The fifth term of the H.P.,2, 2frac{1}{2}, 3f | vedclass

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

Question

(D) The given series $2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ is in $H.P.$Taking the reciprocals,the series $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \do

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) The given series $2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ is in $H.P.$Taking the reciprocals,the series $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \dots$ is in $A.P.$Here,the first term $a = \frac{1}{2}$.The common difference $d = \frac{2}{5} - \frac{1}{2} = \frac{4-5}{10} = -\frac{1}{10}$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.For the $5^{th}$ term of the $A.P.$: $T_5 = \frac{1}{2} + (5-1)(-\frac{1}{10}) = \frac{1}{2} - \frac{4}{10} = \frac{5-4}{10} = \frac{1}{10}$.Since the $H.P.$ is the reciprocal of the $A.P.$,the $5^{th}$ term of the $H.P.$ is the reciprocal of $\frac{1}{10}$,which is $10$.

The fifth term of the $H.P.$,$2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ will be

Similar Questions

If $(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + ... + 10(11)^9 = k(10)^9$,then $k$ is equal to:

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

If the fifth term of a $G.P.$ is $2$,then the product of its first $9$ terms is:

The sum of infinite terms of a $G.P.$ is $x$ and on squaring each term of it,the sum becomes $y$. Then the common ratio of this series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module