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

If $a, b, c, d, e$ are in $A.P.$,then the value of $a + b + 4c - 4d + e$ in terms of $a$,if possible,is:

The third term of a $G.P.$ is the square of the first term. If the second term is $8$,then the $6^{th}$ term is:

The sum of the first $n$ terms of a sequence is given by $S_n = 3n^2 + 4n + 15$. If $T_r$ is the $r^{th}$ term of the sequence,then $T_3 - T_1$ is equal to:

The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

The sum of all the products of the first $n$ natural numbers taken two at a time is

Vedclass Test Series

Exam Paper Generator

Online Exam Module