arithmetic and geometric series?!?! help!!

[catherineeee]

find the sum:
2/3 + 1/3 + 1/6 + 1/12 + ....
--- I got 4/3 is this right at all?!

and one other:
find there arithmetic means between 4 and 324
I got 4, 84, 164, 244 and 324... I dont know what the other two would be.

Please helppp!!!!

 

[skeeter]

find the sum:
2/3 + 1/3 + 1/6 + 1/12 + ....
--- I got 4/3 is this right at all?!
yes

and one other:
find there arithmetic means between 4 and 324
I got 4, 84, 164, 244 and 324... I dont know what the other two would be.
three instead of there? let's see ... 84 would be one, 164 would be two, and 244 would be three ... what do you mean by "the other two"?

Please helppp!!!!

 

[TchrWill]

find the sum:
2/3 + 1/3 + 1/6 + 1/12 + ....
--- I got 4/3 is this right at all?!

2/3 + 1/3 + 1/6 + 1/12 +... =
1/3[2 + 1 + 1/2 + 1/4 +... =
1/3[3 + 1/2 + 1/4 + 1/8 +...=
1/3[3 + (1/2 + 1/4 = 1/8 +...)

The limit of Sn = 1/2^n + 1/2^(n+1) + 1/2^(n+3) + .....as n---> inf. = 1.

Therefore, 1/3[3 + 1) = 4/3

 

[soroban]

Hello, catherineeee!

\(\displaystyle \text{Find the sum: }\;\frac{2}{3} + \frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \hdots\)
\(\displaystyle \text{I got }\frac{4}{3}\quad\hdots\text{ is this right?}\)

\(\displaystyle \text{We have a geometric series with: first term }a =\frac{2}{3},\:\text{common ratio }r = \frac{1}{2}\)

\(\displaystyle \text{Its sum is: }\; S\;=\;\frac{a}{1-r} \;=\;\frac{\frac{2}{3}}{1-\frac{1}{2}} \;=\;\frac{\frac{2}{3}}{\frac{1}{2}} \;=\;\frac{4}{3}\)

Find three arithmetic means between 4 and 324.

\(\displaystyle \text{We have: }\;4,\;a,\;b,\;c,\;324\)


\(\displaystyle \text{Then: }\;\begin{array}{ccccccc}a &=&\frac{4+b}{2} & \Rightarrow & 2a - b &=& 4 \\ \\[-3mm] b &=&\frac{a+c}{2} & \Rightarrow & a - 2b + c &=& 0 \\ \\[-3mm] c &=&\frac{b+324}{2} & \Rightarrow & b - 2c &=& \text{-}324 \end{array}\)


\(\displaystyle \text{Solve the system of equations: }\;\boxed{a \:=\:84,\;b \:=\:164,\;c \:=\:244}\)