Question

How do you find a1 in a geometric series when Sn=218.4; an=1.4; r=.2?

Answer 1

\[Sn=a \frac{1- r ^{n} }{ 1-r },\]

Answer 2

@surjithayer I have the equation set up, but I have no n. I'm not sure how to find the n, but once I find that, I know how to solve the equation.

Answer 3

\[also take Tn=a r ^{n-1},Tn=an=1.4\]

Answer 4

Hint
\[Sn=a \frac{ 1-r ^{n-1}r }{ 1-r }\]

Answer 5

Manders and surjithayer: I, too, see some ambiguity here. \[s _{n}\] represents the nth sum of a geometric series, which here is supposedly 218.4.
\[a _{n}\] represents the nth term of said series, which supposedly is 1.4.

surjithayer: My version of the formula for the nth partial sum is \[a \frac{ 1-r ^{n} }{ 1-r }.\] .

Whether we use your version or mine, the fact remains that either involves TWO unknowns, n and a. To solve this system, it appears to me that we need one more equation. Any ideas?

Answer 6

Another question: how would we use the info that \[a _{n} = 1.4\] ??

Answer 7

1.4=ar^(n-1)

Answer 8

Well, wouldn't we have to set up 1.4=a1+(n-1)^.2 in arithmetic form to attempt to find a1 and n? Truthfully, I'm stumped.

Answer 9

no ,it is a geometric series.

Answer 10

\[218.4= \frac{ a \left( 1-r ^{n-1}r \right) }{ 1-r }=\frac{ a-ar ^{n-1}*r }{ 1-r }\]

Answer 11

\[218.4=\frac{ a-1.4r }{ 1-r },put r=.2 and find a\]

Answer 12

surji: What an elegant solution! Thanks. Manders, if questions remain for you, please ask them and I'll provide you with verbal explanations of the reasoning and steps involved.

Answer 13

yw