How do you find the sum of the infinite geometric series given $a_{1} = 12$ $a_1=12$ , r=-0.6?

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Answer 1 · 2017-10-13T17:51:32

The sum of the infinite geometric series is $7.50$ $7.50$ .

Let us consider an infinite geometric series whose first term is $t_{1}$ $t_1$ & common multiplier is $r$ $r$ .

Then,

$t_{1} = a, \frac{t_{2}}{t_{1}} = r$ $t_1=a,t_2/t_1=r$ .

$S_{\infty} = \frac{a}{1 - r}$ $S_(oo)=a/(1-r)$ ..........(1).

Now given that,

$a = 12$ $a=12$ & $r = - 0.6$ $r=-0.6$

$:.r=-6/10$

$:.r=-3/5$

Now, applying formula (1) $rarr$

$S_(oo)=12/(1-(-3/5))$

$:.S_oo=12/(1+3/5)$

$:.S_oo=12/(8/5)$

$:.S_oo=(12xx5)/8$

$:.S_oo=15/2$

$:.S_oo=7(1/2)$

$:.S_oo=7.50$

Therefore, the sum of the infinite geometric series is $7.50$ . (Answer).

Hope it Helps:)

How do you find the sum of the infinite geometric series given a_1=12a1=12a_1=12, r=-0.6?