Sir please post the solution of the following problem _

Q:-Two distinct,real,infinite geometric series each have a sum of 1 and have the same second term. Third term of one of the series is 1/8. If the second term of both series can be written in the form ((√m) -n)/p , where m,n and p are positive integers and m is not divisible by the square of any prime,find the value of 100m+10n+p.

Dear Tapasranjan

let first series is

S1 =a +ar1  +ar1²  +ar1³ +ar1^{4  ...............................}

and second series is

S1 =b +br2  +br2²  +br2³ +br2^{4  ...............................}

given   ar1 = br2

 S1 =1 =a/1-r1

  or a=1-r1

third term  ar1²  =1/8

put value of a

 (1-r1)r1² =1/8

or  r1³ +r1² +1/8=0

by inspection one value of r1 = 1/2

devide above equation be r1 -1/2 =0

 remaining 2 roots are the roots of equation  4r1² -2r1-1=0

 r1= (1±√5)/4

second term  =ar1 =(1-r1)r1

                           =r1 - r1²

                                  =r1-(r1 /2  +1/4)

                             =r1/2 -1/4

               put value of r1=(1+√5)/4

    second term  = (√5 -1)/8    other value of r will not give this form

 so m=5   , n=1  ,p=8

100m+10n+p = 518

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