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The concept of arithmetic mean has been already discussed in the previous sections. This concept can be extended till the mth power for terms and such mean is termed as the arithmetic mean of mth power. Mathematically, the arithmetic mean for mth power is defined as
If a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number.
However if m ∈ (0, 1), then a1m + a2m+...+ anm/n < (a1+a2+...+an/n)m.
Obviously if m ∈ {0, 1}, then a1m + a2m+...+anm/n = (a1+a2+...+an/n)m.
Illustration:
If a, b, c are positive real numbers such that a + b + c = 1, then prove that a/b + c + b/c+a + c/a+b > 3/2.
Solution: We have to show that (a/b+c+1)+(b/c+a + 1)+(c/a+b + 1) > 3/2 + 3.
i.e. 1/b+c + 1/c+1 + 1/a+b > 9/2.
Now A.M. of mth power > mth power arithmetic mean (m = – 1 here)
⇒ 1/3 [(b+c)–1 + (c+a)–1 + (a+b)–1] > [(b+c) + (c+a) + (a+b)/3]–1
⇒ 1/b+c + 1/c+a + 1/a+b > 3.3/2(a+b+c) = 9/2.
Illustration: Prove that
1m + 2m + 3m + ….. + nm > n[1/2(n+1)]m.
Solution: Taking ‘n’ quantities into consideration we have
Then as stated above, arithmetic mean of mth powers > mth power of the arithmetic mean
(1m + 2m + 3m + ….. + nm) / n > [(1+2+3+…. n) / n]m
or (1m + 2m + 3m + ….. + nm) > n [ {n(n+1)/2}/n]m
{Since the sum of first n natural numbers is [n(n+1)/2]}
Hence, (1m + 2m + 3m + ….. + nm) > n [ (n+1)/2}]m.
askIITians offers extensive study material which covers all the important topics of IIT JEE Mathematics. The concepts of arithmetic mean of an A.P. or the arithmetic formula for the power have also been covered in the material. It is vital for the aspirants to be familiar with the concepts in order to remain competitive in the JEE.
Related resources:
Click here for the Entire Syllabus of IIT JEE Mathematics.
Look into the Solved Papers of Previous Years to get a hint of the kinds of questions asked in the exam.
You can get the knowledge of Important Books of Mathematics here.
You may also like to refer Geometric Mean and Harmonic Mean.
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