1

=>d=+_1

Hence, the numbers are...

5-3,5-1,5-1,5-3

or 5-3,5-1,5-1,5-3

i.e. 2,4,6,8 or 8,6,4,2.
___________________
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