Question:
For any three positive real numbers, \(a\), \(b\) and \(c\), \(9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)\). Then prove that \(a\), \(b\) and \(c\) are in AP.
Solution:
\(9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)\)
or,
\(225a^2+9b^2+25c^2=45ab+15bc+75ac\)
or,
\((15a)^2+(3b)^2+(5c)^2-(15a)(3b)-(3b)(5c)-(15a)(5c)=0\)
We know that,
\(\boxed{x^2+y^2+z^2-xy-yz-xz=\frac{1}{2}[(x-y)^2+(y-z)^2+(x-z)^2]}\)
Learn more about this formula.
So,
(\(15a-3b)^2+(3b-5c)^2+(15a-5c)^2=0\)
According to “SOS rule” (know more)
\(15a=3b\)
\(3b=5c\)
\(15a=5c\)
Therefore, \(15a=3b=5c\)
Now let, \(15a=3b=5c=k\)
\(k=15a \implies a=\frac{k}{15}\)
Also, \(b=\frac{k}{3}\)
And, \(c=\frac{k}{5}\)
Now, we will observe \(a\), \(b\) and \(c\).
In question, we are asked to prove that they are in AP. It means sum of two terms will be equal to twice of third term.
When we observe carefully, we see:
\(\frac{k}{15}+\frac{k}{3}=2 \cdot \frac{k}{5}\)
So, \(\boxed{a+b=2c}\)
Hence, \(a\), \(b\) and \(c\) are in AP.
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