Question

Urgent algebra help ;c
Write the 10th term of the sequence.

Answer 1

okay

Answer 2

what's the common ratio?

Answer 3

5 ?

Answer 4

well.. \(\large \begin{array}{llll}
625&125&25&...\\
\hline\\
&\frac{625}{5}&\frac{125}{5}
\end{array}\)

Answer 5

\(\large \begin{array}{llll}
a_1\\
\downarrow \\
625&125&25&...\\
\hline\\
&625 \cdot \frac{1}{5}&125\cdot \frac{1}{5}\\
&&\qquad \ \uparrow \\
&&r=\textit{common ratio}
\end{array}\qquad \bf \large a_n=a_1\cdot r^{n-1}\\ \quad \\
\Large a_{{\color{red}{ 10}}}=a_1\cdot r^{{\color{red}{ 10}}-1}\)

Answer 6

okay im kinda confused

Answer 7

ok.... where?

Answer 8

do I plug in 1/5 in 'r'?

Answer 9

yes, that's the common ratio in the geometric sequence
notice that to get the next term, you just multiply the previous one by 1/5

Answer 10

Notice that each successive term of the sequence 625,125,25,.. is 1/5 of the preceding term, so, yes, 1/5 is the common ratio.

Can you now write a formula a sub n for the nth term of this sequence? Once you have that, you can easily predict the 10th term by letting n=10.

Answer 11

I got \[\frac{ 1 }{ 3125 }\] ?

Answer 12

is it A?

Answer 13

@jdoe0001 @mathmale

Answer 14

Tima-Shorty,

I did suggest this earlier:

Notice that each successive term of the sequence 625,125,25,.. is 1/5 of the preceding term, so, yes, 1/5 is the common ratio.

Can you now write a formula a sub n for the nth term of this sequence? Once you have that, you can easily predict the 10th term by letting n=10.

Coming up with such a formula is an important skill in sequences and series.

Answer 15

is it A? \(\large \checkmark\)

Answer 16

With all due respect to y our wishes, Tima-Shorty, I'd much rather help you understand the underlying concept here, which is prediction of the nth term of a sequence. Twice you have not responded to my suggestion that you try writing a formula for that.