Question 10 The sum of the squares of three positive numbers that are consecutive multiples of 5 is 725. Find the three numbers. Given three positive numbers that are consecutive multiples of 5
Since the number is a multiple of 5
Let First Number = 5x
Now,
Since other 2 numbers are consecutive multiples of 5, they will have a difference of 5
∴ Second Number = 5x + 5
Third Number = (5x + 5) + 5 = 5x + 10
Given that
Sum of the squares of three positive numbers that are consecutive multiples of 5 is 725
(5x)2 + (5x + 5)2 + (5x + 10)2 = 725
52 × x2 + 52 (x + 1)2 + 52 (x + 2)2 = 725
52 × [x2 + (x + 1)2 + (x + 2)2] = 725
25 × [x2 + (x + 1)2 + (x + 2)2] = 725
x2 + (x + 1)2 + (x + 2)2 = 725/25
x2 + (x + 1)2 + (x + 2)2 = 29
x2 + x2 + 12 + 2x + x2 + 22 + 4x = 29
x2 + x2 + x2 + 2x + 4x + 1 + 4 = 29
3x2 + 6x + 5 = 29
3x2 + 6x + 5 − 29 = 0
3x2 + 6x − 24 = 0
3(x2 + 2x − 8) = 0
x2 + 2x − 8 = 0
Solving by splitting the middle term
x2 + 4x − 2x − 8 = 0
x(x + 4) − 2(x + 4)= 0 (x − 2) (x + 4) = 0
∴ x = 2, x = −4
Since we need a positive number
∴ x = 2
Thus,
First Number = 5x = 5 × 2 = 10
Second Number = 5x + 5 = 5 × 2 + 5 = 15
Third Number = 5x + 10 = 5 × 2 + 10 = 20

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.