Question

What is the sum of the first five terms of a geometric series with a sub 1 = 6 and r = 1/3?

Express your answer as an improper fraction in lowest terms without using spaces.

Answer 1

use the standard formula
\(\large S_n = a_1 \dfrac{1-r^n}{1-r}\)
where n= 5

Answer 2

do you think you could walk me through the steps? I've never really done this type of problem before :o

Answer 3

ok, plug in all known values in that formula, what do u get ?
(try simplifying the denominator atleast ? its simple :))

Answer 4

Sn=a1 1-5^n/-4 ? :o

Answer 5

a1 is =6 , is given
we want sum of 5 terms, so n =5
r= 1/3 is given, not 5
\(\large S_n = a_1 \dfrac{1-r^n}{1-r} \\ \large \large S_5 = 6 \dfrac{1-(1/3)^5}{1-(1/3)}\)

can you simplify a bit ?

Answer 6

oh okay! so then would I multiply 6 with the numerator and the denominator??

Answer 7

(1/3)^5 = 1/3^5 = 1/243
\( \large \large S_5 = 6 \dfrac{1-1/243}{2/3}= 9 \dfrac{243-1}{243}= \dfrac{242}{27}\)
just simple algrbraic steps....

Answer 8

oh I see, but I can't really see what was done with the 6 or the 9. do you mind explaining to me what you did? I'm sorry I know it must be frustrating, I'm frustrated! Math in general is like a foreign language to me!

Answer 9

no problem!
we had 6 in numerator and 2/3 in the denominator