Respuesta :
if he wants to pay 1.2 times the amount of the previous month, then 1.2 becomes a multiplier from the current month's amount, to get the "next" month's amount, namely 1.2 is the "common ratio", that simply means is a geometric sequence, with a first term value of 200
[tex]\bf \textit{sum of a finite geometric sequence}\\\\ \begin{array}{llll} S_n\implies \sum\limits_{n=1}^{30}\ a_1\cdot r^{n-1}\\\\ S_n=a_1\left( \cfrac{1-r^n}{1-r} \right) \end{array} \qquad \begin{cases} n=n^{th} \ term\\ a_1=\textit{value of first term}\\ r=\textit{common ratio}\\ ----------\\ r=1.2\\ a_1=200 \end{cases} \\\\\\ \sum\limits_{n=1}^{30}\ 200(1.2)^{n-1}\iff 200\left( \cfrac{1-1.2^{30}}{1-1.2} \right)[/tex]
is convergent, because is a finite sum
[tex]\bf \textit{sum of a finite geometric sequence}\\\\ \begin{array}{llll} S_n\implies \sum\limits_{n=1}^{30}\ a_1\cdot r^{n-1}\\\\ S_n=a_1\left( \cfrac{1-r^n}{1-r} \right) \end{array} \qquad \begin{cases} n=n^{th} \ term\\ a_1=\textit{value of first term}\\ r=\textit{common ratio}\\ ----------\\ r=1.2\\ a_1=200 \end{cases} \\\\\\ \sum\limits_{n=1}^{30}\ 200(1.2)^{n-1}\iff 200\left( \cfrac{1-1.2^{30}}{1-1.2} \right)[/tex]
is convergent, because is a finite sum
Answer: The series is divergent.
Step-by-step explanation:
The geometric series is given by
[tex]\sum_{n=0}^{\infty }ar^n=a+ar+ar^2+...[/tex]
- If |r| < 1 then the geometric series converges to.
- If |r| ≥1 then the geometric series diverges.
The given sentences a geometric series,
where the first payment represents the first term a.
The number of times he wants to pay back for succeeding payments represent the common ration i.e. r=1.2.
The first 30 payments represents the number of terms to use in finding the sumi.e. n=30.
Since the common ratio, r>1, the series is divergent.