If n arithmetic means a_1, a_2, dots, a_n are | vedclass

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

Question

(A) Let $50, a_1, a_2, \dots, a_n, 100$ be in an arithmetic progression $(A.P.)$.The common difference $d = \frac{100 - 50}{n + 1} = \frac{50}{n + 1}$

Vedclass · Accepted Answer

$5000$

Vedclass · Answer

(A) Let $50, a_1, a_2, \dots, a_n, 100$ be in an arithmetic progression $(A.P.)$.The common difference $d = \frac{100 - 50}{n + 1} = \frac{50}{n + 1}$.Then $a_2 = 50 + 2d = 50 + \frac{100}{n + 1} = \frac{50n + 50 + 100}{n + 1} = \frac{50(n + 3)}{n + 1}$.Let $50, h_1, h_2, \dots, h_n, 100$ be in a harmonic progression $(H.P.)$.Then $\frac{1}{50}, \frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}, \frac{1}{100}$ are in an $A.P.$The common difference $d' = \frac{\frac{1}{100} - \frac{1}{50}}{n + 1} = \frac{-1}{100(n + 1)}$.The $(n-1)$-th harmonic mean $h_{n-1}$ corresponds to the $(n-1+1)$-th term in the $A.P.$ of reciprocals,which is the $n$-th term.$\frac{1}{h_{n-1}} = \frac{1}{50} + (n-1)d' = \frac{1}{50} - \frac{n-1}{100(n + 1)} = \frac{2(n + 1) - (n - 1)}{100(n + 1)} = \frac{2n + 2 - n + 1}{100(n + 1)} = \frac{n + 3}{100(n + 1)}$.Thus,$h_{n-1} = \frac{100(n + 1)}{n + 3}$.Finally,$a_2 h_{n-1} = \left( \frac{50(n + 3)}{n + 1} \right) \cdot \left( \frac{100(n + 1)}{n + 3} \right) = 50 \cdot 100 = 5000$.

If $n$ arithmetic means $a_1, a_2, \dots, a_n$ are inserted between $50$ and $100$ and $n$ harmonic means $h_1, h_2, \dots, h_n$ are inserted between the same two numbers,then $a_2 h_{n-1}$ is equal to

Similar Questions

Two sequences ${t_n}$ and ${s_n}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then:

If the sum of the series $1 + \frac{2}{x} + \frac{4}{x^2} + \frac{8}{x^3} + \dots \infty$ is a finite number,then

If $\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3 = \frac{k}{21}$,then $k$ equals

The sum of an infinite $G.P.$ with common ratio $r$ can be found if:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module