Question

a1=10
ak+1=(.7)ak

what is the value for:
S(1)
S(10)
S(infinity)

Answer 1

arithmetic progression

Answer 2

\begin{array}l\color{red}{\normalsize\text{Q}}\color{orange}{\normalsize\text{u}}\color{#9c9a2e}{\normalsize\text{e}}\color{green}{\normalsize\text{s}}\color{blue}{\normalsize\text{t}}\color{purple}{\normalsize\text{i}}\color{purple}{\normalsize\text{o}}\color{red}{\normalsize\text{n}}\color{orange}{\normalsize\text{ }}\color{#9c9a2e}{\normalsize\text{l}}\color{green}{\normalsize\text{o}}\color{blue}{\normalsize\text{o}}\color{purple}{\normalsize\text{k}}\color{purple}{\normalsize\text{s}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{r}}\color{#9c9a2e}{\normalsize\text{e}}\color{green}{\normalsize\text{a}}\color{blue}{\normalsize\text{l}}\color{purple}{\normalsize\text{l}}\color{purple}{\normalsize\text{y}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{u}}\color{#9c9a2e}{\normalsize\text{n}}\color{green}{\normalsize\text{f}}\color{blue}{\normalsize\text{a}}\color{purple}{\normalsize\text{m}}\color{purple}{\normalsize\text{i}}\color{red}{\normalsize\text{l}}\color{orange}{\normalsize\text{i}}\color{#9c9a2e}{\normalsize\text{a}}\color{green}{\normalsize\text{r}}\color{blue}{\normalsize\text{}}\end{array}

Answer 3

first term=10

Answer 4

Ohhhhhhhhh

Answer 5

(k+1) th term = 0.7 (kth term)

Answer 6

I'd have to read up on it

Answer 7

What happened to the chat box?

Answer 8

Actually Geometric progression

Answer 9

Somebody must've baptized it

Answer 10

\[a _{k+1}/ a _{k} = 0.7 = common ratio\]

Answer 11

wait a second, did something similar to this in physics with finding ratios and such

Answer 12

so what is the variable here? a ? we replace a with 1, 10, and infinity?

Answer 13

first term=a1=10
common ratio = r = a(k+1)/a(k) = 0.7
Sum of n terms,
\[\S _{n} = a[1-r ^{n}]/ [1-r]\]

Answer 14

S(1) = 10 = sum of first 1 terms
S(10)= first 10 terms = 10[1-(0.7)^10]/[1-0.7]
S(infinity) = a/[1-r] = 10/0.3 = 33.33

Answer 15

Interesting....I'll have to read more about it.