Geometric Sequences: Practice

The middle term satisfies $a_2^{2} = a_1 a_3$, so it is the geometric mean of the two.

$a_2 = \sqrt{4 \cdot 36} = \sqrt{144} = 12$

Both $12$ and $-12$ square to $144$, but the terms are stated positive, so $a_2 = 12$.

Here $a_n = 3 \cdot 3^{n-1} = 3^{n}$. Set it equal to $729$ and match powers of $3$.

$3^{n} = 729 = 3^{6} \Rightarrow n = 6$

Isolate the power, then take a logarithm and use the power law.

$5 \cdot 2^{n-1} = 100 \Rightarrow 2^{n-1} = 20 \Rightarrow n = 1 + \frac{\log(20)}{\log(2)}$

The change-of-base quotient gives the exponent, and adding $1$ recovers the position.

The terms are $a_n = 3 \cdot 2^{n-1}$. Test successive terms past $1000$.

$a_9 = 3 \cdot 256 = 768, \quad a_{10} = 3 \cdot 512 = 1536$

The 9th term is still below $1000$, so the first term over $1000$ is $1536$.

Use $a_n = a_1 r^{n-1}$ with $n = 5$, so the exponent is $4$.

$a_5 = 2 \cdot (-3)^{4} = 2 \cdot 81 = 162$

An even power of a negative number is positive, so the term is positive.

From the two terms an even distance apart, $r^{2} = \tfrac{45}{5} = 9$, so $r = 3$ or $r = -3$.

$a_2 = 5r < 0 \Rightarrow r = -3$

The condition $a_2 < 0$ pins the sign, ruling out $r = 3$.

First find $r$ over the $6 - 3 = 3$ steps between the known terms.

$r^{3} = \frac{64}{8} = 8 \Rightarrow r = 2$

Then step from $a_6$ up two more places, $a_8 = a_6 r^{2} = 64 \cdot 4 = 256$.

The terms are $4 - 2 = 2$ steps apart, an even gap, so $r^{2} = \tfrac{54}{6} = 9$ gives $r = 3$ or $r = -3$.

$r^{2} = 9 \Rightarrow r = 3 \ (\text{terms positive})$

The positive-terms condition rules out $r = -3$.

The inserted term is the geometric mean, satisfying (middle)$^2 = 3 \cdot 48$.

$\text{middle} = \sqrt{3 \cdot 48} = \sqrt{144} = 12$

The positive root is required, so the number is $12$ (giving $3, 12, 48$ with $r = 4$).

The common ratio is $r = -2$, so $a_6 = a_1 r^{5}$ with an odd exponent.

$a_6 = 1 \cdot (-2)^{5} = -32$

An odd power of a negative ratio is negative, so the 6th term is $-32$.

The 4th term is the geometric mean of the 3rd and 5th, since $a_4^{2} = a_3 a_5$.

$a_4 = \sqrt{20 \cdot 80} = \sqrt{1600} = 40$

The terms are positive, so the positive root $40$ is the answer, not the arithmetic mean $50$.

Set $300 \cdot 2^{d} = 30{,}000$, isolate the power, then take a logarithm.

$2^{d} = 100 \Rightarrow d = \frac{\log(100)}{\log(2)}$

The change-of-base quotient gives the exponent directly.