High School Math: 5 Geometric Sequence Examples to Master Your Exam

Problem: A geometric sequence starts with a first term of 3. Its fourth term is 81. If the common ratio is positive, what is the second term, $u_2$?

Step 1: Write down your clues.

We have $u_1 = 3$ and $u_4 = 81$. Our goal is to find $u_2$. To do that, we first need to uncover the common ratio, $r$.

Step 2: Use the master formula to find $r$.

We’ll plug the information we have about the fourth term into the formula $u_n = u_1 \cdot r^{(n-1)}$:

$u_4 = u_1 \cdot r^{(4-1)}$

$81 = 3 \cdot r^3$

Step 3: Solve for $r$.

Now it’s just algebra. Let’s isolate $r$.

$81 / 3 = r^3$

$27 = r^3$

Taking the cube root of both sides gives us:

$r = \sqrt[3]{27} = 3$

So, our common ratio is 3.

Step 4: Find the target term, $u_2$.

We can use the formula again, or just use logic! The second term is just the first term multiplied by the ratio.

$u_2 = u_1 \cdot r$

$u_2 = 3 \cdot 3 = 9$

And there we have it. The second term is 9.

Problem: In a geometric sequence, the second term is 10 and the fifth term is 80. Find the first term ($u_1$), the common ratio ($r$), and the seventh term ($u_7$).

Step 1: Set up two equations.

We’ll use our master formula for both pieces of information:

For $u_2=10$: $10 = u_1 \cdot r^{(2-1)} \implies 10 = u_1 \cdot r$ (Equation 1)

For $u_5=80$: $80 = u_1 \cdot r^{(5-1)} \implies 80 = u_1 \cdot r^4$ (Equation 2)

Step 2: Use the division trick.

This is the magic move. When you have two equations like this, divide the one with the higher power of $r$ by the one with the lower power. This will make the $u_1$ terms vanish!

$\frac{u_1 \cdot r^4}{u_1 \cdot r} = \frac{80}{10}$

The $u_1$ on the top and bottom cancel out. Using exponent rules for the $r$ terms ($r^4 / r^1 = r^{4-1}$), we get:

$r^3 = 8$

Taking the cube root, we find $r=2$. Easy peasy!

Step 3: Substitute back to find $u_1$.

Now that we have $r=2$, we can plug it into either of our original equations. Equation 1 looks simplest:

$10 = u_1 \cdot (2)$

$u_1 = 10 / 2 = 5$

Step 4: Calculate the final target, $u_7$.

We have $u_1=5$ and $r=2$. Let’s find $u_7$:

$u_7 = u_1 \cdot r^{(7-1)}$

$u_7 = 5 \cdot 2^6$

$u_7 = 5 \cdot 64 = 320$

So, our final answers are $u_1=5$, $r=2$, and $u_7=320$.

Problem: In a geometric series, the second term ($u_2$) is 15. The sum of the first two terms ($S_2$) is 20. Find the first term ($u_1$) and the common ratio ($r$).

Step 1: Translate the sum information.

What does $S_2 = 20$ actually mean? It’s just a fancy way of saying:

$u_1 + u_2 = 20$

Step 2: Use your other clue to find $u_1$.

We were given that $u_2 = 15$. Let’s substitute that directly into our sum equation:

$u_1 + 15 = 20$

Solving for $u_1$ is now trivial:

$u_1 = 20 – 15 = 5$

Step 3: Find the common ratio $r$.

Remember the basic definition of how a geometric sequence works: to get to the next term, you multiply by $r$.

$u_2 = u_1 \cdot r$

We know $u_1$ and $u_2$, so we can plug them in:

$15 = 5 \cdot r$

$r = 15 / 5 = 3$

So, we’ve found our unknowns: $u_1=5$ and $r=3$.

Problem: The numbers 5, $y$, and 45 are three consecutive terms in a geometric sequence. Find all possible values of $y$.

Step 1: Set up the ratio equation.

The definition of a common ratio means that (second term / first term) must equal (third term / second term).

$\frac{y}{5} = \frac{45}{y}$

Step 2: Solve the equation.

To solve for $y$, we can cross-multiply:

$y \cdot y = 45 \cdot 5$

$y^2 = 225$

Step 3: Find ALL possible values.

Here’s where many students trip up! When you take the square root to solve for $y$, you must remember both the positive and negative roots.

$y = \pm\sqrt{225}$

$y = \pm15$

Both $y=15$ and $y=-15$ are valid answers. Let’s see why:

• If $y=15$, the sequence is 5, 15, 45… The common ratio is $r=3$.
• If $y=-15$, the sequence is 5, -15, 45… The common ratio is $r=-3$.

Problem: Express the repeating decimal $0.454545…$ as a fraction in its simplest form.

Step 1: Rewrite the decimal as a series.

Let’s break the decimal down into its component parts:

$0.454545… = 0.45 + 0.0045 + 0.000045 + …$

Look at that! It’s an infinite geometric series.

Step 2: Identify $u_1$ and $r$.

The first term is obvious: $u_1 = 0.45$.

To find the common ratio, divide the second term by the first:

$r = \frac{0.0045}{0.45} = 0.01$

Step 3: Check the condition and use the formula.

Is $|r| < 1$? Yes, $|0.01| < 1$, so the series converges and we can use our formula.

$S_\infty = \frac{u_1}{1-r} = \frac{0.45}{1 – 0.01}$

$S_\infty = \frac{0.45}{0.99}$

Step 4: Convert to a clean fraction and simplify.

To get rid of the decimals, multiply the numerator and denominator by 100:

$S_\infty = \frac{45}{99}$

Now, we simplify. Both numbers are divisible by 9.

$S_\infty = \frac{45 \div 9}{99 \div 9} = \frac{5}{11}$

And there you have it. The endlessly repeating $0.4545…$ is just a neat $5/11$.