Math: Geometric Series ?

20 + 20 * (1/6) + 20 * (1/6)^2 + 20 * (1/6)^3 + ...

How could this ever be less than 20?  I'm not telling you that 24 is right, but I can tell you with 100% certainty that 19.83 is wrong.

20, 10/3, 10/18, 10/108, ...

a_n = 5 2^(3 - n) 3^(1 - n) 

20 + 10/3 + 10/18 + 10/108, ...

sum_(n=1)^∞ 5 2^(3 - n) 3^(1 - n) = 24

Your answer 24 is correct.

The formula for the infinite sum of a geometric series is:

S = a₁ / (1 - r), when |r| < 1

Plug in your values:

S = 20 / (1 - 1/6)

S = 20 / (5/6)

Change that into a multiplication by the reciprocal:

S = 20 * 6/5

S = 120/5

S = 24

By the way, if you write a few terms, you'll see why 19.8333... doesn't make sense as the answer.

20 + 20/6 + 20/36 + ...

Clearly the first term is 20 and each term after that is positive so the sum has to be at least 20. You should be able to immediately reject the first answer as not correct.

I think you accidentally calculated:

a₁ - r

= 20 - 1/6

= 19 5/6

= 19.8333...

Read more in the link below about geometric sequences and sums.

Answer:

24

S(infinity) = a/(1 - r) 

S(inf) = 20 / ( 1 - 1/6) 

S(inf) = 20/(5/6)

S(inf) = 20 X ( 6/5) 

S(inf) = 120/5

S(inf) = 24 

The answer!!!!