Geometric Sequence worksheet with four problems to find specific terms in sequences, including formulas and answer spaces.

Let’s solve each problem step by step using the geometric sequence formula:

Formula reminder:
> \( a_n = a_1 \cdot r^{n-1} \)

Where:
- \( a_1 \) = first term
- \( r \) = common ratio (what you multiply by to get the next term)
- \( n \) = position of the term you want
- \( a_n \) = the value of the nth term

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Problem 1: Find \( a_{11} \) of 3, 6, 12, 24, ...

Step 1: Identify \( a_1 \) → First term is 3

Step 2: Find \( r \) → Divide any term by the one before it:
\( 6 ÷ 3 = 2 \), so \( r = 2 \)

Step 3: We want the 11th term → \( n = 11 \)

Step 4: Plug into formula:
\( a_{11} = 3 \cdot 2^{11-1} = 3 \cdot 2^{10} \)

Calculate \( 2^{10} \):
\( 2^1 = 2 \)
\( 2^2 = 4 \)
\( 2^3 = 8 \)
\( 2^4 = 16 \)
\( 2^5 = 32 \)
\( 2^6 = 64 \)
\( 2^7 = 128 \)
\( 2^8 = 256 \)
\( 2^9 = 512 \)
\( 2^{10} = 1024 \)

So, \( a_{11} = 3 × 1024 = 3072 \)

✔ Final Answer for Problem 1: 3072

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Problem 2: Find 9th term of 4, –12, 36, –108, ...

Step 1: \( a_1 = 4 \)

Step 2: Find \( r \) → \( -12 ÷ 4 = -3 \), so \( r = -3 \)

Step 3: \( n = 9 \)

Step 4: Use formula:
\( a_9 = 4 \cdot (-3)^{9-1} = 4 \cdot (-3)^8 \)

Note: Even power → negative becomes positive
\( (-3)^8 = 3^8 \)

Calculate \( 3^8 \):
\( 3^1 = 3 \)
\( 3^2 = 9 \)
\( 3^3 = 27 \)
\( 3^4 = 81 \)
\( 3^5 = 243 \)
\( 3^6 = 729 \)
\( 3^7 = 2187 \)
\( 3^8 = 6561 \)

So, \( a_9 = 4 × 6561 = 26244 \)

✔ Final Answer for Problem 2: 26244

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Problem 3: In sequence \( \frac{1}{5}, \frac{1}{10}, \frac{1}{20}, \frac{1}{40}, ... \), find \( a_{10} \)

Step 1: \( a_1 = \frac{1}{5} \)

Step 2: Find \( r \) → \( \frac{1/10}{1/5} = \frac{1}{10} × \frac{5}{1} = \frac{5}{10} = \frac{1}{2} \)
So, \( r = \frac{1}{2} \)

Step 3: \( n = 10 \)

Step 4: Formula:
\( a_{10} = \frac{1}{5} \cdot \left( \frac{1}{2} \right)^{10-1} = \frac{1}{5} \cdot \left( \frac{1}{2} \right)^9 \)

\( \left( \frac{1}{2} \right)^9 = \frac{1}{512} \)

So, \( a_{10} = \frac{1}{5} × \frac{1}{512} = \frac{1}{2560} \)

✔ Final Answer for Problem 3: \(\frac{1}{2560}\)

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Problem 4: What term is 2048 in 2, 4, 8, 16, ...?

We are given \( a_n = 2048 \), and we need to find \( n \).

Step 1: \( a_1 = 2 \)

Step 2: \( r = 4 ÷ 2 = 2 \)

Step 3: Use formula:
\( 2048 = 2 \cdot 2^{n-1} \)

Simplify right side:
\( 2 \cdot 2^{n-1} = 2^{1 + n - 1} = 2^n \)

So:
\( 2^n = 2048 \)

Now, what power of 2 equals 2048?

Try powers of 2:
\( 2^1 = 2 \)
\( 2^2 = 4 \)
\( 2^3 = 8 \)
\( 2^4 = 16 \)
\( 2^5 = 32 \)
\( 2^6 = 64 \)
\( 2^7 = 128 \)
\( 2^8 = 256 \)
\( 2^9 = 512 \)
\( 2^{10} = 1024 \)
\( 2^{11} = 2048 ← YES!

So, \( n = 11 \)

✔ Final Answer for Problem 4: 11

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Final Answer:
1) 3072
2) 26244
3) \(\frac{1}{2560}\)
4) 11