An initial investment of $3 is worth $108 after 5 years. If the annual growth reflects a geometric sequence, approximately how much will the

Question

An initial investment of $3 is worth $108 after 5 years. If the annual growth reflects a geometric sequence, approximately how much will the investment be worth after 11 years?

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Audrey 1 week 2021-09-13T15:15:47+00:00 2 Answers 0

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    0
    2021-09-13T15:17:44+00:00

    Answer:

    The investment be worth $23328 after 11 years.

    Step-by-step explanation:

    It is given that the annual growth reflects a geometric sequence.

    An initial investment of $3 is worth $108 after 5 years.

    It means the initial value of first term of the gp, a₁ = 3

    The 5th term of the gp, a₅ = 108

    The nth term of a gp is

    a_n=ar^{n-1}              …. (1)

    where, a is first term and r is common ratio.

    The 5th term of the gp is

    a_5=ar^{5-1}

    From the given information it is clear that the 5th term of the gp is 108. Substitute a₅ = 108 and a=3.

    108=(3)r^{4}

    Divide both sides by 3.

    \frac{108}{3}=r^{4}

    36=r^{4}

    Taking fourth root on both the sides.

    \sqrt{6}=r

    Substitute r=√6, a=3 and n=11 to find the investment worth after 11 years.

    a_{11}=(3)(\sqrt{6})^{11-1}

    a_{11}=3(\sqrt{6})^{10}

    a_{11}=23328

    Therefore the investment worth $23328 after 11 years.

  1. Charlotte
    0
    2021-09-13T15:17:45+00:00

    \bf \begin{array}{ll} \stackrel{year}{term}&value\\ \cline{1-2} a_1&3\\ a_2&3r\\ a_3&3rr\\ a_4&3rrr\\ a_5&3rrrr\\ &3r^4 \end{array}\qquad \qquad \stackrel{\textit{5th year}}{108}=3r^4\implies \cfrac{108}{3}=r^4\implies 36=r^4 \\\\\\ \sqrt[4]{36}=r\implies \sqrt[4]{6^2}=r\implies 6^{\frac{2}{4}}=r\implies 6^{\frac{1}{2}}=r\implies \sqrt{6}=r

    \bf n^{th}\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} r=\sqrt{6}\\ a_1=3\\ n=11 \end{cases}\implies a_{11}=3(\sqrt{6})^{11-1} \\\\\\ a_{11}=3(\sqrt{6})^{10}\implies a_{11}=3\left(6^{\frac{1}{2}} \right)^{10}\implies a_{11}=3\cdot 6^{\frac{10}{2}} \\\\\\ a_{11}=3\cdot 6^5\implies a_{11}=3\cdot 7776\implies a_{11}=23328

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