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 April 13th, 2012, 04:12 AM #1 Member   Joined: Apr 2012 Posts: 90 Thanks: 0 Geometric Progression Hello World, I am urgently looking for solutions to these questions. Would appreciate the help! 10 15 1.)Prove that a.) ? 8(0.5)^n and b.) ? 2-(5^(-n)) is geometric and find each sum. n=1 n=1 How do you prove that it is geometric and I think I know how to find each sum, if proving that the series is geometric gives me the common ratio. 2.) Nine numbers are in GP and are alternately positive and negative. If the first number is 24 and the last number is 3/32, find the sum of the numbers. I got U1 = 24 and ar^8 = 3/32 Therefore, r^8 = (3/32)/(24) and I got 1/256 Since it is alternately positive and negative, I believe the common ratio should be a negative. But then how do I continue, or am I even doing it right? 100 3.) Evaluate ? (0.01n+0.09^n) Thank you much! n=1
 April 13th, 2012, 04:48 AM #2 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Geometric Progression 1. Use the definition of a geometric serie. 2. What you have done so far is correct, note that $r^8=\frac{1}{256} \Rightarrow r=\frac{1}{2}$, but indeed because you have an alternating sequence $r=\frac{-1}{2}$ From this point, you should be able to find the sum of the numbers. 3. $\sum_{n=1}^{100} [0.01n+(0.09)^n]$ Note that you have a sum that means you can split the summation as: $\sum_{n=1}^{100} (0.01n)+\sum_{n=1}^{100}(0.09)^n$ $=0.01 \sum_{n=1}^{100} n + \sum_{n=1}^{100}(0.09)$ The first summation is an arithmetic serie and the second is a geometric serie.
April 13th, 2012, 05:34 AM   #3
Math Team

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Re: Geometric Progression

Hello, Ter!

Quote:
 $\text{2) Nine numbers are in GP and are alternately positive and negative.} \text{If the first number is 24 and the last number is }\frac{3}{32}, \;\;\;\;\text{find the sum of the numbers.$

$\text{W\!e have: }\:a_{_1} \,=\,24,\;a_{_9} \,=\,\frac{3}{32}$

$\text{And }a_{_9} \,=\,a_{_1}r^{^8}$

$\text{So we have: }\:24\,\!r^{^8} \:=\:\frac{3}{32} \;\;\;\Rightarrow\;\;\;r^{^8} \:=\:\frac{1}{256} \;\;\;\Rightarrow\;\;\;r \:=\:\pm\,\!\frac{1}{2}$

$\text{Since the signs alternate: }\,r \,=\,-\,\!\frac{1}{2}$

$\text{The sum of the first 9 terms is: }\:S_{_9} \:=\:a\,\cdot\,\frac{1\,-\,r^{^9}}{1\,-\,r}$

$\text{Therefore: }\:S_{_9} \;=\;24\,\cdot\,\frac{1\,-\,\left(-\frac{1}{2}\right)^9}{1-\left(-\frac{1}{2}\right)} \;=\;24\,\cdot\,\frac{\frac{513}{512}}{\frac{3}{2} } \;=\;\frac{513}{32}$

Quote:
 $\text{3) Evaluate: }\:\sum^{100}_{k=1}\left(0.01n\,+\,0.09^n\right)$

$\text{W\!e have two sums:}
\;\;\;A \:=\:0.01\sum^{100}_{k=1}k \;\;\text{ and }\;\;B \:=\,\sum^{100}_{k=1}0.09^k$

$A \;=\;0.01\,\cdot\,\frac{(100)(101)}{2} \;=\;\frac{101}{2}$

$B \;=\;0.09\,\cdot\,\frac{1-(0.09)^{100}}{1-0.09} \;\approx\;\frac{9}{{91}$

$\text{Therefore: }\:A\,+\,B \;=\;\frac{101}{2}\,+\,\frac{9}{91} \;=\;\frac{9209}{182}$

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