Question

Given the sequence 2, 6, 18, 54, ..., which expression shown would give the fifteenth term?
2^15
2·3^15
2·3^14

Answer 1

I think it's the second one..

Answer 2

Right?

Answer 3

we can rule a out easily now work out b and let me know when you finish

Answer 4

28,697,814 is what I got

Answer 5

Okay, you need to know one thing to answer this question.
1) Geometric Sequence

The Geometric Sequence Formula is \[a_n = a _{1} * r ^{n-1}\]

an = number we're looking for
a1 = 2
r = ration = 3
n = number of terms = 15

\[a_n = 2 * 3 ^{14}\]

Answer 6

2,6,18,54

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 3 gives the next term. In other words, an=r*an-1.
Geometric Sequence: r=3

Remove the parentheses around the expression 15.
a15=2*3^(15-1)

Subtract 1 from 15 to get 14.
a15=2*3^(14)

Simplify the result.
a15=2*(4782969)

Multiply 2 by each term inside the parentheses.
a15=9565938

Answer 7

so D

Answer 8

Generally, as regards to "geometric sequences"
with "common ratio" r, and the first term \(a_1\)
we can write the equation for "any nth term"
which is denoted as, \(a_n\).

\(\color{#000000 }{ \displaystyle a_n=a_1\times r^{n-1} }\)

Or, you know that any "(n-1)st term" multiplied
times the common ratio "r", gives you the "nth term"
(a term after (n-1)st term). Or, equation-wise,
\(\color{#000000 }{ \displaystyle a_n=a_{n-1}\times r }\)

For example, when this last formula is applied,
\(\color{#000000 }{ \displaystyle a_2=a_1\times r }\)

and that can be rearranged to
\(\color{#000000 }{ \displaystyle a_{2}/ r=a_2\cdot r^{-1}=a_1 }\)

So you can write,
\(\color{#000000 }{ \displaystyle a_n=a_2\cdot r^{-1}\times r^{n-1}=a_2 \cdot r^{n-2} }\)

In fact, if you think about it,
\(\color{#000000 }{ \displaystyle a_n=a_k\cdot r^{-1}\times r^{n-k} }\)
for any integer k.

So in summary,
\(\color{#000000 }{ \displaystyle a_n=a_{n-1}\times r }\)

\(\color{#000000 }{ \displaystyle a_n=a_1\times r^{n-1} }\)

\(\color{#000000 }{ \displaystyle a_n=a_k\cdot r^{-1}\times r^{n-k} }\)

Answer 9

Oh, when I said "in fact ..."
the equation there should be just
\(\color{#000000 }{ \displaystyle a_n=a_k\cdot r^{n-k} }\)

Answer 10

And so is the last equation in the post, just,
\(\color{#000000 }{ \displaystyle a_n=a_k\cdot r^{n-k} }\)

Answer 11

Thanks everyone for the help! I get it now :)

Answer 12

I think that its the second one.