given that the common ratio is 1/3 and the first term is 81 .write down the first 3 terms of a GP

In a geometric progression (GP), each term is obtained by multiplying the previous term by the common ratio. 

Given:
- First term \( a = 81 \)
- Common ratio \( r = \frac{1}{3} \)

The first three terms of the GP are:

1. \( a_1 = 81 \)
2. \( a_2 = a_1 \times r = 81 \times \frac{1}{3} = 27 \)
3. \( a_3 = a_2 \times r = 27 \times \frac{1}{3} = 9 \)

Thus, the first three terms are **81, 27, and 9**.

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the formula for the sum of the first \(n\) terms in a geometric progression?
2. How would the terms change if the common ratio was negative?
3. What is the sum to infinity for this GP, and how is it calculated?
4. Can you determine the 10th term of this GP?
5. How do geometric progressions differ from arithmetic progressions?

**Tip:** In a GP, the terms grow (or shrink) exponentially based on the common ratio, unlike arithmetic progressions which grow linearly.

Given that the common ratio is 1/3 and the first term is 81, write down the first 3 terms of a GP.

Find the first three terms of a geometric progression (GP) where the first term is 81 and the common ratio is 1/3. This problem covers key concepts in sequences and geometric progressions. Ideal for students in Grades 7-9.

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