Question

Given the terms $a_{10}=\frac{3}{512}$ and $a_{15}=\frac{3}{16384}$ of a geometric sequence, find the exact value of the term $a_{30}$ of the sequence.

Answer 1

We first use the formula for the n th term to write $a_{10}$ and $a_{15}$ as follows

$a_{10}$ $=a_1\ast r^{10-1}=\frac{3}{512}$

$a_{15}=a_1\ast r^{15-1}=\frac{3}{16384}$

We now divide the terms $a_{10}$ and $a_{15}$ to write

$\frac{a_{15}}{a_{10}}=\left(\frac{a_1\ast r^{14}}{a_1\ast r^9}\right)=\frac{\left(\frac{3}{16384}\right)}{\left(\frac{3}{512}\right)}$

Solve for $r$ to obtain.

$r^5=\frac{1}{32}$ which gives $r=\frac{1}{2}$

We now use $a_{10}$ to find $a_1$ as follows.

$a_{10}=\frac{3}{512}=a_1(\frac{1}{2}{)}^9$

Solve for $a_1$ to obtain.

$a_1=3$

We now use the formula for the n th term to find {a}_{30} as follows.

$a_{30}=3(\frac{1}{2}{)}^{29}=\frac{3}{536870912}$