Question

three numbers are in AP their sum is 24 and sum of their square is 200 find the numbers ​

Answer 1

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Let the three numbers be (x-a),x(x+a)

Now,

x - a + x + x + a = 24 \\ => 3x = 24 \\ => x = 8 \\ Again, \\ => (x-a)^{2} + x^{2} + (x+a)^{2} = 200 \\ => x^{2} + a^{2} = 100 \\ => 8^{2} + a^{2} = 100 \\ => a^{2} = 36 \\ a = 6

According to first condition,

a - d + a + a + d = 24

=> 3a = 24

=> a = 8

Now,

According to second condition,

(a-d)^2 + a^2 + (a+d)^2 = 200

=> ( 8 - d)^2 + (8)^2 + (8 +d)^2 = 200

=> 64 + d^2 - 16d + 64 + 64 + d^2 + 16d = 200

=> 64 × 3 + 2d^2 = 200

=> 192 + 2d^2 = 200

=> 2d^2 = 8

=> d^2 = 4

d=root 4

=> d = ±2

First number = 8- 2 = 6

Second number = 8

Third number = 8 + 2 = 10

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Answer 2

let the numbers be :

a-d, a, a+d

sum= 24

a-d+a+a+d=24

3a=24

a=8

Sum of the square

(a-d)^2+ a^2 + (a+d)^2=200

a^2+d^2-2d +a^2 +a^2+d^2+2d= 200

putting the value of 'a'

we get,

3×64+2d^2=200

200-192=2d^2

8/2=d^2

d^2=4

d=√4

d=±2

So numbers are:

when d= +2

6,8,10

when d= -2

10,8,6

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