Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: R

There are no items in this cart.
Continue Shopping
Tapasranjan Das Grade: 12

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.

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points

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

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.
Askiitians Experts

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details