1 Answer

Prabah · Answer 1 · 2023-09-19T20:37:09+0000

The given geometric series is

$120+20+(10)/(3)+(5)/(9)+\cdots$
$The\text{ common difference r=}(20)/(120)=(1)/(6)$
$\text{The first term is a=120.}$

Recall that the formula for the sum of the infinite geometric series is

$\sum ^(\infty)_(k\mathop=0)ar^k=(a)/(1-r)$

Substitute a=120 and r=1/6, we get

$\sum ^(\infty)_(k\mathop=0)120((1)/(6))^k=(120)/(1-(1)/(6))$
=(120)/((6-1)/(6))

$\text{ Use }(a)/((b)/(c))=a*(c)/(b)\text{.}$

Hence the sum of the given infinite geometric series is 144.

Hence the third option is correct.

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