Question

A geometric series has the form ______ and equals _____ when ∣r∣<1. What happens when a does not equal 0 and ∣r∣≥1?

A. a+ar+ar^2+…; a/1−r; Diverges.

B. a+ar+ar^2+…; a/1−r; Converges to zero.

C. ar^n;a(1−r^n)/(1−r); Diverges.

D. ar^n; a(1−r^n)/(1−r); Converges to a/(1−r).

Answer 1

A geometric series with a form of a+ar+ar^2+... equals a/(1-r) if |r|<1. When a does not equal 0 and |r|≥1, the series diverges, meaning it doesn't sum to a finite number.

Step-by-step explanation:

A geometric series has the form a+ar+ar^2+... and equals a/(1-r) when |r|<1. If a does not equal 0 and |r|≥1, then the series diverges. This means the sum of the series does not approach a finite value as the number of terms increases. The correct answer is A. a+ar+ar^2+…; a/1−r; Diverges.