Question

find 3 consecutive even natural number such that the sum of their squares is 200 ​

Answer 1

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• Sum of squares of 3 consecutive even natural number is 200

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$\sf\large\bold{\orange{\underline{\blue{ To \: Find :-}}}}$

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• The 3 consecutive even natural number whose sum of squares is 200

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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Let the first even natural number be "x"

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$\underline{\bold{\texttt{1st even natural number :}}}$

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➠ x ⚊⚊⚊⚊ ⓵

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$\underline{\bold{\texttt{2nd consecutive even natural number :}}}$

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➠ x + 2 ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{3rd consecutive even natural number :}}}$

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➠ x + 4 ⚊⚊⚊⚊ ⓷

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$\underline{\bold{\texttt{Square of 1st even natural number :}}}$

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➠ x² ⚊⚊⚊⚊ ⓸

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$\underline{\bold{\texttt{Square of 2nd consecutive even natural number :}}}$

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➠ (x + 2)² ⚊⚊⚊⚊ ⓹

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$\underline{\bold{\texttt{Square of 3rd consecutive even natural number :}}}$

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➠ (x + 4)² ⚊⚊⚊⚊ ⓺

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Given that , Sum of squares of 3 consecutive even natural number is 200

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Thus ,

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⓸ + ⓹ + ⓺ = 200

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➜ x² + (x + 2)² + (x + 4)² = 200

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➜ x² + x² + 4 + 4x + x² + 16 + 8x = 200

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➜ 3x² + 12x + 20 = 200

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➜ 3x² + 12x + 20 - 200 = 0

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➜ 3x² + 12x - 180 = 0

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➜ x² + 4x - 60 = 0

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➜ x² + 10x - 6x - 60 = 0

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➜ x(x + 10)-6(x + 10) = 0

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➜ (x + 10)(x - 6) = 0

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• x = -10
• x = 6 ⚊⚊⚊⚊ ⓻

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We know that natural numbers starts from 1 i.e they are always positive

So x = -10 [ Not possible ]

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∴ Hence x = 6

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• Hence the first consecutive even natural number in this case is 6

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⟮ Putting x = 6 from ⓻ to ⓶ ⟯

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➜ x + 2

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➜ 6 + 2

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➨ 8

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• Hence the second consecutive even natural number in this case is 8

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⟮ Putting x = 6 from ⓻ to ⓷ ⟯

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➜ x + 4

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➜ 6 + 4

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➨ 10

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• Hence the third consecutive even natural number in this case is 10

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∴ The 3 consecutive even natural number whose sum of square is 200 are 6 , 8 , 10