The problem is, you can only use the formula for the sum of the geometric serie if but is this given?
Q.: A geometric series has first term a and common ratio r. Its sum to infinity is 12. The sum to infinity of the squares of the terms of this geometric series is 48. Find the values of a and r.
Ans.: From text book: a = 6, r = 1/ 2
Eq.: S¥ = a/ 1 - r
Attempt:
S¥ = a/ (1 - r) = 12
a = 12 ( 1 - r)
a = 12 - 12r
S¥ = a/ 1 - r
a^2/ (1 - (ar^2/ a^2)) = 48
(12 - 12r)/ (1 - ((r(12 - 12r))^2/ (12 - 12r)^2)) = 48
(12 - 12r)/ (1 - (12r - 12r^2)^2/ (144 - 288r + 144r^2)) = 48
(12 - 12r)/ (1 - (144r^2 - 288r^3 + 144r^4)/ (r^2 - 2r +1)) = 48
(12 - 12r)/ (1 - (r^4 - 2r^3 + r^2)/ (r^2 - 2r + 1)) = 48
(12 - 12r)/ (1 - (r^2 - r^2 + r^2) = 48
(12 - 12r)/ (1 - r^2) = 48
(12 - 12r) = 48 (1 - r^2)
(12 - 12r) = 48 - 48r^2
1 - r = 4 - 4r^2
4r^2 - r - 3
(4r + 3)(r - 1)
r = -3/ 4 or r = 1 … from answer above, though, r should = 1/ 2.
Can anyone help me spot where I have gone wrong? Thank you.
Actually, I have already factored that in the attempt; see the bold reference following:Here is your mistake.
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S¥ = a/ (1 - r) = 12
a = 12 ( 1 - r)
a = 12 - 12r
S¥ = a/ 1 - r
a^2/ (1 - (ar^2/ a^2)) = 48
(12 - 12r)/ (1 - ((r(12 - 12r))^2/ (12 - 12r)^2)) = 48
Is this still incorrect?
Thank you for responding.
First of all you really need learn to post is symbols. You can use LaTeX tags
[tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives .
In the sum of the squares the first term is the ratio is .
Therefore if is the geometric series then
the series of sum of its squares is .
In the second sum, the first term is
and the ratio is .
Therefore, the sum of the second series is
.
I really think that your lack of ability to use symbols is a hindrance to your understanding.