Let be in GP. What is equal to?

. In a GP, . The product . Raising this to the power of gives , which is exactly the formula for the term, .

Let t_1, t_2, t_3, \dots be in GP. What is (t_1t_3\dots t_{21})^{\frac{1}{11}} equal to?

A. t_{10}
B. t_{10}^2
C. t_{11} ✓
D. t_{11}^2

Correct Answer: C. t_{11}

Explanation

In a GP, t_n = ar^{n-1}. The product t_1t_3\dots t_{21} = (a)(ar^2)\dots(ar^{20}) = a^{11} r^{2+4+\dots+20} = a^{11} r^{110}. Raising this to the power of \frac{1}{11} gives ar^{10}, which is exactly the formula for the 11^{th} term, t_{11}.

Let t_1, t_2, t_3, \dots be in GP. What is (t_1t_3\dots t_{21})^{\frac{1}{11}} equal to?

Explanation

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