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Sequences And Series

Question
CBSEENMA11015627

If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 + y)m are in A.P., then m and r satisfy the equation

  • m2 – m(4r – 1) + 4r2 – 2 = 0

  • m2 – m(4r+1) + 4r2 + 2 = 0

  • m2 – m(4r + 1) + 4r2 – 2 = 0

  • m2 – m(4r – 1) + 4r2 + 2 = 0

Solution

C.

m2 – m(4r + 1) + 4r2 – 2 = 0

Given space to the power of straight m straight C subscript straight r minus 1 end subscript space comma space to the power of straight m straight C subscript straight r space plus straight C presuperscript straight m subscript straight r plus 1 end subscript space are space in space straight A. straight P.
2 to the power of space straight m end exponent straight C subscript 2 space equals space fraction numerator blank to the power of straight m straight C subscript straight r minus 1 end subscript over denominator straight C presuperscript straight m subscript straight r end fraction space plus fraction numerator straight C presuperscript straight m subscript straight r plus 1 end subscript over denominator straight C presuperscript straight m subscript straight r end fraction
space equals space fraction numerator straight r over denominator straight m minus straight r plus 1 end fraction space plus fraction numerator straight m minus straight r over denominator straight r plus 1 end fraction
rightwards double arrow space straight m squared minus straight m space left parenthesis 4 straight r plus 1 right parenthesis space plus 4 straight r squared space minus 2 space equals space 0 space