What was the formula to calculate the sum of an infinite geometric sequence ?

Question

Like this one> 4+8/3+16/9+32/27+.....
or this one> -60+20-20/3+..... ?
Thank you.

Answer 1

t1/(1-r)
t1 is the first number
r is the rate

for example for the first sequence:
t1=4 and r=2/3

Answer 2

The sum of n terms in geometric series , a, ar, ar^2, ar^(n-1) is

given by

S(n) = a(1 - r^n)/(1-r) , where r <1

since r<1, when n->∞ r^n ->0

S(∞) = a/(1-r)

1)

4 + 8/3 + 16/9 -----------∞

here a = 4, r = 2/3

so S(∞) = (4)/(1 - (2/3)) = (4)/(1/3) = (4)(3) = 12

S(∞) = 12

2)

-60 + 20 -20/3 ----------∞

here a = -60 r = -1/3

S(∞) = -60/(1- (-1/3) = -60/(4/3) = -60(3/4) = -45

s(∞) = -45

Answer 3

The sum of a geometric series of the form
a + ar + ar^2 + ar^3 + .... is a/(1-r).

For the first one, a=4 and r=(2/3), so it's 4/(1-2/3) = 4/(1/3) = 12.

For the second one, a=-60 and r=-1/3, so it's -60/(1-(-1/3)) = -60/(4/3) = (3/4)60 = 45.

Answer 4

The first S = 4+8/3( 1/(1-2/3) ) =12
In general S = 1/(1-r) if you start with 1
2) -40 -20/3(1-1/3+1/9++) = -40-20/3*3/4= -45

Answer 5

S1 = a/(one million - r) = 4 => (S1)^3 = a^3 / (one million - r)^3 = sixty 4 ... (one million) S2 = a^3 / (one million - r^3) = 192 ... (2) (2) ÷ (one million) = (one million - r)^3 / (one million - r^3) = 3 => one million - 2r + r^2 = 3 + 3r + 3r^2 => 2r^2 + 5r + 2 = 0 => r = -2 OR -one million/2. yet |r| < one million => r = -one million/2 => a = 4(3/2) = 6. => GP = 6, -3, 3/2, -3/4, ...