need help solving geometric series questions

Question

A geometric sequence has its first term as $10000$ and a fourth term as $−7290$. If the pattern continues forever, find the sum of the terms in the sequence.

I know that the $n^{th}$ term is found by $$t_n=a_1r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio.

Thus the fourth term is given by $$10000 * r^3 = -7290$$

so $$r^3 = -0.729$$ $$r= -0.9$$

So the sum of the series is: $$S_ \infty = \frac{10000}{1-(-0.9)}$$

Is this correct?

2. The sum of the first 2018 terms of a geometric sequence is 200. The sum of the first 4036 terms is 380. Find the sum of the first 6054 terms

Answer 1

For your second question, let $S_n$ be sum of first $nth$ terms and r be the common ratio, then we have $ S_{2018} = 200$ and $ S_{4036} = 380$ Note that $$ S_{4036} = S_{2018} + r^{2018} * S_{2018}$$ $$ 380 = 200 + r^{2018} * 200$$ $$ r^{2018} = 0.9$$

Also note that $$ S_{6054} = S_{2018} + r^{2018} * S_{2018} + r^{4036} * S_{2018} $$ $$ S_{6054} = 200+ 0.9 * 200+ 0.9^2 * 200 $$ $$ S_{6054} = 542$$

Answer 2

For #1 Yes. You correctly used the formula for the sum of an infinite geometric series. It works because $|-0.9|<1.$

For #2, let's try to write it out.

The first 2018 terms would be $a+ar^1+ar^2...ar^{2017}$

The first 4036 terms would be $a+ar^1...ar^{2017}...ar^{2018}+ar^{2019}..ar^{4035}$

Notice that this is simply $(a+ar^1+ar^2...ar^{2017})(1+r^{2018})$.

You can figure out that $1+r^{2018}=1.9$ because the first $2018$ terms add up to $200$ and the first $4036$ to $380$.

Therefore, $r^{2018} = 0.9$.

Finally, the first 6054 terms, as you may be able to tell, is equal to $(a+ar^1+ar^2...ar^{2017})(1+r^{2018}+r^{4036})=200(1+0.9+0.9^2)=200(2.71)=\boxed{542}$

Answer 3

Your first answer is correct.

For the second, you need to do some sum manipulations. You know if the first term is $a$ and the ratio is $r$, and if we call the known sums $S_1,S_2$,

$$S_1 = \sum_{n=1}^{2018} ar^{n-1} = 200 \;\;\;\;\; S_2 = \sum_{n=1}^{4036} ar^{n-1} = 380$$

Now, we want to find

$$S = \sum_{n=1}^{6054} ar^{n-1}$$

Note that we can split up $S$ as so:

$$S = \sum_{n=1}^{6054} ar^{n-1} = \sum_{n=1}^{4036} ar^{n-1} + \sum_{n=4037}^{6054} ar^{n-1} = S_2 + \sum_{n=4037}^{6054} ar^{n-1}$$

From the latter, notice what happens if we factor out $r^{4036}$ and then reindex the sum with $m=n-4036$:

$$\sum_{n=4037}^{6054} ar^{n-1} = r^{4036} \sum_{n=4037}^{6054} ar^{n-1-4036} = r^{4036} \sum_{m=1}^{2016} ar^{m-1} = r^{4036}S_1$$

Thus,

$$S = S_2 + r^{4036}S_1$$

We still need $r$. We can do similar trickery with $S_2$ though:

$$S_2 = \sum_{n=1}^{4036} ar^{n-1} = 380 = \sum_{n=1}^{2018} ar^{n-1} + r^{2018} \sum_{n=1}^{2018} ar^{n-1} = r^{2018}S_1 + S_1$$

Thus,

$$r^{2018} = \frac{380 - S_1}{S_1} = \frac{380 - 200}{200} = \frac 9 {10} \implies (r^{2018})^2 = r^{4036} = \left(\frac 9 {10} \right)^2 = \frac{81}{100}$$

Then

$$S = S_2 + \frac{81}{100} S_1 = 380 + \frac{81}{100} \cdot 200 = 542$$