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# Let p be the first of the n arithmetic means between two numbers and q the first of n harmonic means between the same numbers. Show that q does not lie between p and(n + 1)/n – 1)2 p.

6 years ago
Hello Student,
Let a and b be two numbers and A1, A2, A3, ………. An be n A. M’s between a and b
Then a, A1, A2, ……. An, b are in A. P. There are (n + 2) terms in the series, so that
a + (n + 1)d = b ⇒ d = b – a/n + 1
∴ A1 = a + b – a/n + 1 = an + b/n + 1
∴ ATQ p = an + b/n + 1 ….. (1)
The first H. M. between a and b, when nHM’s are inserted between a and b can be obtained by replacing a by 1/a and b by 1/b in eq. (1) and then taking its reciptocal.
Therefore, q = 1/(1/a)n + 1/b/n + 1 = (n + 1)ab/bn + a
∴ q = (n + 1)ab/a + bn ……… (2)
We have to prove that q cannot lie between p and (n + 1)2/(n – 1)2p.
Now, n + 1 > n – 1 ⇒ n + 1/n – 1 > 1
⇒ (n + 1/n – 1)2 > 1 or p(n + 1/n - 1)2 > p
⇒ p < p(n + 1/n – 1)2 ………….. (3)
Now to prove the given, we have to show that q is less than p.
For this, let, p/q = (na + b) (nb + a)/(n + 1)2 ab
⇒ p/q – 1 = n(a2 + b2) + ab (n2 + 1) – (n + 1)2 ab/(n + 1)2 ab
⇒ p/q – 1 = n(a2 + b2 – 2ab)/(n + 1)2 ab
⇒ p/q – 1= n/(n + 1)2 (a – b/√ab)2
= n/(n + 1)2 ($\sqrt{a/b}-\sqrt{b/a}$)2 ⇒ p/q – 1 > 0
⇒ ( provided a and b and hence p and q are +ve )
p > q …….. (4)
From (3) and (4), we get,
q < p <(n + 1/n + 1)2 p
∴ q can not lie between p and (n + 1/n + 1)2 p, if a and b are +ve numbers.
ALTERNATE SOLUTION :
After getting equations (1) and (2) as in the previous method, substitute b = p(n + 1) – an [from (1)] in equation (2) to get aq + nq [p(n + 1) – an] = (n + 1)a [p(n + 1) – an ]
⇒ a2n (n + 1) + a [q(1 – n2) – p(n + 1)2] + npq (n + 1) = 0
⇒ na2 – [(n + 1)p +(n – 1)q] a + npq = 0
⇒ D ≥ 0 ( ∵ a is real)
⇒ [n + 1)p + (n – 1)q]2 – 4n2pq ≥ 0
⇒ (n – 1)2q2 + {2(n2 – 1) – 4n2} pq + (n + 1)2p2 ≥ 0
⇒ q2 – 2 n2 + 1/(n – 1)2 pq + (n + 1/n – 1)2 p2 ≥ 0
⇒ [q – p(n + 1/n – 1)2 [q – p] ≥ 0
[ On factorizing by discriminent method] ⇒ q can not lie between p and p(n + 1/n – 1)2.

Thanks