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For any three positive real numbers a, b and c,

9(25a²+b²)+25(c²−3ac)=15b(3a+c). Then :

Option 1)
b, c and a are in A.P.

Option 2)
a, b and c are in A.P.

Option 3)
a, b and c are in G.P.

Option 4)
b, c and a are in G.P.

Answers (1)

best_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

Option 1)

b, c and a are in A.P.

Correct option

Option 2)

a, b and c are in A.P.

Incorrect option

Option 3)

a, b and c are in G.P.

Incorrect option

Option 4)

b, c and a are in G.P.

Incorrect option

Posted by

Himanshu

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