Question

*Hi! I need help solving this problem, it’s from the calculus portion of my ACT prep guide.*

Answer 1

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))`

`=(120)/((5)/(6))`
`\text{ Use }(a)/((b)/(c))=a*(c)/(b)\text{.}`

`=120*(6)/(5)`
`=144`

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

Hence the third option is correct.