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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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} =

Vedclass · Accepted Answer

$1\frac{10}{17}$

Vedclass · Answer

(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

Similar Questions

The mean of the series $a, a + nd, a + 2nd$ is

The sum of the first three terms of a $G.P.$ is to the sum of the first six terms as $125: 152$. Find the common ratio of the $G.P.$ (in $/5$)

The sum of the series $\frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \frac{5}{3! + 4! + 5!} + \dots + \frac{2008}{2006! + 2007! + 2008!}$ is equal to

Find the geometric mean between $\frac{3}{2}$ and $\frac{27}{2}$.

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module