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Question

If a, b, c are in A.P then a open parentheses fraction numerator 1 over denominator b end fraction plus fraction numerator 1 over denominator c end fraction close parentheses, b open parentheses 1 over c plus 1 over a close parentheses comma c open parentheses 1 over a plus 1 over b close parentheses are in

  1. A.P.    
  2. G.P.    
  3. H.P.    
  4. A.G.P    

Hint:

If a, b , c are in AP. We know that in AP
cross times second term = first term + third term
 

The correct answer is: A.P.


     Given : a, b , c are in AP.
    We know that in AP
    cross times second term = first term + third term
    2b = a + c
    Now, 
    (1) t subscript 1 =  a open parentheses fraction numerator 1 over denominator b end fraction plus fraction numerator 1 over denominator c end fraction close parentheses  a open parentheses 1 over b plus 1 over c space plus space 1 over a space minus space 1 over a close parentheses space left curly bracket space a d d i n g space a n d space s u b t r a c t i n g 1 over a right curly bracket 
    Let 1 over b plus 1 over c space plus space 1 over a space equals space m
    rightwards double arrow a open parentheses m space minus space 1 over a close parentheses space equals space a m space minus space 1

rightwards double arrow space T e r m space 1 left parenthesis t subscript 1 right parenthesis space equals space a m space minus space 1 space........... space left parenthesis 1 right parenthesis

    (2) Term2 (t subscript 2) =  b open parentheses 1 over c plus 1 over a close parentheses
     Term2 (t subscript 2) =  b open parentheses 1 over c plus 1 over a space plus 1 over b space minus space 1 over b close parentheses space left curly bracket space a d d i n g space a n d space s u b t r a c t i n g space 1 over b space o n space b o t h space s i d e s right parenthesis
    Term2 (t subscript 2) =  b open parentheses m space minus space 1 over b close parentheses
    Term2 (t subscript 2) =  b m space minus space 1
    Similarly,
    (3) Term3 (t subscript 3) =  c m space minus space 1
    Let t subscript 1 , t subscript 2 , t subscript 3 be in AP
    We can check if they satisfy the condition 2b = a + c
    2t subscript 2 = t subscript 1 + t subscript 3
    Substituting the values
    rightwards double arrowcross timesleft parenthesis b m space minus space 1 right parenthesis = a m space minus space 1 space plus space c m space minus 1
    rightwards double arrow 2 b m space minus space 2 = a m space plus space c m space minus space 2
    rightwards double arrow 2 b m space = a m space plus space c m space
    Taking m common and cancelling out from both sides
    rightwards double arrow 2 b space = a space plus space c space
    Thus,  a, b, c are in A.P then a open parentheses fraction numerator 1 over denominator b end fraction plus fraction numerator 1 over denominator c end fraction close parenthesesb open parentheses fraction numerator 1 over denominator c end fraction plus fraction numerator 1 over denominator a end fraction close parentheses comma c open parentheses fraction numerator 1 over denominator a end fraction plus fraction numerator 1 over denominator b end fraction close parentheses are in A.P.

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