Question

Find three consecutive odd positive integers the sum of whose squares is 371.

Answer 1

$Let \: (x-2), x \:and \: (x+2) \:are \: three \\consecutive \: odd \: positive \:integers$

/* According to the problem given */

$Sum \:of \: the \: squares \: of \:three \\numbers = 371$

$\implies (x-2)^{2} + x^{2} + (x+2)^{2} = 371$

$\implies [ (x-2)^{2} + (x+2)^{2}] + x^{2} = 371$

$\implies 2(x^{2} + 2^{2}) + x^{2} = 371$

$\implies 2x^{2} + 8 + x^{2} = 371$

$\implies 3x^{2} = 371 - 8$

$\implies 3x^{2} = 363$

$\implies x^{2} = \frac{363}{3}$

$\implies x^{2} = 121$

$\implies x = \pm\sqrt{11^{2}}$

$\implies x = \pm 11$

/* But x should be odd positive integer */

$\implies x =11$

Therefore.,

$\red { Required \: 3 \: consecutive \: odd } \\\red{integers\: are \: } \green {\: 9, \: 11 \:and \: 13 }$

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