Question

What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

(I have no idea what the common ratio is for this sequence.)

Answer 1

Wait nvm

Answer 2

Well, you start at 10, and multiply by x to get to the second term, and x again to get to the third...you'll eventually multiply by x 7 times.

10x^7 = 78250

That should be a start to find the common ratio.

Answer 3

So let the first term be 'a' = 10
The last, that is the 8th term = \(T_8\) = 781250

Now, the nth term of a GP series = \(\large ar^{n-1}\), where 'r' is the common ratio.

So, here,

\[T_8 = ar^{n-1} = 781250\]

You have 'a', and 'n' as well, can you find 'r' from this?

Answer 4

Once you have found out the common ratio 'r', you can find the sum of the series using the following relation:
Sum 'S' of the first 'n' terms of a GP with common ratio 'r', and first term 'a' is:
\[S = \frac{a(r^n-1)}{r-1}\]

Answer 5

Thanks but I had already figured it out. "nvm" = nevermind lol

Answer 6

Lol no worries ;)

Answer 7

@apoorvk and I believe the formula is \[S _{n} = a _{1}(1 - r ^{n}) \div 1-r\]

Answer 8

(Not really sure how to use the equation button on here correctly.)

Answer 9

@workin_daily ahaaa...

\[\frac{(r^n−1)}{r−1} = \frac{(1-r^n)}{1-r}\]

If you look carefully, both the terms mean the same things - understand this? :p

Answer 10

For adding fractions use this code --> \frac{<numerator>}{<denominator>} within the starting syntax '/[' and the ending syntax '\]'.

Answer 11

11(1-(5)^8) -3906240
---------- = ---------- = -976560
1-5 4

Answer 12

Yeahhhh, -976560 doesn't seem right.

Answer 13

So where did I go wrong? o.O

Answer 14

11(1-(5^7) , not "...5^8.." ---> since for the ntyh term we use "r^(n-1)"
I just wrote that equality to relate that the two are the one and the same, don't use that!
Use the equation posted above that (or the one that you know, same thing)