Question

How do I find a1 for a geometric series when I'm only given Sn, an, and r?

Answer 1

\(\Large S_n = a_1\frac{1-r^n}{1-r}\) ---- (1)

\(\Large a_n = a_1 r^{n-1}\) ----- (2)

Answer 2

From (2): \(\Large a_n * r = a_1 r^{n}; ~~~r^n = \frac{a_nr}{a_1}\). Put this in (1) and solve for \(a_1\).

Answer 3

Thanks, this helps a lot! One more question though, how did you get:
an * r = a1rn

Answer 4

I multiplied both sides of (2) by r so that I can find \(\large r^n\) that occurs in (1)

Answer 5

\(\Large a_n = a_1 r^{n-1}\)
Multiply both sides by r:
\(\Large a_n * r = a_1 r^{n-1} * r = a_1r^n\)

Answer 6

So when you multiply \[r^n-1 * r\] , the -1 cancels out?

Answer 7

\(\large a^m * a^n = a^{m+n}\)

\(\large r^{n-1} * r^1 = r^{n-1+1} = r^n\)