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Sagot :
Answer:
[tex]\begin{gathered} a)a_n=-36\cdot\frac{1}{3}^{n-1} \\ b)\text{ }a_n=\frac{1}{3}\cdot a_{n-1} \\ c)\text{ converges to -54} \\ d)\text{ s=-54} \end{gathered}[/tex]Step-by-step explanation:
The explicit and recursive formula for a geometric sequence is represented by the following:
[tex]\begin{gathered} \text{ Explicit formula:} \\ a_n=a_1\cdot r^{n-1} \\ \text{ Recursive formula:} \\ a_n=r\cdot a_{n-1} \\ \text{where,} \\ r=\text{ common ratio} \end{gathered}[/tex]The common ratio of the pattern is:
[tex]\begin{gathered} \frac{-12}{-36}=\frac{1}{3} \\ \frac{-4}{-12}=\frac{1}{3} \end{gathered}[/tex]Then, for the explicit formula:
[tex]a_n=-36\cdot\frac{1}{3}^{n-1}[/tex]Recursive formula:
[tex]a_n=\frac{1}{3}\cdot a_{n-1}[/tex]Now, to determine if the pattern converge or diverge:
[tex]\begin{gathered} \lvert r\rvert<1,\text{ the series converge to }\frac{a_1}{1-r} \\ \lvert r\rvert\ge1,\text{ the series diverges} \end{gathered}[/tex]Since the common ratio is less than 1, the series converges to:
[tex]\text{converges to }\frac{-36}{1-\frac{1}{3}}=-54[/tex]A sum of an infinite geometric series can be determined if it converges since this pattern converges, the sum would converge to;
[tex]\begin{gathered} S=\frac{a_1}{1-r} \\ S=\frac{-36}{1-\frac{1}{3}}=-54 \end{gathered}[/tex]
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