Given geometric progression: $b_3 = 16$ and $b_6 = 1$ how to find common ratio ($q$) and start value?

Question

Given geometric progression sequence $b_3 = 16$ and $b_6 = 1$

How can I find common sequence $q$ and $b_1$

my try: $q = \sqrt[6 - 3]{\frac{b_6}{b_3}} = \sqrt[3]{\frac{1}{16}}$

Answer 1

$b_3=16=b_1q^2$.

$b_6=1=b_1q^5$.

Divide.

Therefore, $q^3={1\over16}$, which gives $q={1\over{\sqrt[3]{16}}}$. You may now calculate $b_1$.

Answer 2

Hint:

$$b_3=ar^2=16\tag{a is first term and r : common ratio}$$ $$b_6=ar^5=1$$

Divide both to get $$r^3=\frac{1}{16}$$

I hope you can carry on from this?

Answer 3

A geometric progression is a sequence of the form $$a,ar,ar^2,\dots$$

So $b_3=ar^2$ and $b_6=ar^5$ (if we label the first term to be $b_1$).

Hence $16=ar^2$ and $ar^5=1=ar^2r^3=16r^3$, hence $r^3=\frac{1}{16}$. Now you can solve for $a$ as well.

Answer 4

$N^{\text{th}}_{}$ term ($b_{n}^{}$) of a Geometric progression with common ratio $q$ and first term $b_{1}^{}$ is : $$b_{n}^{}=b_{1}^{}q_{}^{n-1}.$$ Using this : $$b_{3}^{}=b_{1}^{}q_{}^{2}=16 \text{ and } b_{6}^{}=b_{1}^{}q_{}^{5}=1$$ Hence $q^{3}_{}=\frac{1}{16} \Rightarrow q=(\frac{1}{16})^{\frac{1}{3}}$ and $b_{1}^{}=16_{}^{\frac{5}{3}}.$