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# 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. Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear Tapasranjan

let first series is

S1 =a +ar1  +ar1+ar13 +ar14  ...............................

and second series is

S1 =b +br2  +br2+br23 +br24  ...............................

given   ar1 = br2

S1 =1 =a/1-r1

or a=1-r1

third term  ar1=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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