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Geometric sequences

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Find the 5th term of the geometric sequence with a1 = 2 and common ratio r = 3.

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From the 1st term to the 5th term, how many times is the ratio 3 multiplied?

Common mix-up

In this example, a student wrote 486.

The student used exponent n instead of n - 1 and wrote a5 = 2 * 3^5 = 486.

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See why this works

What is the 5th term of 2, 6, 18, ...?

In a geometric sequence, each term multiplies the last by the same ratio. That ratio is the engine of the whole sequence.

a_n = a1 * r^(n - 1): the ratio is multiplied (n - 1) times.

A geometric sequence grows by a multiplier, not an addend. In 2, 6, 18, ..., each term is the previous one times 3, so the common ratio r is 3 and the first term a1 is 2. The ratio is what turns a sequence into a giant multiplication problem.

Dry-erase diagram: Geometric sequences — a_n = a1 * r^(n - 1): the ratio is multiplied (n - 1) times.
See the ideaGeometric sequences: a_n = a1 * r^(n - 1): the ratio is multiplied (n - 1) times.

The formula a_n = a1 * r^(n - 1) counts the multiplications. The first term is already given, so the ratio is applied once per jump, not once per term. The 5th term has had the ratio applied four times: 2 * 3 * 3 * 3 * 3. The exponent is therefore 4, exactly one less than 5.

So a5 = 2 * 3^4 = 2 * 81 = 162. Using exponent 5 by mistake gives 486, which is the 6th term. Listing a short version of the sequence catches any doubt: 2, 6, 18, 54, 162, and the count of multiplications is visible in the exponents you visited.

The same exponent logic applies to decimals, negative ratios, and ratios less than one. The term number always counts one more than the exponent. Find the ratio by dividing any term by the one before it, then apply the formula. The growth may go up or down, but the pattern of exponent plus one never changes.

Keep this idea

In a_n = a1 * r^(n - 1), the exponent is one less than the term number. Confirm with a short list when unsure.

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