An infinite geometric series has 1 and 1/5 as its frist two terms:1, 1/5' 1/25' 1/125' what is the sum, s, of the infinite series?

A 1/4

B1/25

c 5/4

d 1

you have a geometric series

since you are summing the powers of 1/5, this converges

(1/5 < 1)

it's equal to 1 / (1-1/5)

=1 / (4/5) = 5/4

for geometric series, the sum is always

x = 1 / (1-r)

where r is the ratio of successive terms.

if you set the sum equal to x:

x = 1+r+r^2+r^3 ...

and multiply each term by r

rx = r+r^2+r^3 + ...

then subtract

x-rx = 1+r-r+r^2-r^2+r^3-r^3 + ...

x-rx = 1

x (1-r) = 1

x=1 / (1-r)