Question

the sum of these numbers in A.P is 24 and the sum of their squares is 194. find the number

Answer 1

Hey

Here is your answer,

let, the numbers be a-d,a,a+d.

By given condition,

(a-d)+a+(a+b)=24
3a = 24
.°. a = 8.

By second given condition,
(a-d)²+a²+(a+d)²=194
(8-d)²+8²+(8+d)²=194
2(64+d²)= 194-64=130
64+d^2 =65
d² = 1,
d = ±1

When a=8 and d=1
The numbers are 7,8,9

When a=8 and d=-1
The numbers are 9,8,7

Hope it helps you!

Answer 2

Let's assume that the three numbers be (a-d), a and (a+d)

Now, sum of the numbers
(a-d)+a+(a+d)=24
3a=24
a=8

sum of the squares
(a-d)²+a²+(a+d)²=194
3a²+2d²=194 after substituting the value of a we get
(3x64)+2d²=194
2d²=2
d²=1
d=1

So, the three numbers are a-d, a and a+d
 7, 8 and 9

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