Question

The sum of squares
I natural no. is 1455
of five consecutive
Find the numbers.
Lille​

Answer 1

Step-by-step explanation:

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 five consecutive natural numbers are 15 , 16 , 17 , 18 , 19

Correct question : The sum of the squares of five consecutive natural numbers is 1455. find the numbers

Given :

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

To find :

The five consecutive natural numbers

Formula Used :

$\displaystyle \sf {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

Solution :

Step 1 of 2 :

Form the equation to find the numbers

Let five consecutive natural numbers are n - 2 , n - 1 , n , n + 1 , n + 2

Since sum of the squares of five consecutive natural numbers is 1455

So by the given condition

$\displaystyle \sf {(n - 2)}^{2} + {(n - 1)}^{2} + {n}^{2} + {(n + 1)}^{2} + {(n + 2)}^{2} = 1455$

Step 2 of 2 :

Find the five consecutive natural numbers

$\displaystyle \sf {(n - 2)}^{2} + {(n - 1)}^{2} + {n}^{2} + {(n + 1)}^{2} + {(n + 2)}^{2} = 1455$

$\displaystyle \sf{ \implies }{(n - 2)}^{2} + {(n + 2)}^{2} + {n}^{2} + {(n - 1)}^{2} + {(n + 1)}^{2} = 1455$

$\displaystyle \sf{ \implies }2\bigg[ {n}^{2} + {2}^{2} \bigg] + {n}^{2} + 2\bigg[ {n}^{2} + {1}^{2} \bigg] = 1455\: \: \bigg[ \: \because \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} ) \bigg]$

$\displaystyle \sf{ \implies }2\bigg[ {n}^{2} + 4 \bigg] + {n}^{2} + 2\bigg[ {n}^{2} + 1 \bigg] = 1455$

$\displaystyle \sf{ \implies }2{n}^{2} + 8 + {n}^{2} + 2{n}^{2} + 2= 1455$

$\displaystyle \sf{ \implies }5{n}^{2} + 10= 1455$

$\displaystyle \sf{ \implies }5{n}^{2} = 1455 - 10$

$\displaystyle \sf{ \implies }5{n}^{2} = 1445$

$\displaystyle \sf{ \implies }{n}^{2} = 289$

$\displaystyle \sf{ \implies }{n}^{2} = {17}^{2}$

$\displaystyle \sf{ \implies }n = 17$

Hence five consecutive numbers are 15 , 16 , 17 , 18 , 19

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