Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

i got a 4

Answer 5

ok! correct!

Answer 6

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

Answer 7

I got -33554432

Answer 8

that's right!

Answer 9

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

Answer 10

ok!

Answer 11

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