# Can any one help me?

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#1
A geometric progression has first term logbase2 27 and common difference logbase2 x.

i) Find the set of values of y for which the geometric progression has a sum to infinity.
ii) Find the exact value of y for which the sum to infinity of the geometric progression is 3.
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1 month ago
#2
(Original post by iusama0)
A geometric progression has first term logbase2 27 and common difference logbase2 x.

i) Find the set of values of y for which the geometric progression has a sum to infinity.
ii) Find the exact value of y for which the sum to infinity of the geometric progression is 3.
Do you have notes or a textbook? If so read the relevant section if not Google it. In particular, what is the requirement on the common ratio in order for there to be a sum to infinity? For part ii, if necessary, remind yourself of the formula for the sum to infinity and then form an equation and solve it to find y (or is it x?).
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#3
(Original post by Plücker)
Do you have notes or a textbook? If so read the relevant section if not Google it. In particular, what is the requirement on the common ratio in order for there to be a sum to infinity? For part ii, if necessary, remind yourself of the formula for the sum to infinity and then form an equation and solve it to find y (or is it x?).
Y is equal to logbase2 x. x=y.
1
1 month ago
#4
(Original post by iusama0)
Y is equal to logbase2 x. x=y.
Can you make any progress now?
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#5
(Original post by Plücker)
Can you make any progress now?
What is the solution, if geometric progression is 3 is not given?
0
1 month ago
#6
(Original post by iusama0)
What is the solution, if geometric progression is 3 is not given?
Have you done part one? Do you know the required formula?
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#7
(Original post by Plücker)
Have you done part one? Do you know the required formula?
Yes!
If the geometric progression has a sum to infinity, the common ratio, r must be less than 1 and more than -1, so -1<r<1 and r=log2(y) so -1<log2(y)<1

log2(y)>-1 is same as 2^(-1)<y i.e. 1/2<y

log2(y)<1 is the same as 2^1>y i.e. 2>y

So y must be in between 1/2 and 2 i.e. 1/2<y<2
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1 month ago
#8
(Original post by iusama0)
Yes!
If the geometric progression has a sum to infinity, the common ratio, r must be less than 1 and more than -1, so -1<r<1 and r=log2(y) so -1<log2(y)<1

log2(y)>-1 is same as 2^(-1)<y i.e. 1/2<y

log2(y)<1 is the same as 2^1>y i.e. 2>y

So y must be in between 1/2 and 2 i.e. 1/2<y<2
I thought that your common ratio was .

Anyway, you know the first term and the sum to infinity. Can you write an equation and solve it to find the common ratio?
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#9
(Original post by Plücker)
I thought that your common ratio was .

Anyway, you know the first term and the sum to infinity. Can you write an equation and solve it to find the common ratio?
first term is logbase2 27 and sum to infinity are unknow.
equation is:
S = a/1-r
a = first term
r = common ratio
S = sum to infinity
0
1 month ago
#10
(Original post by iusama0)
first term is logbase2 27 and sum to infinity are unknow.
equation is:
S = a/1-r
a = first term
r = common ratio
S = sum to infinity
S = 3.
0
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