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Suppose $a, b, c >0$ are in geometric progression and $a^p = b^q =c^r \neq 1$. Which one of the following is always true?

  1. $p, q, r$ are in geometric progression
  2. $p, q, r$ are in arithmetic progression
  3. $p, q, r$ are in harmonic progression
  4. $p=q=r$
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given a,b,c are in gp means b/a = c/b (common ratio)

                                            => b2 = ac ...............@1

now given ap =bq = cr

           p loga = q logb = r logc = k(say)   [taking log on every values]

         so,  a = 10 k/p  b= 10 k/q    c= 10 k/r

putting these values in @1.

             10 2k/q = 10 ( k/p + k/r)

                2k/q  = k/p + k/r

               2/q = 1/p + 1/r (cancelling k from both sides)

              1/q -1/p  = 1/r -1/q (writing above eqn in other way)

 which means that 1/p , 1/q , 1/r is in AP.

now we know that

if p,q,r is in AP than i/p,1/q,1/r is in HP.

so, as 1/p,1/q,1/r is in AP than p,q,r is in HP..

answer is option C.

        

          

               

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