Illustration 1: Let ‘a’ and ‘b’ be positive real numbers. If a, A_{1}, A_{2}, b are in arithmetic progression, a, G_{1}, G_{2}, b are in geometric progression and a, H_{1}, H_{2}, b are in harmonic progression, show that
G_{1}G_{2}/ H_{1}H_{2} = (A_{1} + A_{2})/ (H_{1} + H_{2}) = (2a+b)(a+2b)/9ab. (2002)
Solution: We are given that a, A_{1}, A_{2}, b are in A.P
Hence, A_{1}+A_{2} = a+b
Now, a, G_{1}, G_{2}, b are in G.P
So, G_{1}G_{2} = ab
Similarly, as a, H_{1}, H_{2}, b are in H.P
So, H_{1} = 3ab/ (2b+a)
And H_{2} = 3ab/ (b+2a)
Hence, 1/H_{1} + 1/H_{2} = 1/a + 1/b
So, (H_{1} + H_{2})/ H_{1}.H_{2} = (A_{1}+A_{2})/G_{1}.G_{2 }= 1/a + 1/b
Now, (G_{1}G_{2})/ H_{1}.H_{2} =
= (2a+b) (a+2b)/ 9ab
Thus combining the obtained results, we get
(G_{1}G_{2})/ H_{1}.H_{2} = (A_{1}+A_{2})/(H_{1} + H_{2}) = (2a+b) (a+2b)/ 9ab
Illustration 2: Let A_{1}, H_{1} denote the arithmetic and harmonic means of two distinct positive numbers. For n ≥ 2, let A_{n1}, H_{n1 }have arithmetic and harmonic means as An and H_{n }respectively. Then which of the following statements is correct? (2007)
(a) 1. H_{1 }> H_{2} > H_{3} > ….
2. H_{1 }< H_{2} < H_{3 }< ….
3. H_{1 }> H_{3} > H_{5} > …. and H_{2} < H_{4} < H_{6} < ….
4. H_{1} < H_{3 }< H_{5 }< …. and H_{2 }> H_{4} > H_{6} > ….
(b) 1. A_{1} > A_{2} > A_{3} > ….
2. A_{1 }< A_{2} < A_{3} < ….
3. A_{1 }> A_{3} > A_{5} > …. and A_{2} < A_{4} < A_{6 }< ….
4. A_{1} < A_{3} < A_{5} < …. and A_{2} > A_{4} > A_{6 }> ….
Solution: H1 is the harmonic mean between two numbers ‘a’ and ‘b’.
A_{1 }= (a+b)/2
H_{1 }= 2ab/ (a+b)
H_{n} = 2A_{n1}H _{n1} / (A_{n1 }+ H_{ n1})
A_{n} = (A_{n1} + H _{n1)}/ 2
A_{2} is A.M of A_{1} and H_{1} and A_{1} > H_{1}
Hence, A_{1 }> A_{2} > H_{1}
A_{3} is A.M of A_{2 }and H_{2} and A_{2 }> H_{2}
So, A_{2} > A_{3} > H_{2}
…… …… …….
Hence, A_{1} > A_{2} > A_{3} > ……
Now, as we have stated above A_{1 }> H_{2} > H_{1}, A_{2} > H_{3} > H_{2}
So, H_{1} < H_{2} < H_{3 }< ….
Illustration 3: Prove that three quantities a, b, c are in A.P., G.P., or H.P. iff
(ab)/(bc) = a/a, a/b or a/c respectively.
Solution: If a, b and c are in A.P, then
ba = cb
This can be written as a, b and c are in A.P. iff ba = cb
iff (ab)/(bc) = 1 = a/a
Similarly, when a, b and c are in G.P,
iff b/a = c/b i.e
iff 1  b/a = 1  c/b
i.e. iff (ab)/a = (bc)/b
i.e. iff (ab)/(bc) = a/b
Similarly a, b, c are in H.P.
iff 1/b  1/a = 1/c  1/b
i.e. iff (ab)/ab = (bc)/bc
i.e. iff (ab)/(bc) = ab/bc = a/c
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