Question

25. If a, b, c are positive real numbers, then Vatbx Votex Vota is equal to
1
(d)
(a) 1
(b) abc
(c) Vabc
abc​

Answer 1

Answer:

use the concept

Arithmetic mean of two numbers (AM) -

A=\frac{a+b}{2}

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

Geometric mean of two numbers (GM) -

GM= \sqrt{ab}

- wherein

It is to be noted that a,G,b are in GP and a,b are two non - zero numbers.

Harmonic mean (HM) of two numbers a and b -

HM= \frac{2ab}{a+b}

-

9\left ( 25a^{2}+b^{2} \right )+25\left ( c^{2}-3ac \right )=15b\left ( 3a+c \right )

=225a^{2}+9b^{2}+25c^{2}-75ac= 45ab+15bc

225a^{2}+9b^{2}+25c^{2}-45ab-15bc-75ac=0

multiply and divide by 2

\frac{1}{2}\left [ 450a^{2}+18b^{2}+50c^{2}-90ab-30bc-150ac \right ]=0

225a2+9b2+25c2+225a2+9b2+25c2-2x15ax3b-2x3bx5c-2x5cx15a

\therefore \left ( 15a-3b\right )^{2}+\left ( 3b-5c \right )^{2}+\left ( 5c-15a \right )^{2}= 0

\because \: \left ( a-b \right ) ^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}

= 2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ac

So that 5a- 3b=0

b=\frac{5a}{3}

and 3b =5c

c=\frac{3b}{5}

So that

15a=3b=5c=K

\therefore \: a=\frac{K}{15}

b=\frac{K}{3}

C=\frac{K}{5}

Now

2\times b=\frac{2K}{3}

=a+c=\frac{K}{15}+\frac{K}{5}

=\frac{\left ( 1+3 \right )k}{15}

=\left ( \frac{4K}{15} \right )

\therefore a, b,c are not in A.p

Now

2c=a+b

\therefore \:\frac{2K}{5}=\frac{K}{15}+\frac{K}{3}

=\frac{6K}{15}=\frac{2K}{5}

So b, c,a are in A.P

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