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?

Answer 1

`\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^{(2)/(4)}=r\implies 6^{(1)/(2)}=r\implies √(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=√(6)\\ a_1=3\\ n=11 \end{cases}\implies a_(11)=3(√(6))^(11-1) \\\\\\ a_(11)=3(√(6))^(10)\implies a_(11)=3\left(6^{(1)/(2)} \right)^(10)\implies a_(11)=3\cdot 6^{(10)/(2)} \\\\\\ a_(11)=3\cdot 6^5\implies a_(11)=3\cdot 7776\implies a_(11)=23328`

Answer 2

The investment be worth $23328 after 11 years.

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.

`(108)/(3)=r^(4)`

`36=r^(4)`

Taking fourth root on both the sides.

`√(6)=r`

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

`a_(11)=(3)(√(6))^(11-1)`

`a_(11)=3(√(6))^(10)`

`a_(11)=23328`

Therefore the investment worth $23328 after 11 years.