If $A_1, A_2$ are two arithmetic means,$G_1, G_2$ are two geometric means,and $H_1, H_2$ are two harmonic means between two numbers,then $\frac{A_1 + A_2}{H_1 + H_2} \cdot \frac{H_1 H_2}{G_1 G_2} = \dots$

  • A
    $0$
  • B
    $4$
  • C
    $1$
  • D
    $8$

Explore More

Similar Questions

Let $a_i = i + \frac{1}{i}$ for $i = 1, 2, \ldots, 20$. Let $p = \frac{1}{20} \sum_{i=1}^{20} a_i$ and $q = \frac{1}{20} \sum_{i=1}^{20} \frac{1}{a_i}$. Then,

Three unequal positive numbers $a, b, c$ are such that $a, b, c$ are in $G.P.$ while $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in $A.P.$ Then $a, b, c$ are the lengths of the sides of

If the $(m + 1)^{th}$,$(n + 1)^{th}$,and $(r + 1)^{th}$ terms of an arithmetic progression are in geometric progression,and $m, n, r$ are in harmonic progression,then what is the ratio of the common difference to the first term of the arithmetic progression?

Difficult
View Solution

Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

If $e^{(\cos^{2} x + \cos^{4} x + \cos^{6} x + \dots \infty) \log_{e} 2}$ satisfies the equation $t^{2} - 9t + 8 = 0$,then the value of $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$ for $0 < x < \frac{\pi}{2}$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo