Question
The sum of the squares of three consecutive natural numbers is 2030. Then, what is the middle number?
Answer: Option C
Answer: (c)Let the three consecutive natural numbers be x, x + 1 and x + 2. According to question, $x^2 + (x + 1)^2 + (x + 2)^2$ = 2030or $x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$ = 2030 or $3x^2 + 6x + 5 = 2030 $or $3x^2 + 6x – 2025$ = 0 or $x^2 + 2x – 675$ = 0 or $x^2 + 27x – 25x – 675$ = 0 $x (x + 27) – 25 (x + 27)$ = 0 or $(x – 25) (x + 27)$ = 0$x = 25 and – 27$∴ Required number = $x$ + 1 = 25 + 1 = 26
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Answer: (c)Let the three consecutive natural numbers be x, x + 1 and x + 2. According to question, $x^2 + (x + 1)^2 + (x + 2)^2$ = 2030or $x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$ = 2030 or $3x^2 + 6x + 5 = 2030 $or $3x^2 + 6x – 2025$ = 0 or $x^2 + 2x – 675$ = 0 or $x^2 + 27x – 25x – 675$ = 0 $x (x + 27) – 25 (x + 27)$ = 0 or $(x – 25) (x + 27)$ = 0$x = 25 and – 27$∴ Required number = $x$ + 1 = 25 + 1 = 26
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