Geometric Series, determining the value of the first term.

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tug187

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Question : If the 3rd and 9th term of a geometric series with a positive common ratio are -3 and -192 respectively, determine the value of the first term, a.

I was thinking that you have to find out r first so i did this :

an = a3 x r^n-1
a9 = -3 x r^3-1
-192 = -3r^2
/-3 /-3
64=r^2
r=8

Using this, I don't know how to find out the first term still.. I tried this:

a1 = -3x8^1-1
a1 = -3x8^0

I don't really know what to do at this point but if its 8^0 then thats 0 so the a1=-3 but the 3rd term is -3 :shock:
 
Your common ratio is a little steep.

The formula for a geometric series is:

\(\displaystyle \L\\a_{n}=\underbrace{a_{1}}_{\text{first term}}r^{n-1}\)

Set up a system of equations and solve for the needed variables.

You need r=common ratio and a_1=first term.

\(\displaystyle \L\\-3=a_{1}r^{2}\)
\(\displaystyle \L\\-192=a_{1}r^{8}\)

Solve for r and a_1.
 
Re: Geometric Series, determining the value of the first ter

tug187 said:
Question : If the 3rd and 9th term of a geometric series with a positive common ratio are -3 and -192 respectively, determine the value of the first term, a.

I was thinking that you have to find out r first so i did this :

an = a3 x r^n-1
a9 = -3 x r^3-1
-192 = -3r^2
/-3 /-3
64=r^2
r=8

Using this, I don't know how to find out the first term still.. I tried this:

a1 = -3x8^1-1
a1 = -3x8^0

I don't really know what to do at this point but if its 8^0 then thats 0 so the a1=-3 but the 3rd term is -3 :shock:
Click to expand...

So...let's call -3 the first term. Then, -192 would be the 7th term, right?

a<SUB>n</SUB> = a<SUB>1</SUB>*r<SUP>(n - 1)</SUP>

-192 = -3 * r<SUP>(7 - 1)</SUP>

-192 = -3*r<SUP>6</SUP>

Divide both sides by -3.....

64 = r<SUP>6</SUP>
2<SUP>6</SUP> = r<SUP>6</SUP>, so

2 = r

Ok...

Now, if we know that term 3 is -3,
a<SUB>3</SUB> = a<SUB>1</SUB>*r<SUP>3-1</SUP>
-3 = a<SUB>1</SUB>*2<SUP>2</SUP>
-3 = 4*a<SUB>1</SUB>
-3/4 = a<SUB>1</SUB>
 
Re: Geometric Series, determining the value of the first ter

Hello, tug187!

Another approach . . .

If the 3rd and 9th term of a geometric series
with a positive common ratio are -3 and -192 respectively,
determine the value of the first term, a.
Click to expand...

We're expected to know that the \(\displaystyle n^{th}\) term is: \(\displaystyle \:a_n\:=\:ar^{n-1}\)
. . where \(\displaystyle a\) is the first term and \(\displaystyle r\) is the common ratio.

We are told that the \(\displaystyle 3^{rd}\) term is -\(\displaystyle 3.\)
. . We have: \(\displaystyle \:a_3\:=\:ar^2 \:=\:-3\;\) [1]

We are told that the \(\displaystyle 9^{th}\) term is -\(\displaystyle 192.\)
. . We have: \(\displaystyle \:a_9 \:=\:ar^8 \:=\:-192\;\) [2]


Divide [2] by [1]: \(\displaystyle \L\:\frac{ar^8}{ar^2} \:=\:\frac{-192}{-3}\)\(\displaystyle \;\;\Rightarrow\;\;r^6 \:=\:64\;\;\Rightarrow\;\;r\:=\:\pm2\)

Substitute into [1]: \(\displaystyle \:a(\pm2)^2\:=\:-3\;\;\Rightarrow\;\;\fbox{a\:=\:-\frac{3}{4}}\)

 
I like the "money" approach:

$-3 is deposited at i% periodically, and is worth $-192 after 6 periods:

-3(1 + i)^6 = -192
(1 + i)^6 = 64
1 + i = 64^(1/6)
1 + i = 2

So "r" = 2
 
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