How do I find the Nth term?

Question

So, the first terms are 16, 8, 4...

(The process of going smaller is continued indefinitely)

What is the 11th term?

What is the sum of all those terms?

Answer 1

This is a geometric series, as each term is the previous one multiplied by a fixed number (1 / 2).

If the first term in a geometric series is a, the common ratio is r, and the number of terms is n, the k-th term is ar^(k - 1), and the n-th term is ar^(n - 1).

The terms are, in succession:

a, ar, ar^2, ar^3 ... ar^(k - 1) ... ar^(n - 1).

For the given series: a = 16 and r = 1 / 2.

Putting k = 11, the 11th term is:

16(1 / 2)^(11 - 1)

= 16 / 2^10

= 2^4 / 2^10

= 2^(- 6).

When | r | < 1 sum to infinity is:

lim (n -> infty) [ a(1 - r^n) / (1 - r) ]

= a / (1 - r).

Putting a = 16, r = 1 / 2 gives the sum:

16 / (1 / 2)

= 32.