For the series, identify *a*, *r*, and *n*. Find

1. *S*_{7} for the geometric series with *a* = 3 and *r* = 0.2.

2.

3.

4.

5.

Answer

1. This problem directly tells us everything we need to know. We're given *a* = 3 and *r* = 0.2. Since we're asked to find *S*_{7} we use *n* = 7.

2. We're adding terms with denominators from 7^{0} to 7^{12}. This means we're adding 13 terms together, so we must be finding *S*_{13}. The first term is *a* = 2 and the ratio is .

3. If we expand this series we get

The first term is *a* = 1, the ratio is , and we're adding *n* = 10 terms (since the exponents on go from 0 to 9). Using the formula, we get

4. The first term is and the ratio is . If we rewrite the denominators, we can see that we're adding 8 terms, so *n* = 8:

We get

5. If we expand the sum we get

2(0.9)^{6} + ... + 2(0.9)^{30}.

The first term is *a* = 2(0.9)^{6} and the ratio is *r* = 0.9. We're adding up terms with exponents from 6 to 30, so we're missing the exponents from 1 to 5.

This means we have

*n* = 30 – 5 = 25

terms. Applying the formula,