Solve the following mathematical problems:

The sum of three numbers in Arithmetic Progression is 30, and the sum of their square 318.Find the numbers.

Created: 3 months ago | Updated: 3 months ago

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Let, second term = a

Common difference = d

First term will be=a-d

And third term will be = a+d

according to question,

$a - d + a + a + d = 30 \Rightarrow 3 a = 30 \Rightarrow a = \frac{30}{3} ∴ a = 10$

So, second term = 10

First term = (10-d)

and third = (10+d)

According to question ,

${(10 - d)}^{2} + {(10)}^{2} + {(10 + d)}^{2} = 318 \Rightarrow {(10)}^{2} - 2 \times 10 \times d + d^{2} + 100 + {(10)}^{2} + 2 \times 10 \times d + d^{2} = 318 \Rightarrow 100 - 20 d + d^{2} + 100 + 100 + 20 d + d^{2} = 318 \Rightarrow 2 d^{2} + 300 = 318 \Rightarrow 2 d^{2} = 18 ∴ d = \sqrt{9} = 3$

so, 1^st term = 10-3=7; 2^nd term= 10 & 3^rd term = 10+3=13

ans. the number is 7,10,13

3 months ago

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