Algebra 2?

when r = 1/2, and n = 4

n = number of term = 4

we can find a

Last term = ar^(n - 1)

Last term = ar^(4 - 1)

Last term = ar³

Last term = a(1/2)³

Last term = a(1/8)

Let a = 1

∴ Last term = 1(1/8)

∴ Last term = 1/8

The geometric series is therefore,

first term, a = 1

second term, ar = 1(1/2) = 1/2

third term, ar² = 1(1/2)² = 1/4

forth term, ar³ = 1(1/2)³ = 1/8

∴ The series are:

1, 1/2, 1/4, 1/8 ......

where, a = 1, r = 1/2, n = 4

A series is the sum of a sequence or progression

For a geometric sequence, the series is:

Sn = a( 1 + r^n)/(1 - r) where a is the first term

For n =4 and r = 1/2

Sn = a( 1 + (1/2)^4)/( 1 - 1/2)

= a( 1 + 1/16)/(1/2)

= 2a(17/16)

= 17a/8

As you have presented the problem:
The common ratio (r) is ½
The number of terms (n) is 4

We can start with any starting number and then keep multiplying it by ½. Do that until you have 4 terms.

Example:
24, 12, 6, 3

Or:
16, 8, 4, 2

Or:
4, 2, 1, ½

P.S. If n was supposed to represent the starting number, then use the last sequence (4, 2, 1, ½, ...)

32, 16, 8, 4
or
.1, .05, .025, .0125
or
8000 4000 2000 1000

Start with anything and then keep dividing by 2.

Let a = 1st term

a + ar + ar^2 + ar^3

a + a/2 + a/4 + a/8