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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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