Geometric Sequences and Series: Practice

The ratio is $6/2 = 3$, so the last term satisfies $2 \cdot 3^{\,n-1} = 1458$.

$3^{\,n-1} = \frac{1458}{2} = 729 = 3^6 \quad\Longrightarrow\quad n - 1 = 6$

So $n = 7$. Listing them out: $2, 6, 18, 54, 162, 486, 1458$, which is seven terms. The exponent $6$ is one less than the count, as always.

Each term is $a_n = 800\left(\tfrac{1}{2}\right)^{n-1} = \dfrac{800}{2^{\,n-1}}$, so the term drops below $1$ exactly when $2^{\,n-1} > 800$.

$2^9 = 512 < 800, \qquad 2^{10} = 1024 > 800$

So the first exponent that works is $n - 1 = 10$, giving $n = 11$. Checking: $a_{10} = 800/512 = 1.5625$ is still above $1$, while $a_{11} = 800/1024 = 0.78125$ is below it.

A constant ratio means $\dfrac{x}{4} = \dfrac{9}{x}$, and cross-multiplying gives $x^2 = 36$.

$x = \sqrt{36} = 6$

The problem asks for the positive value, so $x = 6$, and the sequence $4, 6, 9$ has ratio $\tfrac{3}{2}$ throughout. The arithmetic middle $6.5$ would give the unequal ratios $1.625$ and about $1.38$.

Growth of $20$ percent means the new value is the old value plus a fifth of it, so each year the value is multiplied by

$r = 1 + \frac{20}{100} = 1.2 .$

Multiplying by $0.2$ instead would shrink the quantity to a fifth of its size, and $0.8$ is the ratio for a $20$ percent decrease.

A ratio of $1$ makes every term equal to $a_1 = -4$. The standard formula would divide by $1 - r = 0$, so it cannot be used at all; add the identical terms instead.

$S_{25} = 25 \cdot (-4) = -100$

The sum exists and is perfectly ordinary. It is only the formula that breaks down when $r = 1$.

The terms are divided by $3$ each time, so $a = 81$, $r = \tfrac{1}{3}$, and $n = 5$.

$S_5 = \frac{81\left(1 - \left(\tfrac{1}{3}\right)^5\right)}{1 - \tfrac{1}{3}} = \frac{81 \cdot \tfrac{242}{243}}{\tfrac{2}{3}} = \frac{242}{3} \cdot \frac{3}{2} = 121$

Adding by hand confirms it: $81 + 27 + 9 + 3 + 1 = 121$. Using $r = 3$ by mistake would give the wildly large $9801$.

Annual compounding multiplies the balance by $1 + 0.05 = 1.05$ each year, so the yearly balances form a geometric sequence with $r = 1.05$, starting at $2000$ dollars.

$2000 \cdot (1.05)^3 = 2000 \cdot 1.157625 = 2315.25$

The balance is $2315.25$ dollars. Simple interest would have added $100$ dollars a year, giving only $2300$ dollars, and $2205$ dollars is the balance after just two years.

The summand $3 \cdot 2^{\,k-1}$ is geometric with $a = 3$ and $r = 2$, and $k$ runs from $1$ to $4$, so there are four terms.

$\sum_{k=1}^{4} 3 \cdot 2^{\,k-1} = \frac{3(2^4 - 1)}{2 - 1} = 3 \cdot 15 = 45$

Writing the terms out: $3 + 6 + 12 + 24 = 45$. Running $k$ up to $5$ instead would give $93$.

The fourth term is three multiplications past the first.

$a_4 = a_1 r^3 \quad\Longrightarrow\quad 54 = a_1 \cdot 27 \quad\Longrightarrow\quad a_1 = 2$

The sequence is $2, 6, 18, 54$. Dividing by $r$ only once, instead of three times, would leave you at $18$, which is $a_3$.

First get the ratio from the two terms you have.

$r = \frac{a_3}{a_2} = \frac{4}{6} = \frac{2}{3}$

Going back one term means dividing by $r$ rather than multiplying, so $a_1 = 6 \div \tfrac{2}{3} = 6 \cdot \tfrac{3}{2} = 9$. The sequence is $9, 6, 4$, and each quotient is $\tfrac{2}{3}$.

With $a = 3$ and $r = 2$, set the general term equal to $3072$.

$3 \cdot 2^{\,n-1} = 3072 \quad\Longrightarrow\quad 2^{\,n-1} = 1024 = 2^{10} \quad\Longrightarrow\quad n - 1 = 10$

So $3072 = a_{11}$. The exponent is $10$, but the term number is one more, which is the off-by-one to watch.

Twenty hours is five half-lives, since $20 \div 4 = 5$. Each half-life multiplies the amount by $\tfrac{1}{2}$, so reading the concentration once every four hours gives a geometric sequence with $r = \tfrac{1}{2}$.

$320 \cdot \left(\tfrac{1}{2}\right)^5 = \frac{320}{32} = 10$

So $10$ mg remain. Counting only four halvings would leave $20$ mg, so the number of halvings has to be counted carefully.