Geometric sequences calculator

problem

Find first element ( $ a_1 $ ) and a common ration ( $ r $ ) of a geometic progression if $ a_{ 4 } = \frac{ 4 }{ 3 } ~~ \text{and} ~~ a_{ 8 } = \frac{ 64 }{ 243 } $.

solution

$$ r = \frac{ 2 }{ 3 } ~~ \text{ and } ~~ a_1 = \frac{ 9 }{ 2 } $$

explanation

To find $ r $ we use formula for the $ n^{th} $ term $\color{blue}{a_n = a_1 \cdot r^{n-1}}$

$$ \begin{aligned} \text{ for n = 4 } \rightarrow a_{ 4} &= a_1 \cdot r^{ 4 - 1} \\ \text{ for n = 8 } \rightarrow a_{ 8} &= a_1 \cdot r^{ 8 - 1} \end{aligned}$$

To find $ r $ we will divide above equations:

$$ \begin{aligned} \frac{ a_{ 4}}{a_{ 8}} &= \frac { a_1 \cdot r^{ 4 - 1}} { a_1 \cdot r^{ 8 - 1}} \\ \frac{ \frac{ 4 }{ 3 }}{ \frac{ 64 }{ 243 } } &= \frac{ r^{ 3}}{ r^{ 7} } \\ \frac{ 81 }{ 16 } &= \frac{1}{r^{ 4 }} \\ r^{ 4 } &= \frac{ 16 }{ 81 } \\ r &= \sqrt[ 4 ]{ \frac{ 16 }{ 81 } } \\ r &= \frac{ 2 }{ 3 } \end{aligned}$$

Now we can find $ a_1 $ using first equation

$$ \begin{aligned} a_{ 4} &= a_1 \cdot r^{ 4 - 1} \\ a_1 &= \frac{ a_{ 4 }}{r^{ 3 }} \\ a_1 &= \frac{ \frac{ 4 }{ 3 } }{ \frac{ 8 }{ 27 } } \\ a_1 &= \frac{ 9 }{ 2 } \end{aligned} $$