The 9^{th} term of the series 27 + 9 + 5frac{ | vedclass

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(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$Rewriting the terms: $27, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$This c

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$1\frac{10}{17}$

Vedclass · Answer

(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$Rewriting the terms: $27, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$This can be expressed as $\frac{27}{1}, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$The denominator follows an arithmetic progression: $1, 3, 5, 7, \dots$ where the $n^{th}$ term is $a_n = 1 + (n-1)2 = 2n - 1$.Thus,the $n^{th}$ term of the series is $T_n = \frac{27}{2n - 1}$.For the $9^{th}$ term $(n = 9)$:$T_9 = \frac{27}{2(9) - 1} = \frac{27}{18 - 1} = \frac{27}{17} = 1\frac{10}{17}$.

The $9^{th}$ term of the series $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$ is:

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