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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesG.P.Medium2 minQB

MathematicsMediumsingle choice

If $a, b, c$ are in G.P. and the fractions exist, what progression contains $1/ (b - a), 1/ (2 b), 1/ (b - c)$ ?

Options:

Answer:

Solution:

The G.P. relation $b^{2} = a c$ implies $(b - a) (b - c) = 2 b^{2} - b (a + c)$ . Rearrangement gives $1/ (b - a) + 1/ (b - c) = 1/ b$ , which is twice the middle term.

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:G.P.

⏱ 2mℹ️ Source: QB

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