Geometric Sequence??? Please help

Dec 2015
9
0
LA
I am also stuck on this question. Let {b sub k} be a geometric sequence. Given that (b sub 4)= 80, (b sub 9)=1280, and (S sub 9)=2555. Find the first term (b sub 1).

I have tried it every way I know how but I am getting nowhere. No amount of searching on the internet has helped either. Thanks!!!

Picture of problem:
math1.png
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,216
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North Texas
Recheck the given values and subscripts ... as written, the problem doesn't work.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I think there is a typo ... If $b_5=80$, then $b_1=5$, common ratio is $r=2$, and $S_9=2555$.
 
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Dec 2015
9
0
LA
I think there is a typo ... If $b_5=80$, then $b_1=5$, common ratio is $r=2$, and $S_9=2555$.
Can you explain how you arrived at that? I have been unable to find any example where geometric sequences have 2 letters (b & S). Thanks.
 

romsek

MHF Helper
Nov 2013
6,646
2,994
California
Can you explain how you arrived at that? I have been unable to find any example where geometric sequences have 2 letters (b & S). Thanks.
$S_9 = \displaystyle{\sum_{n=1}^9} b_n$
 
Feb 2015
2,255
510
Ottawa Ontario
Only way for B4 = 80 is B1 (or a) = 5/4 and ratio = 4:
5/4 * 4 = 5 * 4 = 20 * 4 = 80
which of course ain't what the doctor ordered...
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I assume you understand the form of a geometric sequence ...

$\{b_k\} = b_1,b_1r,b_1r^2, b_1r^3, \, ...$, where $b_1$ is the first term & $r$ is the common ratio.

$b_5=b_1r^4=80$

$b_9=b_1r^8=1280$

$\dfrac{b_1r^8}{b_1r^4}=\dfrac{1280}{80} \implies r^4=16 \implies r = \pm 2$

either value of $r$ makes $b_1=5$

now for the sum. you should know this formula & how it was derived ...

$S_n=\dfrac{b_1(1-r^n)}{1-r}$

for $r=2$ ...

$S_9=\dfrac{5(1-2^9)}{1-2}=2555$

for $r=-2$ ...

$S_9=\dfrac{5[1-(-2)^9]}{1-(-2)}=855$

which tells us the common ratio is $r=2$
 
Dec 2015
9
0
LA
I assume you understand the form of a geometric sequence ...

$\{b_k\} = b_1,b_1r,b_1r^2, b_1r^3, \, ...$, where $b_1$ is the first term & $r$ is the common ratio.

$b_5=b_1r^4=80$

$b_9=b_1r^8=1280$

$\dfrac{b_1r^8}{b_1r^4}=\dfrac{1280}{80} \implies r^4=16 \implies r = \pm 2$

either value of $r$ makes $b_1=5$

now for the sum. you should know this formula & how it was derived ...

$S_n=\dfrac{b_1(1-r^n)}{1-r}$

for $r=2$ ...

$S_9=\dfrac{5(1-2^9)}{1-2}=2555$

for $r=-2$ ...

$S_9=\dfrac{5[1-(-2)^9]}{1-(-2)}=855$

which tells us the common ratio is $r=2$
Thank you very much. This was very helpful. Honestly I'm in way over my head here and barely understand any of this. But I can follow along with your equations. Thanks again.