+145 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
\(\displaystyle a, \ ar,\ ar^{2}, \ ar^{3}, \dots\)
Its sum to infinity is 2000, so
\(\displaystyle \frac{a}{1-r}=2000\dots(1)\)
For the second series, each term of the first series is squared, so the second series will be
\(\displaystyle a^{2}, \ a^{2}r^{2}, \ a^{2}r^{4}, \ a^{2}r^{6}, \dots\)
Its sum to infinity is 16 times the sum of the first series, so
\(\displaystyle \frac{a^{2}}{1-r^{2}}=32000\dots(2)\)
Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic
\(\displaystyle 63r^{2} - 125r +62 = 0,\)
from which
\(\displaystyle (63r-62)(r-1)=0,\)
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
\(\displaystyle a, \ ar,\ ar^{2}, \ ar^{3}, \dots\)
Its sum to infinity is 2000, so
\(\displaystyle \frac{a}{1-r}=2000\dots(1)\)
For the second series, each term of the first series is squared, so the second series will be
\(\displaystyle a^{2}, \ a^{2}r^{2}, \ a^{2}r^{4}, \ a^{2}r^{6}, \dots\)
Its sum to infinity is 16 times the sum of the first series, so
\(\displaystyle \frac{a^{2}}{1-r^{2}}=32000\dots(2)\)
Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic
\(\displaystyle 63r^{2} - 125r +62 = 0,\)
from which
\(\displaystyle (63r-62)(r-1)=0,\)
so r = 62/63, (r can't equal 1).
Substitution in (1) shows that a = 2000/63.
