There are three consecutive numbers whose sum is 129. Find the square of the middle one.

1849

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1764

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1829

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1936

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Solution

The correct option is A 1849
Let numbers be x, x + 1, x + 2
According to question,
(x) + (x + 1) + (x + 2) = 129
$\Rightarrow$ (3x+3) = 129
$\Rightarrow$ 3x = 129-3
$\Rightarrow$ 3x = 126
$\Rightarrow$ x = 42
$\Rightarrow$ x+1 = 43
$∴$ x $^{2}$ = 43 $\times$ 43 = 1849.

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There are three consecutive numbers whose sum is 129. Find the square of the middle one.

The correct option is A 1849Let numbers be x, x + 1, x + 2 According to question, (x) + (x + 1) + (x + 2) = 129 ⇒ (3x+3) = 129 ⇒ 3x = 129-3 ⇒ 3x = 126 ⇒ x = 42 ⇒ x+1 = 43 ∴ x2 = 43 × 43 = 1849.

The correct option is A 1849
Let numbers be x, x + 1, x + 2
According to question,
(x) + (x + 1) + (x + 2) = 129
$\Rightarrow$ (3x+3) = 129
$\Rightarrow$ 3x = 129-3
$\Rightarrow$ 3x = 126
$\Rightarrow$ x = 42
$\Rightarrow$ x+1 = 43
$∴$ x $^{2}$ = 43 $\times$ 43 = 1849.