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Question Number 46923 by 786786AM last updated on 02/Nov/18

a, b, c are in AP or GP or HP according  as ((a−b)/(b−c))  is equal to

a,b,careinAPorGPorHPaccordingasabbcisequalto

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

  1) when a,b,c are in A.P  then a−b=b−c    so ((a−b)/(b−c))=1  2) when a,b,c are in GP  b=ar   c=ar^2   ((a−b)/(b−c))=((a−ar)/(ar−ar^2 ))=((a(1−r))/(ar(1−r)))=(1/r)=reciprocal of   common ratio  3)when a,b,c are in GP  (2/b)=(1/a)+(1/c)  (1/b)−(1/a)=(1/c)−(1/b)  ((a−b)/(ab))=((b−c)/(bc))  ((a−b)/(b−c))=(a/c)

1)whena,b,careinA.Pthenab=bcsoabbc=12)whena,b,careinGPb=arc=ar2abbc=aararar2=a(1r)ar(1r)=1r=reciprocalofcommonratio3)whena,b,careinGP2b=1a+1c1b1a=1c1babab=bcbcabbc=ac

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