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Question

asked Nov 21, 2019 163k views

1 Answer

Skm · Answer 1 · 2019-11-28T09:21:33+0000

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

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.