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

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(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$We can rewrite the terms as:$27 = \frac{27}{1}$$9 = \frac{27}{3}$$5\frac{2}{5} =

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$1\frac{10}{17}$

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(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$We can rewrite the terms as:$27 = \frac{27}{1}$$9 = \frac{27}{3}$$5\frac{2}{5} = \frac{27}{5}$$3\frac{6}{7} = \frac{27}{7}$Thus,the series is $\frac{27}{1} + \frac{27}{3} + \frac{27}{5} + \frac{27}{7} + \dots$The numerator is constant $(27)$,and the denominators form an arithmetic progression: $1, 3, 5, 7, \dots$The $n^{th}$ term of the denominator is $a + (n-1)d = 1 + (n-1)2 = 2n - 1$.Therefore,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 $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$ will be

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