# What is the sum of a 6–term geometric series if the first term is 20 and the last term is 336,140?

Given:

Then general form of a term of a geometric series is:

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To find the sum of a geometric series, you can use the formula:

[S_n = \frac{{a(1 - r^n)}}{{1 - r}}]

Where:

- (S_n) is the sum of the first (n) terms,
- (a) is the first term,
- (r) is the common ratio, and
- (n) is the number of terms.

Given (a = 20) and the last term (336,140), you can find the common ratio (r) using the formula for the nth term of a geometric sequence:

[a_n = a \times r^{(n-1)}]

Since the last term is the 6th term, you have:

[336,140 = 20 \times r^{(6-1)}]

Solving for (r):

[r^5 = \frac{336,140}{20}] [r^5 = 16,807]

[r = \sqrt[5]{16,807}]

Now that you have (r), you can find the sum (S_6) using the formula for the sum of a geometric series:

[S_6 = \frac{{20(1 - \sqrt[5]{16,807}^6)}}{{1 - \sqrt[5]{16,807}}}]

[S_6 = \frac{{20(1 - 9)}}{{1 - 9}}]

[S_6 = \frac{{20 \times (-8)}}{{-8}}]

[S_6 = -20 \times 8]

[S_6 = -160]

Therefore, the sum of the 6-term geometric series is (-160).

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