The AM and GM of two numbers are 17 and 8 res | vedclass

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

Question

(C) Given $AM = 17$ and $GM = 8$.Let the two numbers be $a$ and $b$.We know that $AM = \frac{a+b}{2} = 17 \implies a+b = 34$ $(1)$.We know that $GM =

Vedclass · Accepted Answer

$\frac{64}{17}$

Vedclass · Answer

(C) Given $AM = 17$ and $GM = 8$.Let the two numbers be $a$ and $b$.We know that $AM = \frac{a+b}{2} = 17 \implies a+b = 34$ $(1)$.We know that $GM = \sqrt{ab} = 8 \implies ab = 64$ $(2)$.Using the relation $(a-b)^2 = (a+b)^2 - 4ab$,we get:$(a-b)^2 = (34)^2 - 4(64) = 1156 - 256 = 900$.Thus,$a-b = 30$ $(3)$.Solving $(1)$ and $(3)$,we get $2a = 64 \implies a = 32$ and $b = 2$.The Harmonic Mean $(HM)$ is given by $HM = \frac{2ab}{a+b} = \frac{2(64)}{34} = \frac{128}{34} = \frac{64}{17}$.The $AP$ is $2, 17, 32$.The $GP$ is $2, 8, 32$.The $HP$ is $2, \frac{64}{17}, 32$.

The $AM$ and $GM$ of two numbers are $17$ and $8$ respectively. Find their Harmonic Mean $(HM)$ and write the $AP, GP,$ and $HP$.

Similar Questions

The sum of $n$ terms of the following series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots$ is:

Determine $k$ so that $\frac{2}{3}, k$ and $\frac{5}{8} k$ are the three consecutive terms of an $A.P.$

If the $19^{th}$ term of a non-zero $A.P.$ is zero,then the ratio of its ($49^{th}$ term) to ($29^{th}$ term) is:

The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0, (abc \neq 0)$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module