The 9^{th} term of the sequence 27, 9, 5frac{ | vedclass

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Question

(A) The given sequence is $27, 9, \frac{27}{5}, \frac{27}{7}, \dots$ This can be written as $\frac{27}{1}, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \

Vedclass · Accepted Answer

$1\frac{10}{17}$

Vedclass · Answer

(A) The given sequence is $27, 9, \frac{27}{5}, \frac{27}{7}, \dots$ This can be written as $\frac{27}{1}, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$ The numerator is constant at $27$. The denominator follows the arithmetic progression $1, 3, 5, 7, \dots$ The $n^{th}$ term of the denominator is $a_n = 1 + (n-1)2 = 2n - 1$. Thus,the $n^{th}$ term of the sequence is $t_n = \frac{27}{2n-1}$. For the $9^{th}$ term,substitute $n = 9$: $t_9 = \frac{27}{2(9) - 1} = \frac{27}{18 - 1} = \frac{27}{17} = 1\frac{10}{17}$.

The $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$

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