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Hi friends,

 

I have stumbled upon a problem which I believe is not solveable, am I doing something wrong or is it one of those sums?

 

Above the Sigma notation we have "m"

Bottom we have k=1

 

The expression is: \((2)^{8-k}=255{1 \over2}\)

Determine "m"

 

Okay, I calculated T1, and got 128

T2, and got 64

 

Since it is stated that this is an Arithmetic sequence, I know that d = -64

 

Using \(Sn={n \over2}[2a+(n-1)d]\)

 

I get to a step where it says 

 

\(64n^2-320n+511=0\)

 

This is not solveable...I'm I doing something wrong here?..Please help. Thank you.

 Mar 12, 2023
 #1
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The answer is 15.

 Mar 12, 2023
 #2
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Hi guest,

 

The answer sheet says it's 9

juriemagic  Mar 12, 2023
 #3
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a=listfor(k, 1, 9, (2^(8 - k))

 

The above is your sequence. It is NOT arithmetic sequence but a GEOMETRIC sequence, where:

F==128 - this the first term, m ==9 - this is the number of terms, r ==1/2 - this is the common ratio.

 

So, your GS looks like this:

 

(128, 64, 32, 16, 8, 4, 2, 1, 0.5)==255.5  {As you can see here, r ==1/2}

 

You sum it up using the sum formula for GS:

 

{128 *  [( 1 - 1/2^9) / (1 - 1/2)]}==S

{128 * [(1 -  0.001953125) / (1/2)]}==S

{128 * [0.998046875 / 0.5]} ==S

{128 *  1.99609375} ==255.5 - which is the answer you want.

 Mar 12, 2023
 #6
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Hi guest,

 

Thank you...yes the paper indicated this was an Arithmetic sequence...no wonder I got stuck...Thank you.

juriemagic  Mar 13, 2023
 #4
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Hi Juriemagic,

 

\(\displaystyle \sum _{k=1}^m\;\;2^{8-k}=255.5\)

 

First I did a little spreadsheet to see if it worked

You can see that the answer is 9.

Now I will try it without the help of Excel.

 

The sequence is      128, 64, 32, 16 ,....

This is a GP with r=-0.5 and a=128

 

\(S_n=\frac{a(1-r^n)}{1-r}\\ S_n=\frac{128(1-(0.5)^n)}{0.5}=255.5\\~\\ 256(1-(0.5)^n)=255.5\\~\\ 1-(0.5)^n=\frac{255.5}{256}\\~\\ 1-\frac{255.5}{256}=(0.5)^n\\~\\ \frac{0.5}{256}=\frac{1}{2^n}\\~\\ \frac{1}{512}=\frac{1}{2^n}\\~\\ 2^n=512\\~\\ n=9 \)

 Mar 13, 2023
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Hello Melody,

 

mmm, I know I would have gotten it right if it was'nt for the misleading info that this was an Arithmetic sequence. Let this be a lesson to ALWAYS calculate the 3rd term as well and MAKE SURE the sequence is indeed what the paper says it is. Thank you very much Melody...I do appreciate..

juriemagic  Mar 13, 2023
 #7
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You are always welcome. :)

 

I could see straight off that it wasn't an AP becasue of the power., 

You will get more used to seeing such hints too. 

Melody  Mar 13, 2023
edited by Melody  Mar 13, 2023
 #8
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laughlaugh..Thank you

juriemagic  Mar 13, 2023

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