Geometric Sequences and Series: Practice

A sequence is geometric only when every quotient of consecutive terms is the same.

$\frac{10}{5} = 2, \qquad \frac{20}{10} = 2, \qquad \frac{40}{20} = 2$

So $5, 10, 20, 40$ has the constant ratio $r = 2$. The sequence $2, 4, 6, 8$ has a constant difference instead, and in $1, 4, 9, 16$ the quotients $4/1 = 4$ and $9/4 = 2.25$ disagree, as do those of $3, 6, 10, 15$.

Reaching the $n$th term from the first means multiplying by $r$ once for each of the $n - 1$ gaps between them.

$a_n = a_1 r^{\,n-1} = 3 \cdot 5^{\,n-1}$

Check it at $n = 1$: the formula gives $3 \cdot 5^0 = 3$, the first term, as it must. An exponent of $n$ would wrongly report the first term as $15$.

Divide consecutive terms.

$\frac{48}{96} = \frac{1}{2}, \qquad \frac{24}{48} = \frac{1}{2}, \qquad \frac{12}{24} = \frac{1}{2}$

The terms are halving, so $r = \tfrac{1}{2}$. A ratio smaller than $1$ in size makes the terms shrink, and the ratio is positive because no term changes sign.

Here $a = 2$, $r = 2$, and $n = 5$.

$S_5 = \frac{2(2^5 - 1)}{2 - 1} = 2(32 - 1) = 62$

Checking by hand: $2 + 4 + 8 + 16 + 32 = 62$. Stopping after four terms would give $30$.

The third term sits two multiplications past the first, so the exponent is $3 - 1 = 2$.

$a_3 = a_1 r^2 = 7 \cdot 3^2 = 7 \cdot 9 = 63$

The sequence starts $7, 21, 63$.

The fifth term needs $5 - 1 = 4$ multiplications by $\tfrac{1}{2}$.

$a_5 = 100 \left(\tfrac{1}{2}\right)^4 = \frac{100}{16} = 6.25$

The terms run $100, 50, 25, 12.5, 6.25$, so the fifth is $6.25$ and the fourth is $12.5$.

Test the ratios, not the differences.

$\frac{6}{3} = 2, \qquad \frac{9}{6} = 1.5$

The two quotients disagree, so the sequence is not geometric. Its terms do all differ by the same amount ($3$ each time), which makes it arithmetic, so the reason offered in the third option is false.

Divide the second term by the first, remembering that dividing by a fraction means multiplying by its reciprocal.

$r = \frac{1/6}{1/2} = \frac{1}{6} \cdot \frac{2}{1} = \frac{1}{3}$

The next pair agrees, since $\dfrac{1/18}{1/6} = \tfrac{1}{3}$ as well, so $r = \tfrac{1}{3}$.

Count the multiplications rather than guessing. Starting at $3$, one doubling gives $6$, two give $12$, and three give $24$.

$a_4 = 3 \cdot 2^{4-1} = 3 \cdot 8 = 24$

Three multiplications land on the fourth term, because the exponent is always one less than the term number.

Here $a = 5$, $r = 3$, and $n = 3$.

$S_3 = \frac{5(3^3 - 1)}{3 - 1} = \frac{5 \cdot 26}{2} = 65$

Adding directly: $5 + 15 + 45 = 65$.

The fifth term is four multiplications past the first, so the exponent is $4$, not $5$.

$a_5 = 2 \cdot 3^{5-1} = 2 \cdot 81 = 162$

The sequence runs $2, 6, 18, 54, 162$. Using $3^5$ by mistake gives $486$, which is the sixth term.

The common ratio is any term divided by the one before it.

$r = \frac{a_2}{a_1} = \frac{20}{4} = 5$

Each term is five times the last, so the sequence continues $4, 20, 100, 500, \ldots$. The value $16$ is the difference, which is what an arithmetic sequence would use.