Question

Solve for … Find the sum of the series.Round to the nearest whole number. (No decimals)

Answer 1

Given:

`-12-4-(4)/(3)-\ldots\text{..-}(4)/(243)`

Find the general representation of the given series,

`\begin{gathered} a_n=ar^(n-1) \\ a=\text{ first term =-12} \\ r=\text{ common ratio = 1/3} \\ a_n=-12((1)/(3))^(n-1) \end{gathered}`

As, the nth term of the series is -4/243.

`\begin{gathered} a_n=-12((1)/(3))^(n-1) \\ -(4)/(243)_{}=-12((1)/(3))^(n-1) \\ (1)/(729)=((1)/(3))^(n-1) \\ \text{Apply exponent rule} \\ -(n-1)=-6 \\ n=7 \end{gathered}`

The sum of the first 7 terms of the given series is,

`\begin{gathered} S_n=(a(1-r^n))/(1-r),|r|<1 \\ S_7=(-12(1-((1)/(3))^7)/(1-(1)/(3)) \\ =(-(8744)/(729))/((2)/(3)) \\ =-(4372)/(243) \\ =-17.99 \\ \approx-18 \end{gathered}`

Answer: the sum of the series is -18.