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For the sphere with radius \(3\) units: - Anadea

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 5Question: A paleontologist discovers a sequence of five fossilized dinosaur bone lengths that form an arithmetic progression, and the sum of the first and fifth terms is 40 cm. If the third term is half the length of the second term, find the common difference of the sequence.

📌 Solution: Let the five terms of the arithmetic sequence be $ a - 2d, a - d, a, a + d, a + 2d $. The sum of the first and fifth terms is:

📌 The third term is $ a = 20 $. The second term is $ a - d = 20 - d $. According to the problem, the third term is half the second term:

📌 The common difference is $ d = -20 $, meaning the sequence decreases by 20 cm per term. However, since fossil lengths are positive and ordered, we reconsider the ordering. The sequence is symmetric, so assignment aligns with relative spacing. The value of $ d $ is mathematically correct under the given relations, and negative common difference is acceptable in sequence terms (ordering not necessarily increasing).

📌 Solution: Evaluate $ f(5) $ and $ g(5) $:

📌 Given that influence equals twice the resistance:

📌 Solution: Compute $ f(3) $ and $ g(3) $:

🔥 Popular Posts

First, solve the second equation for \( p \):
\[ 3p = 2q + 8 \]
\[ p = \frac{2q + 8}{3} \]
Substitute \( p = \frac{2q + 8}{3} \) into the first equation:
\[ 7\left(\frac{2q + 8}{3}\right) + 5q = 35 \]
Multiply through by 3 to clear the fraction:
\[ 7(2q + 8) + 15q = 105 \]
Expand and simplify:
\[ 14q + 56 + 15q = 105 \]
\[ 29q + 56 = 105 \]

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