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.

• A. $a_{30}=3{\left(\frac{1}{2}\right)}^{19}$
• B. $a_{30}=3{\left(\frac{1}{2}\right)}^{59}$
• C. $a_{30}=3{\left(\frac{1}{2}\right)}^{29}$
• D. $a_{30}=3{\left(\frac{1}{2}\right)}^{39}$

Answer 1

The correct option is

D

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

We first use the formula for the n th term to write $a_{10}$ and $a_{15}$ as follows
$a_{10}=a_1\times r^{10-1}=\frac{3}{512}$
$a_{15}=a_1\times r^{15-1}=\frac{3}{16384}$
We now divide the terms $a_{10}$ and $a_{15}$ to write
$\frac{a_{15}}{a_{10}}=\frac{a_1\ast r^{14}}{a_1\times r^9}=\frac{\frac{3}{16384}}{\frac{3}{512}}$
Solving for $r$
$r^5=\frac{1}{32}$ which gives $r=\frac{1}{2}$
$a_{10}=\frac{3}{512}=a_1{\left(\frac{1}{2}\right)}^9$

$\Rightarrow a_1=\frac{3}{512}\times (2{)}^9$

$\Rightarrow$$a_1=3$

So, $a_{30}=a_1\times r^{(30-1)}$

$\Rightarrow$$a_{30}=3{\left(\frac{1}{2}\right)}^{29}$

Hence, correct answer is option $\left(D\right)$