Question

Please help me right now.

Answer 1

`\sum\limits_(k=1)^(\infty)12\left(0.7\right)^(k-1)`

The infinite geometric series is converges if |r| < 1.

We have r = 0.7 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

`S=(a_1)/(1-r)`

We have:

`a_1=12(0.7)^(1-1)=12(0.7)^0=12\\\\r=0.7`

Substitute:

`S=(12)/(1-0.7)=(12)/(0.3)=40`

Answer: c. Converges, 40.