Question

three numbers in ap their sum is 24 and sum of their square is 200 find the numbers

Answer 1

Let the three numbers be (x-a),x(x+a)
Now,
$x - a + x + x + a = 24 \\ => 3x = 24 \\ => x = 8 \\ Again, \\ => (x-a)^{2} + x^{2} + (x+a)^{2} = 200 \\ => x^{2} + a^{2} = 100 \\ => 8^{2} + a^{2} = 100 \\ => a^{2} = 36 \\ a = 6$
Mark brainelist.

Answer 2

Let the number be (a-d), a, (a + d)

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 = 2



First number = 8- 2 = 6

Second number = 8

Third number = 8 + 2 = 10