## anonymous one year ago Identify the 12th term of a geometric sequence where a1 = 8 and a6 = -8,192. 134,217,728 33,554,432 -33,554,432 -134,217,728

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1. Michele_Laino

the general formula, for the n-th term, is: $\Large {a_n} = {a_1}{q^{n - 1}}$ so we can write: $\Large \begin{gathered} {a_6} = {a_1}{q^5} \hfill \\ {a_{12}} = {a_1}{q^{11}} \hfill \\ \end{gathered}$

2. Michele_Laino

now, substituting your data into the first formula, we get: $\Large - 8192 = 8{q^5}$ please solve that equation for q

3. Michele_Laino

hint: $\Large q = \sqrt[5]{{\frac{{ - 8192}}{8}}} = ...?$

4. anonymous

i got a 4

5. Michele_Laino

ok! correct!

6. Michele_Laino

Please wait, I got q=-4 now, substitute q=-4 and a_1=8 into the second equation: $\Large {a_{12}} = 8 \times {\left( { - 4} \right)^{11}} = ...?$

7. anonymous

I got -33554432

8. Michele_Laino

that's right!

9. anonymous

Can you help me with this one: Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth. (2 points) a17 ≈ 123,802.31 a17 ≈ 30,707.05 a17 ≈ 19,684.01 a17 ≈ 216,654.05

10. Michele_Laino

ok!

11. Michele_Laino

we can write this: $\Large {a_5} = {a_1}{q^4}$ so, substituting your data, we have: $\Large q = \sqrt[4]{{\frac{{{a_5}}}{{{a_1}}}}} = \sqrt[4]{{\frac{{150.06}}{{16}}}} = ...?$