Question

help no idea how to solve it
what is the tenth term of the geometric sequence that has a common ratio of '1/3' and 36 as its fifth term

Answer 1

A.4/27
B.27/4
C.4/81
D.1/36

Answer 2

\(\huge a_n = a_1(r)^{n-1}\)
\(a_1\) = 1st term
r = ratio
n = the \(n^{th}\) term

Answer 3

so when you find the n term do you subtract it by 1

Answer 4

so we know that the 5th term is 36, we know that the ratio is 1/3
so what's the first term anyway?

well
the 5th term equals 36 with a ratio of 1/3, thus
$$\bf a_n = a_1(r)^{n-1}\\
a_5 = a_1\left(\cfrac{1}{3}\right)^{5-1}\\
36 = a_1\left(\cfrac{1}{3}\right)^{4} \implies
36 = a_1\left(\cfrac{1^4}{3^4}\right)\\
\cfrac{
\cfrac{36}{1}}{
\left(\cfrac{1^4}{3^4}\right)} = a_1
$$

Answer 5

what do you do in the last part

Answer 6

\(\bf \cfrac{
\frac{36}{1}}{
\left(\frac{1^4}{3^4}\right)} = a_1 \implies
\cfrac{36}{1} \times \cfrac{81}{1}\)

Answer 7

so now we know that the 1st term, \(a_1 = 36 \times 81 = 2916\)

Answer 8

so you want the \(\bf n^{th}\) term
in this case is 10
recall => \(\huge a_n = a_1(r)^{n-1}\)

Answer 9

so, what does that give you?

Answer 10

im still lost

Answer 11

ok, what confused you? :)

Answer 12

the formula

Answer 13

ok, what confuses you in the formula?

Answer 14

like do you multiply everything then subtract the n and 1

Answer 15

hmm, the n-1 is the exponent
say if your \(\bf n^{th}\) term is say 23

then n = 23, thus n- 1 => 23 -1

Answer 16

and \(\bf \large a_n = a_{23}\)

Answer 17

n just stands as a generic variable in the notation for the \(\bf n^{th}\) term
whichever it happens to be, it can be 3, or 5 or 17, or 1,000,000
in this case it's just 10

Answer 18

still confused?