We are given a geometric sequence defined recursively where the first term \( a_1 = 10 \) and the recursive relation \( a_n = 4a_{n-1} \).
To find the next few terms of this sequence:
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Calculate \( a_2 \): \[ a_2 = 4a_1 = 4 \times 10 = 40 \]
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Calculate \( a_3 \): \[ a_3 = 4a_2 = 4 \times 40 = 160 \]
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Calculate \( a_4 \): \[ a_4 = 4a_3 = 4 \times 160 = 640 \]
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Calculate \( a_5 \): \[ a_5 = 4a_4 = 4 \times 640 = 2560 \]
So the first five terms of the sequence are:
- \( a_1 = 10 \)
- \( a_2 = 40 \)
- \( a_3 = 160 \)
- \( a_4 = 640 \)
- \( a_5 = 2560 \)
The common ratio can be calculated as: \[ \text{Common Ratio} = \frac{a_n}{a_{n-1}} = 4 \]
This matches the recursive rule \( a_n = 4a_{n-1} \).
If there are any additional questions or details you'd like to explore, please let me know!