If m is the A.M. of two distinct real numbers | vedclass

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

Question

(C) Given $m = \frac{l+n}{2}$,so $2m = l+n$.Since $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,the sequence $l, G_1, G_2, G_3, n$ is

Vedclass · Accepted Answer

$4lm^2n$

Vedclass · Answer

(C) Given $m = \frac{l+n}{2}$,so $2m = l+n$.Since $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,the sequence $l, G_1, G_2, G_3, n$ is in $G.P.$Let the common ratio be $r$. Then $n = l r^4$,so $r^4 = \frac{n}{l}$.The terms are $G_1 = lr, G_2 = lr^2, G_3 = lr^3$.We need to evaluate $G_1^4 + 2G_2^4 + G_3^4 = (lr)^4 + 2(lr^2)^4 + (lr^3)^4$.$= l^4r^4 + 2l^4r^8 + l^4r^{12} = l^4r^4(1 + 2r^4 + r^8) = l^4r^4(1 + r^4)^2$.Substitute $r^4 = \frac{n}{l}$:$= l^4 \left(\frac{n}{l}\right) \left(1 + \frac{n}{l}\right)^2 = l^3n \left(\frac{l+n}{l}\right)^2 = l^3n \frac{(l+n)^2}{l^2} = ln(l+n)^2$.Since $l+n = 2m$,we have $ln(2m)^2 = 4lm^2n$.

If $m$ is the $A.M.$ of two distinct real numbers $l$ and $n$ $(l, n > 1)$ and $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,then $G_1^4 + 2G_2^4 + G_3^4$ equals:

Similar Questions

If $a, b, c$ are in $G$.$P$. and $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A$.$P$.,then $a, b, c$ are the lengths of the sides of a triangle which is

Let $a, b \in \mathbb{R}$ be such that $a, a + 2b, 2a + b$ are in $A.P.$ and $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$ Then $(a + b)$ equals:

Let $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$ for all $n > 1, n \in N$. Then:

If three unequal non-zero real numbers $a, b, c$ are in $G.P.$ and $b - c, c - a, a - b$ are in $H.P.$,then the value of $a + b + c$ is independent of

What is the solution to the equation $8^{(1 + |\cos x| + |\cos^2 x| + |\cos^3 x| + \dots)} = 4^3$ in the interval $(-\pi, \pi)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module