A Problem in Geometric Progressions

[PeterK]

I am baffled by this problem. I have a geometric progression in which A₁=log65536, A_n=-log(2^1/2) and S_n=log(32*2^1/2) (i.e the sum of the n terms in the progression). We need to find r and n (that is the reason and the number of terms in the progression).

I expressed all the quantities as functions of the log of 2:

A₁=16log2

A_n=-(1/2)log2

S_n=(11/2)log2

From the formula for S_n=(A_n.r-A₁)/(r-1) I can find easily that r=-2 by replacing the above values and solving the resulting equation for r.

From the formula for S_n=(A₁.(rⁿ - 1))/(r-1) I should be able to find n. Substitution of the above values plus the value of r that I just found gives the following equation:

(-2)ⁿ = (-1/32)

which gives n=-5 a result that makes no sense as the number of terms should be a positive number. Even the previous result of r=-2 does not make sense to me. If I calculate A₂ from A₁*r I would get -32log2 then A₃=64log2 and so on. I can't see how I eventually can end up with A_n=-log(2^1/2).

What am I doing wrong? Thanks for any help!

 

[Deleted member 4993]

I am baffled by this problem. I have a geometric progression in which A₁=log65536, A_n=-log(2^1/2) and S_n=log(32*2^1/2) (i.e the sum of the n terms in the progression). We need to find r and n (that is the reason and the number of terms in the progression).

I expressed all the quantities as functions of the log of 2:

A₁=16log2

A_n=-(1/2)log2

S_n=(11/2)log2

From the formula for S_n=(A_n.r-A₁)/(r-1) I can find easily that r=-2 by replacing the above values and solving the resulting equation for r.

From the formula for S_n=(A₁.(rⁿ - 1))/(r-1) I should be able to find n. Substitution of the above values plus the value of r that I just found gives the following equation:

(-2)ⁿ = (-1/32)

which gives n=-5 a result that makes no sense as the number of terms should be a positive number. Even the previous result of r=-2 does not make sense to me. If I calculate A₂ from A₁*r I would get -32log2 then A₃=64log2 and so on. I can't see how I eventually can end up with A_n=-log(2^1/2).

What am I doing wrong? Thanks for any help!

I am getting:

11/2*(r-1) = -r/2 - 16

6r = -16 + 11/2 = - 21/2

r = -21/12

 

[soroban]

Hello, PeterK!

Please check the wording of the problem.
As given, it makes no sense.

I have a geometric progression in which \(\displaystyle A_1 = \log(65,\!536).\)

. . \(\displaystyle A_n \,=\,-\log\left(2^\frac{1}{2}}\right)\) . This is a constant . . . It should be a function of n.

. . \(\displaystyle S_n \,=\,\log\left(32\cdot2^{\frac{1}{2}}\right)\) . Same here!

We need to find \(\displaystyle r\) and \(\displaystyle n.\)

 

[PeterK]

I am getting:

11/2*(r-1) = -r/2 - 16

6r = -16 + 11/2 = - 21/2

r = -21/12

Indeed. You are correct! I made an algebra error.... omitted the r on the right hand side. However, there is still something wrong with that answer (which I also get):

A₂=-(7/4)*16log2=-7*4*log2

A₃=49log2

etc. I can't see how I would get to A_n=-log2^1/2.

The correct answer is r=-1/2 which I got by simple inspection. However I can't get to it algebraically. Thanks anyway!

 

[PeterK]

Hello, PeterK!

Please check the wording of the problem.
As given, it makes no sense.

You are thinking of a generic term An which indeed would be a function of n. In this case however, this is a well defined progression that has a fixed number of terms (which is one of the questions asked) the last of which is A_n=-log2^1/2. I hope this clarifies the meaning of Sn as well. Thanks!

 

[Deleted member 4993]

Given r = -1/2, there is a typo with Sn. it should be

\(\displaystyle S_n\ = \ \frac{21}{2}log(2)\)

and n = 6

 

[PeterK]

Given r = -1/2, there is a typo with Sn. it should be

\(\displaystyle S_n\ = \ \frac{21}{2}log(2)\)

and n = 6

That solved the problem. Thanks!