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 :

$S=\dfrac{a_1}{1-r}$

We have:

$a_1=12(0.7)^{1-1}=12(0.7)^0=12\\\\r=0.7$

Substitute:

$S=\dfrac{12}{1-0.7}=\dfrac{12}{0.3}=40$

Answer: c. Converges, 40.