Quote:
Originally Posted by Peter G
I have to find the sum of a sequence. They don't tell me until what term but they give me the term itself in the sequence, so:
1/3  1/9 + 1/27 ..... 1/729
So I did: 1/729 = 1/3 x 1/3 ^ (n1)
And got: n = 6
Then: Sn = 1/3 (1 + 1/3 ^ 6) / (1 + 1/3)
I get 365/1458 which is slightly different from what I get when I right all down and perform it "manually" (364/1458, thus, 182/729) and very different from the answer both in the book and what I got from a Geometric Sequence calculator online. Can anyone please help me?
Thanks

Hello Peter and everyone,
I think you got your answer right and the book has a misprint. Quite common in maths books.
Anyway, have a look at the following:
The sequence upto the 6th term is as follows:
1/3, 1/9 ,1/27, 1/81, 1/243, 1/729
I got these two extra terms just by using the previous term and mutiplying it by the common ratio as follows:
1st term = 1/3
2nd term = 1/3 * 1/3 = 1/9
3rd term = 1/9 * 1/3 = 1/27
4th term = 1/27 *  1/3 = 1/81
5th term = 1/81 * 1/3 = 1/243
6th term = 1/243 * 1/3 = 1/729
So adding the terms together gives the sum which equals 182/729.
Now using formulae for a GP to find the nth term is as follows:
nth term = a * r^(n1)
where:
a = first term = 1/3
r = common ratio = 1/3
n = number of the term we are trying to find
we get term 6 = 1/3(1/3)^(61) = 1/729
Now the sum of n terms formula is:
a * (1  r^n)/(1  r)
where:
a = first term = 1/3
r = common ratio = 1/3
n = number of terms we want to sum (in our case the 6th term)
So by formula the 6th term of the above Geometic Progression can be found with:
sum of the 6 terms = 1/3(1 (1/3)^6)) / (1  1/3)
= 1/3 * 728/729/ (4/3)
= 1/3 * 728/729 * 3/4 = 182/729
Hope this helps