If ( 1 + 3 + ⋯ + a ) + ( 1 + 3 + ⋯ + b ) = 1 + 3 + ⋯ + c , with a + b + c = 57 and a < 10 , find 2 b − 2 c + a .

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MathematicsSequence & SeriesSpecial sequences/seriesHard2 minQB

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Passage / Comprehension

If $(1 + 3 + \dots + a) + (1 + 3 + \dots + b) = 1 + 3 + \dots + c$ , with $a + b + c = 57$ and $a < 10$ , find $2 b - 2 c + a$ .

Options:

Answer:

5.00

Solution:

Write $a = 2 n_{1} - 1, b = 2 n_{2} - 1, c = 2 n_{3} - 1$ . Then $n_{1}^{2} + n_{2}^{2} = n_{3}^{2}$ and $n_{1} + n_{2} + n_{3} = 30$ . With $a < 10$ , the triple is $(5, 12, 13)$ , so $(a, b, c) = (9, 23, 25)$ and $2 b - 2 c + a = 5$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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