Given that (10)/(7) n6, n7, and n8 are the first three consecutive terms of a geometric sequence (G.P.).

April 3, 2026Mathematics

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April 3, 2026

Answer

\frac{a_3}{a_2}$.

Here are the solutions to question 9:

Given that $\frac{10}{7} \binom{n}{6}$ , $\binom{n}{7}$ , and $\binom{n}{8}$ are the first three consecutive terms of a geometric sequence (G.P.).

Let the terms be $a_1 = \frac{10}{7} \binom{n}{6}$ , $a_2 = \binom{n}{7}$ , and $a_3 = \binom{n}{8}$ . For a geometric progression, the common ratio $r$ is constant, so $\frac{a_2}{a_1} = \frac{a_3}{a_2}$ .

We use the property of combinations: $\binom{n}{k} = \frac

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Given that (10)/(7) n6, n7, and n8 are the first three consecutive terms of a geometric sequence (G.P.).

Asked on April 3, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on April 3, 2026

Here are the solutions to question 9: Given that (10)/(7) n6, n7, and n8 are the first three consecutive terms of a geometric sequence (G.P.). Let the terms be a_1 = (10)/(7) n6, a_2 = n7, and a_3 = n8. For a geometric progression, the common ratio r is constant, so (a_2)/(a_1) = (a_3)/(a_2). We use the property of combinations: $nk =