# Thread: Series and Sequences Last part

1. ## Series and Sequences Last part

The sum of the first 4 terms of a geometric series is 30, and the sum of the infinite series is 32. Find the first 3 terms.

For the geometric sequence a, ar, ar^2,…show that the sequence log a, log (ar), log (ar^2) is an arithmetic sequence

Find a number which, when added to each of 2,6 13 gives three number in geometric sequence

These are the last questions, I apologise for posting up so many, I just really need some help with these, thanks again.

2. Originally Posted by christina
The sum of the first 4 terms of a geometric series is 30, and the sum of the infinite series is 32. Find the first 3 terms.

For the geometric sequence a, ar, ar^2,…show that the sequence log a, log (ar), log (ar^2) is an arithmetic sequence

Find a number which, when added to each of 2,6 13 gives three number in geometric sequence

These are the last questions, I apologise for posting up so many, I just really need some help with these, thanks again.
For the second part

$\displaystyle \log{(ar)} = \log{a} + \log{r}$

$\displaystyle \log{(ar^2)} = \log{a} + \log{r} + \log{r}$

$\displaystyle \log{(ar^3)} = \log{a} + \log{r} + \log{r} + \log{r}$.

I think it should be obvious that you have an arithmetic sequence with $\displaystyle t_1 = \log{a}$ and $\displaystyle d = \log{r}$.

3. Originally Posted by christina
The sum of the first 4 terms of a geometric series is 30, and the sum of the infinite series is 32. Find the first 3 terms.

For the geometric sequence a, ar, ar^2,…show that the sequence log a, log (ar), log (ar^2) is an arithmetic sequence

Find a number which, when added to each of 2,6 13 gives three number in geometric sequence

These are the last questions, I apologise for posting up so many, I just really need some help with these, thanks again.
For the first part:

$\displaystyle S_n = \frac{a(r^n - 1)}{r - 1}$

so $\displaystyle S_4 = 30 = \frac{a(r^4 - 1)}{r - 1}$.

Call this equation (1).

$\displaystyle S_{\infty} = 32 = \frac{a}{1 - r}$

Call this equation (2).

$\displaystyle \frac{(1)}{(2)}$ gives

$\displaystyle 1 - r^4 = \frac{30}{32}$

$\displaystyle r^4 = \frac{1}{16}$

$\displaystyle r = \frac{1}{2}$.

Since you know $\displaystyle 32 = \frac{a}{1 - r}$

$\displaystyle 32 = \frac{a}{1 - \frac{1}{2}}$

$\displaystyle 32 = \frac{a}{\frac{1}{2}}$

$\displaystyle 32 = 2a$

$\displaystyle a = 16$.

Since the first term is $\displaystyle 16$ and the ratio is $\displaystyle \frac{1}{2}$, the first three terms are

$\displaystyle 16, 8, 4$.