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An infinite geometric series has sum 2000. A new series, obtained by squaring each term of the original series, has sum 16 times the sum of the original series. The common ratio of the original series is m/n, where m and n are relatively prime positive integers. Find m+n.

 

Please help me.

 Jul 12, 2018

Best Answer 

 #3
avatar
+3

Curious, different mistakes in #1 and #2 above, both leading to the same incorrect answer.

 

Suppose that the first series is 

a, ar, ar2, ar3,

Its sum to infinity is 2000, so

a1r=2000(1)

 

For the second series, each term of the first series is squared, so the second series will be 

a2, a2r2, a2r4, a2r6,

Its sum to infinity is 16 times the sum of the first series, so

a21r2=32000(2)

 

Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic

63r2125r+62=0,

from which

(63r62)(r1)=0,

so r = 62/63, (r can't equal 1).

Substitution in (1) shows that a = 2000/63.

 Jul 13, 2018
 #1
avatar
+1

1-       

Sum = F / [1 - R], where F= First term, R= Common ratio.

2000 = 16 / [1 - R], solve for R

R = 124 / 125

 

2-

32,000 = 256 / [1 - R], solve for R

R = 124 / 125

Therefore: m/n =124/125 and m + n = 124 + 125 =249

 Jul 12, 2018
 #2
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+1

2000 = F / [1 - R],  ....................................(1)
32000 =F^2 / [1 - R]...................................(2)
From (1) above:
F =2000 - 2000R - Sub this for F in (2) above:
32000 =[2000 - 2000R]^2 / [1 -R] 
Solve for R:
32000 = (2000 - 2000 R)^2/(1 - R)

32000 = (2000 - 2000 R)^2/(1 - R) is equivalent to (2000 - 2000 R)^2/(1 - R) = 32000:
(2000 - 2000 R)^2/(1 - R) = 32000

Multiply both sides by 1 - R:
(2000 - 2000 R)^2 = 32000 (1 - R)

Expand out terms of the right hand side:
(2000 - 2000 R)^2 = 32000 - 32000 R

Subtract 32000 - 32000 R from both sides:
-32000 + (2000 - 2000 R)^2 + 32000 R = 0

Expand out terms of the left hand side:
4000000 R^2 - 7968000 R + 3968000 = 0

The left hand side factors into a product with three terms:
32000 (R - 1) (125 R - 124) = 0

Divide both sides by 32000:
(R - 1) (125 R - 124) = 0
Split into two equations:
R - 1 = 0 or 125 R - 124 = 0

Add 1 to both sides:
R = 1 or 125 R - 124 = 0

Add 124 to both sides:
R = 1 or 125 R = 124

Divide both sides by 125:
R = 1 or R = 124/125

(2000 - 2000 R)^2/(1 - R) ⇒ (2000 - (2000 124)/125)^2/(1 - 124/125) = 32000:
So this solution is correct

(2000 - 2000 R)^2/(1 - R) ⇒ (2000 - 2000 1)^2/(1 - 1) = (undefined):
So this solution is incorrect

The solution is:
R = 124/125

 Jul 12, 2018
 #3
avatar
+3
Best Answer

Curious, different mistakes in #1 and #2 above, both leading to the same incorrect answer.

 

Suppose that the first series is 

a, ar, ar2, ar3,

Its sum to infinity is 2000, so

a1r=2000(1)

 

For the second series, each term of the first series is squared, so the second series will be 

a2, a2r2, a2r4, a2r6,

Its sum to infinity is 16 times the sum of the first series, so

a21r2=32000(2)

 

Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic

63r2125r+62=0,

from which

(63r62)(r1)=0,

so r = 62/63, (r can't equal 1).

Substitution in (1) shows that a = 2000/63.

Guest Jul 13, 2018
 #4
avatar+145 
-1

The answer was 125. I got it wrong.

 Jul 14, 2018
 #5
avatar+118703 
-1

Hi Daniel, it is good that you gave feedback but it would have been a lot nicer if you had said thank you to all the people who allocated their time to your question.

Melody  Jul 14, 2018

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