Question

Find a general formula for the nth term of the geometric series with:

Answer 1

\[a _{9}=1/7.......r=1/7\]

Answer 2

@SolomonZelman

Answer 3

is that \(\Large\color{black}{ \displaystyle a_{9}}\) ?

Answer 4

yes

Answer 5

so you have a common ratio 1/7 and 1st term of 1/7.

now, for any geom. sequence, you know:

\(\Large\color{black}{ \displaystyle a_n=a_{k}\times(r)^{n-k}}\)

WHERE:
(1)
k and n are natural numbers

can be denoted as \(\Large\color{black}{ \displaystyle \left\{ k,~n\right\}\in{\bf N}}\)

(2)
\(\Large\color{black}{ \displaystyle n>k}\)
n is a bigger natural number than k.

Answer 6

\(\Large\color{black}{ \displaystyle a_9=a_{1}\times\left(\frac{1}{7}\right)^{9-1}}\)

Answer 7

so \[a _{n}=7^{6,7,8,9} (1/7) ^{n-1}\]

Answer 8

\(\Large\color{black}{ \displaystyle\frac{1}{7}=a_{1}\times\left(\frac{1}{7}\right)^{9-1}}\)

Answer 9

so, this way you can find the 1st term

Answer 10

then write, \(\Large\color{black}{ \displaystyle a_n=a_1\times (r)^{n-1}}\) formula, BUT FILL IN YOUR VALUES

Answer 11

or do you want a \(\Large\color{black}{ \displaystyle \sum}\) notation ?

Answer 12

well the only i gotta do now is find what the exponent is

Answer 13

also sorryu no i dont

Answer 14

find the exponent ?

Answer 15

so after i fill in my values are you thinking its 6 or 7

Answer 16

well see what i wrote above yours ^

Answer 17

it has to be one of those exonents

Answer 18

and I think its 7

Answer 19

have you found the first term?

Answer 20

dont have to its just \[a\]

Answer 21

\(\Large\color{black}{ \displaystyle \frac{1}{7}=(a_1) \times \left( \frac{1}{7} \right)^{8} }\)

\(\Large\color{black}{ \displaystyle 7^{1}=(a_1) \times \left( 7^{-1} \right)^{8} }\)

\(\Large\color{black}{ \displaystyle 7^{1}=(a_1) \times \left( 7^{-8} \right)}\)

\(\Large\color{black}{ \displaystyle 7^{-1} \div 7^{-8}=a_1 }\)

\(\Large\color{black}{ \displaystyle 7^{-1 -(-8)}=a_1 }\)

\(\Large\color{black}{ \displaystyle 7^{7}=a_1 }\)

Answer 22

this si what I got

Answer 23

is*

Answer 24

and your common ratio is \(\Large\color{black}{ \displaystyle 7^{-1} }\)

Answer 25

just a?

Answer 26

so i was right? I thought it was 7 too

Answer 27

\[a _{n}\]

Answer 28

don't know what you meant by your last comment.

but your terms are

so, as you said your terms will just be

\(\Large\color{black}{ \displaystyle 7^{7},~~~~ 7^{6},~~~~7^{5},~~~~ 7^{4},~~~~}\) etc...

\(\Large\color{black}{ \displaystyle \Downarrow,~~~~~ \Downarrow,~~~~~\Downarrow,~~~~~ \Downarrow,~~~~}\) etc...

\(\Large\color{black}{ \displaystyle a_1,~~~~ a_2~~~~a_3,~~~~ a_4,~~~~}\) etc...

Answer 29

but we agree its \[7^{7}\]

Answer 30

So, your formula \(\Large\color{brown}{ a_{\rm n}= a_{\rm 1} \times ({\rm r})^{{\rm n}-1} }\)

can be written as \(\Large\color{blue }{ a_{\rm n}= \left(7^7 \right) \times ({\rm 7^{-1}})^{~{\rm n}-1} }\)

Answer 31

or you can write 7^(-1) as 1/7, it is all same./

Answer 32

the first term, yes

Answer 33

or, if you want to add all the terms, then write \(\Large\color{blueviolet}{ \displaystyle \sum_{ n=1 }^{ \infty } 7^{(8-n)}}\)

Answer 34

my connection is very bad, sorry for delays

Answer 35

same here dude