Question

The sum of the squares of three consecutive odd numbers is 2,531. Find the numbers.

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume that the Required three consecutive odd numbers are x , (x + 2) and (x + 4).

A/q,

→ x² + (x + 2)² + (x + 4)² = 2531

→ x² + x² + 4x + 4 + x² + 8x + 16 = 2531

→ (x² + x² + x²) + (4x + 8x) + (16 + 4) = 2531

→ 3x² + 12x + 20 = 2531

→ 3x² + 12x + 20 - 2531 = 0

→ 3x² + 12x - 2511 = 0

→ 3(x² + 4x - 837) = 0

→ x² + 4x - 837 = 0

→ x² + 31x - 27x - 837 = 0

→ x(x + 31) - 27(x + 31) = 0

→ (x + 31)(x - 27) = 0

→ x = (-31) & 27.

Since Negative value not Possible.

Hence, Required Three consecutive odd numbers are 27 , 29 and 31.

Answer 2

${ \underline{ \underline{ \red{ \huge{ \tt{Question}}}}}}$



▪ The sum of the squares of three consecutive odd numbers is 2531. Find the numbers.



${ \underline{ \underline{ \red{ \tt{ \huge{Solution}}}}}}$



➡ Let three consecutive odd numbers be



${ \sf{ \purple{a \:, \: (a + 2) \: and \: (a + 4)}}}$



▪ it's given in the question that the sum of square of these above assumed numbers is equal to 2531



${ \green{ \sf{ {a}^{2} + {(a + 2)}^{2} + {(a + 4)}^{2} = 2531}}}$



we know that,



${ \boxed{ \boxed{ \sf{ \pink{ {(m + n)}^{2} = {m}^{2} + \: 2mn \: + {n}^{2} }}}}}$



• using this identity in the L.H.S...



${ \blue{ \sf{ {a}^{2} + {a}^{2} + 2.a.2 + {2}^{2} + {a}^{2} + 2.a.4 + {4}^{2} = 2531}}}$



${ \implies{ \blue{ \sf{ {a}^{2} + {a}^{2} + {a}^{2} + 4a + 8a + 4 + 16 = 2531}}}}$



${ \implies{ \blue{ \sf{ 3 {a}^{2} + 12a + 20 = 2531}}}}$



${ \implies{ \blue{ \sf{3 {a}^{2} + 12a + 20 - 2531 = 0}}}}$



${ \implies{ \green{ \sf{3 {a}^{2} + 12a - 2511 = 0}}}}$



${ \implies{ \green{ \sf{3( {a}^{2} + 4a - 837) = 0 }}}}$



${ \implies{ \green{ \sf{ {a}^{2} + 4a - 837 = 0}}}}$



${ \implies{ \green{ \sf{ {a}^{2} + (31a - 27a) - 837 = 0}}}}$



${ \implies{ \green{ \sf{ {a}^{2} + 31a - 27a - 837 = 0}}}}$



${ \implies{ \green{ \sf{a(a + 31) - 27(a + 31) = 0}}}}$



${ \implies{ \green{ \sf{(a + 31)(a - 27) = 0}}}}$



${ \implies{ \green{ \sf{a = - 31 \: and \: 27}}}}$




we can't consider the negative value of a as the consecutive odd number assumed above...



therefore,



${ \red{ \tt{ \underline{the \: consecutive \: odd \: numbers \: are - }}}}$



${ \star{ \pink{ \sf{ \: \: \: a = 27}}}}$



${ \star{ \pink{ \sf{ \: \: \: (a + 2) = 27 + 2 = 29}}}}$



${ \star{ \pink{ \sf{ \: \: \: (a + 4) = 27 + 4 = 31}}}}$