SOLUTION: I have one last question on my homework and I can�t seem to figure it out. I�ll send the full question and show you what I did throughout. 9. The first two terms of an infinit

Question 1104573: I have one last question on my homework and I can�t seem to figure it out.
I�ll send the full question and show you what I did throughout.
9. The first two terms of an infinite geometric sequence, in order, are 2log(sub2)x, log(sub2)x, where x > 0.
a. Find r.
For this, I did log(sub2)x/2log(sub2)x = 1/2
b. Show that the sum of an infinite sequence is 4log(sub2)x.
I used a1/(1-r) which was 2log(sub2)x/(1-1/2) = 4log(sub2)x
c. The first three terms of an arithmetic sequence, in order, are log(sub2)x, log(sub2)x/2,log(sub2)x/4
Find d, giving your answer as an integer.
(All the ones below are going to be log(sub2), but I�ll fight them as log so it won�t be as messy.
Logx/2=logx-log2=logx-1
Logx/4=logx-log4=logx-2
D=-2
d. Let S(sub12) be the sum of the first 12 terms of the arithmetic sequence. Show that S(sub12)=12log(sub2)x-66.
(I did this part but it�s a lot and would take a long time to type out)

e. THE PART I NEED HELP ON, :)
given that S(sub12) is equal to half the sum of the infinite geometric sequence, find x, giving your answer in the form 2^p, where p has the domain of Q.

Found 3 solutions by greenestamps, Theo, apshu:
Answer by greenestamps(13195)   (Show Source): You can put this solution on YOUR website!

It seems this part is easier than most of the other parts, if I read it correctly.

We have an infinite geometric series with a sum of 4*log2(x); and we have the sum of 12 terms of an arithmetic series with a sum of 12*log2(x)-66; and we are supposed to solve for x if the sum of the arithmetic series is half the sum of the geometric series:

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
i think you're ok for part a and b.
for part c, i think d = -1 would be correct.
reason.

general formula is An = A1 + (n-1) * d

when d = -1 ....

A1 = log2(x) + (0 * -1) = log2(x)

A2 = log2(x) + (1 * -1) = log2(x) - 1 = log2(x) - log2(2) = log2(x/2).

A3 = log2(x) + (2 * -1) = log2(x) - 2 = log2(x) - log2(4) = log2(x/4).

A4 = log2(x) + (3 * -1) = log2(x) - 3 = log2(x) - log2(8) = log2(x/8)

d = -1 satisfies the requirements of the sequence.

for part d, you want to show that S12 = 12 * log2(x) - 66.

the formula for Sn is Sn = n/2 * (A1 + An).

when n = 12, the formula becomes S12 = 12/2 * (A1 + An).

this becomes S12 = 12/2 * (log2(x) + log2(x) - 11) which becomes S12 = 6 * (2 * log2(x) - 11) which becomes 12 * L2(x) - 66

log2(x) is the same as log[sub2](x).

it's just a shorter way to show it.

so, as best i can determine:

you're good for parts a and b.

part c should be d = -1.

part d is shown to be what they say it is by applying the sum of an arithmetic series formula and finding A12 which is equal to log2(x) - 11.

Answer by apshu(1)   (Show Source): You can put this solution on YOUR website!
d is equal to -1, not -2, that is your mistake. divide the second term by the first term and you get log 2 (1/2) which is -1. Proceed by finding the Un, with the formula U1+ (n-1)d, which is log2 of x - 11. Then proceed with the formula of the S12 = 12/2 (log 2 of x - log 2 of x - 11)
you get s12 = 6 (log 2 of x square - 11) = 6 (2 log 2 of x -11) = 12 log 2 of x - 66.
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