Question

the sum of the squares of five consecutive terms natural numbers is 1445 find them

Answer 1

Let the five consecutive natural numbers be x, (x + 1), (x + 2), (x + 3) and (x + 4).
By the given condition, we get:
x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2 = 1455
⇒x2 + x2 + 2x + 1 +x2 + 4x + 4 + x2 + 6x + 9 + x2 + 8x + 16 = 1455
⇒ 5x2 + 20x + 30 = 1455
⇒ 5(x2 + 4x + 6) = 1455
⇒ x2 + 4x + 6 = 291
⇒ x2 + 4x – 285 = 0
On splitting the middle term 4x as 19x – 15x, we get:
x2 + 19x – 15x – 285 = 0
⇒ x(x + 19) – 15(x + 19) = 0
⇒ (x + 19)(x– 15) = 0
⇒x + 19 = 0    or     x – 15 = 0
⇒ x = –19        or     x = 15
Since x is a natural number, which cannot be negative,  x = 15
Thus, the five consecutive numbers are 15, 16, 17, 18 and 19

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Answer 2

The Correct Question is as follows:

The sum of the squares of five consecutive terms of natural numbers is 1455. Find them.

The correct answer to the question is 15, 16, 17, 18, and 19.

Given:

The sum of the squares of five consecutive terms of natural numbers is 1455.

To Find:

The five natural numbers =?

Solution:

To solve the question we will first assume the natural numbers as follows;

Let the five natural numbers be (n-2), (n-1), n, (n+1), and (n+2).

Now, according to the question, the sum of their squares is equal to 1455.

i.e., (n-2)² + (n-1)² + n² + (n+1)² + (n+2)² = 1455

Now, expanding them and adding all we get;

⇒ (n² - 4n + 4) + (n² - 2n + 1) + (n²) + (n² + 2n + 1) + (n² + 4n + 4) = 1455

⇒ n² + n²+ n²+ n²+ n² + 10 = 1455

⇒ 5n² + 10 = 1455

⇒ 5n² = 1455 - 10

⇒ 5n² = 1445

⇒ n² = $\frac{1445}{5}$

⇒ n² = 289

⇒ n = ±17, but a natural number is always greater than 0.

∴ n ≠ -17 i.e., n = 17

The other natural numbers are as follows;

(n - 2) = 17 - 2 = 15

(n - 1) = 17 - 1 = 16

n = 17

(n + 1) = 17 + 1 = 18

(n + 2) = 17 + 2 = 19

Thus, the required natural numbers whose sum of squares is equal to 1455 are 15, 16, 17, 18, and 19.

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