If the first and (2n-1)-th terms of an AP, GP | vedclass

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

Question

(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively, they are the Arithmet

Vedclass · Accepted Answer

$ac-b^2=0$

Vedclass · Answer

(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively, they are the Arithmetic Mean, Geometric Mean, and Harmonic Mean of $x$ and $y$.We know that for any two positive numbers $x$ and $y$, the relationship between their means is $AM \geq GM \geq HM$.Therefore, $a \geq b \geq c$.Also, the relationship between $AM, GM,$ and $HM$ is $AM \cdot HM = GM^2$, which implies $a \cdot c = b^2$ or $ac - b^2 = 0$.Since the question asks for the relationship that is 'always' true, both $a \geq b \geq c$ and $ac = b^2$ are correct. However, in standard competitive mathematics, $ac = b^2$ is the specific identity derived from the definition of these means.

If the first and $(2n-1)$-th terms of an $AP$, $GP$, and $HP$ are equal and their $n$-th terms are respectively $a, b, c$, then always

Similar Questions

If ${A_1}, {A_2}$; ${G_1}, {G_2}$ and ${H_1}, {H_2}$ are two $A.M.s$,$G.M.s$ and $H.M.s$ between two numbers respectively,then $\frac{{{G_1}{G_2}}}{{{H_1}{H_2}}} \times \frac{{{H_1} + {H_2}}}{{{A_1} + {A_2}}} = $

The Geometric Mean $(G.M.)$ and Harmonic Mean $(H.M.)$ of two numbers are $10$ and $8$ respectively. The numbers are:

If the ratio of $H.M.$ and $G.M.$ of two quantities is $12:13$,then the ratio of the numbers is

What is the minimum value of the sum of the real numbers $a^{-5}, a^{-4}, 3a^{-3}, 1, a^8$,and $a^{10}$ where $a > 0$?

The ratio of the $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ is $m: n.$ Show that $a: b = (m + \sqrt{m^{2} - n^{2}}) : (m - \sqrt{m^{2} - n^{2}}).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module