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.
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
a1−r=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
a21−r2=32000…(2)
Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic
63r2−125r+62=0,
from which
(63r−62)(r−1)=0,
so r = 62/63, (r can't equal 1).
Substitution in (1) shows that a = 2000/63.
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
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
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
a1−r=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
a21−r2=32000…(2)
Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic
63r2−125r+62=0,
from which
(63r−62)(r−1)=0,
so r = 62/63, (r can't equal 1).
Substitution in (1) shows that a = 2000/63.