Geometric Sequences and Series: Practice

With $a = 4$ and $r = 3$ the sum of the first $n$ terms simplifies, because the denominator $r - 1 = 2$ halves the first term.

$S_n = \frac{4(3^{\,n} - 1)}{3 - 1} = 2(3^{\,n} - 1)$

Requiring $2(3^{\,n} - 1) > 20{,}000$ reduces to $3^{\,n} > 10{,}001$. Taking logarithms preserves the inequality, because the logarithm is an increasing function.

$n > \frac{\log 10{,}001}{\log 3} \approx \frac{4.0000}{0.4771} \approx 8.38$

The number of terms is a whole number, so $n = 9$. Checking the boundary: $S_8 = 2(6561 - 1) = 13{,}120$ falls short, while $S_9 = 2(19{,}683 - 1) = 39{,}364$ clears the target.

Terms three places apart differ by three multiplications, so $a_6 = a_3 r^3$.

$r^3 = \frac{160}{20} = 8 \quad\Longrightarrow\quad r = 2$

Stepping back from $a_3$ to $a_1$ is two steps, so divide by $r$ twice.

$a_1 = \frac{a_3}{r^2} = \frac{20}{4} = 5$

The sequence runs $5, 10, 20, 40, 80, 160$.

The fifth term is four multiplications past the first, so $12 = 12 r^4$, giving $r^4 = 1$ and $r = \pm 1$. A ratio of $-1$ would make the terms alternate in sign, and every term here is positive, so $r = 1$.

$S_{10} = 10 \cdot 12 = 120$

Every term equals $12$, so the sum is ten copies of it. The usual formula cannot be used, since it would divide by $1 - r = 0$, but the sum itself is perfectly well defined.

With $a = 1$ and $r = 2$, the sum formula gives a power of two, minus one.

$S_{10} = \frac{1(2^{10} - 1)}{2 - 1} = 1024 - 1 = 1023$

The tenth term is $a_{10} = 2^9 = 512$, not $2^{10}$, which is the off-by-one to keep straight. The value $2047$ would be the sum of eleven terms.

Write the sum of four terms with an unknown first term $a$.

$S_4 = \frac{a(2^4 - 1)}{2 - 1} = 15a$

Setting $15a = 30$ gives $a = 2$. The sequence is $2, 4, 8, 16$, and those four terms do add to $30$.

The exponent here is $k$, not $k - 1$, so the first term is $2^1 = 2$, not $1$. That makes $a = 2$, $r = 2$, and $n = 6$.

$\sum_{k=1}^{6} 2^{\,k} = \frac{2(2^6 - 1)}{2 - 1} = 2(64 - 1) = 126$

Writing the terms out: $2 + 4 + 8 + 16 + 32 + 64 = 126$. Starting the sum at $1$ instead would give $127$, which is the trap here.

The count doubles hourly, and the starting count is the value at hour zero, so after $h$ hours the population is $250 \cdot 2^{\,h}$.

$250 \cdot 2^{\,h} > 1{,}000{,}000 \quad\Longrightarrow\quad 2^{\,h} > 4000$

Taking logarithms of both sides brings the exponent down.

$h > \frac{\log 4000}{\log 2} \approx \frac{3.6021}{0.3010} \approx 11.97$

So the first whole hour that works is $h = 12$. Checking: $250 \cdot 2^{11} = 512{,}000$ is under a million, while $250 \cdot 2^{12} = 1{,}024{,}000$ is over.

The terms halve, so $a = 64$, $r = \tfrac{1}{2}$, and $n = 7$.

$S_7 = \frac{64\left(1 - \left(\tfrac{1}{2}\right)^7\right)}{1 - \tfrac{1}{2}} = 128\left(1 - \tfrac{1}{128}\right) = 128 - 1 = 127$

Adding by hand: $64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$.

The first term is $1$ and the ratio is $\tfrac{1}{2}$, so the twentieth term needs $20 - 1 = 19$ halvings, not $20$.

$a_{20} = 1 \cdot \left(\tfrac{1}{2}\right)^{19} = \frac{1}{2^{19}}$

The exponent is always one less than the term number, because the first term has not been halved at all.

With $a = 2$ and $r = 3$ the sum formula simplifies, because the denominator $r - 1 = 2$ cancels the first term.

$S_n = \frac{2(3^{\,n} - 1)}{3 - 1} = 3^{\,n} - 1$

Setting this equal to $242$ gives $3^{\,n} = 243 = 3^5$, so $n = 5$. The five terms $2 + 6 + 18 + 54 + 162$ do add to $242$.

Divide the second term by the first, using the rule that dividing powers of the same base subtracts the exponents.

$r = \frac{2^{\,x+2}}{2^{\,x}} = 2^{\,(x+2) - x} = 2^2 = 4$

The next pair gives the same thing, $2^{\,x+4} / 2^{\,x+2} = 2^2 = 4$, so the ratio is $4$ no matter what $x$ is.

Write the first three terms as $3$, $3r$, and $3r^2$, and set their total to $21$.

$3 + 3r + 3r^2 = 21 \quad\Longrightarrow\quad r^2 + r - 6 = 0 \quad\Longrightarrow\quad (r + 3)(r - 2) = 0$

The roots are $r = 2$ and $r = -3$. The problem asks for the positive ratio, so $r = 2$, and the terms $3 + 6 + 12$ do add to $21$.