Geometric Sequences

[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 ^ (n-1)
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

[Lisasmith111]

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 ^ (n-1)
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^(n-1)

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)^(6-1) = -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

[MAS1]

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 ^ (n-1)
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

1/3 - 1/9 + 1/27 - 1/81 + 1/243 - 1/729 ...
= 243/729 - 81/729 + 27/729 - 9/729 + 3/729 - 1/729
= 182/729

The sum of a geometric sequence is:

Sn = a(1 - r^n)/(1 - r) where a = the first term and r = common ratio

In this case a = 1/3 and r = -1/3

So
Sn = 1/3 (1 - (-1/3) ^ 6) / (1 - (- 1/3))
Sn = 1/3(1 - (1/729)) / (1 + 1/3)
Sn = 1/3(728/729) / (4/3)
Sn = (1/3)(728/729)(3/4) = 728/(4 x 729) = 182/729

I think the difference in your answers is due to the switched sign in the numerator of the sum formula.