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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.
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 +ar14 ............................... and second series is S1 =b +br2 +br22 +br23 +br24 ............................... 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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
Dear Tapasranjan
let first series is
S1 =a +ar1 +ar12 +ar13 +ar14 ...............................
and second series is
S1 =b +br2 +br22 +br23 +br24 ...............................
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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards,Askiitians ExpertsBadiuddin
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