Question

Let A < B < C and C is multiple of B. If A, B, C in HP and A=20 then the number of values of B is what? PLEASE ANSWER AS SOON AS POSSIBLE

Answer 1

VMC test

you cheater

chjaghcscghjgascjnasjcjhgcjgsc

Answer 2

We can find the required number of values of B by putting n= 1,2,3... in the expression $B=\frac{20\times (2n-1)}{n}$.

Given:

A, B, and C are in HP.

A<B<C

C=multiple of C

A=20

To find:

the number of values of B.

Solution:

We know that when the numbers A, B, and C are in harmonic progression, the numbers $\frac{1}{A} ,\frac{1}{B} ,\frac{1}{C}$ are in arithmetic progression.

(Arithmetic progression indicates that the members in the sequence in arithmetic progression differ from each successive member by a constant term called the common difference.)

(Harmonic progression indicates that the respective reciprocals of the members in the sequence in harmonic progression are in arithmetic progression.)

We can write the following equation as per the given data.

$\frac{1}{B} -\frac{1}{A} =\frac{1}{C}-\frac{1}{B}\\\frac{1}{A} +\frac{1}{C} =2\times\frac{1}{B}\\\frac{A+C}{AC} =\frac{2}{B}\\B =\frac{2AC}{A+C}\\B =\frac{2\times20\times C}{20+C}$

We can suppose $C=n\times B$ where n = 1,2,3....

Therefore, we can rewrite the above equation as under.

$B =\frac{40 C}{20+C}\\\frac{1}{B}=\frac{20+C}{40C}\\\frac{1}{B}=\frac{1}{2C}+\frac{1}{40}\\\frac{1}{B}=\frac{1}{2nB}+\frac{1}{40}\\\frac{2n-1}{2nB}=\frac{1}{40}\\B=\frac{20\times (2n-1)}{n}$

Thus, we can find the required number of values of B by putting n=1,2,3... in the expression $B=\frac{20\times (2n-1)}{n}$.

#SPJ3