# 37. AM, GM, HM SolutionsΒΆ

Let the \(n\) means be \(x_1, x_2, ..., x_n\),.

Then \(20, x_1, x_2, ..., x_n, 80\) are in A. P.

Now, \(80 = (n + 2)\) th term of A. P.

\(80 = 20 + (n + 1)d\) where \(d\) is the common difference.

\(d = \frac{60}{n + 1}\)

\(x_1 = 20 + d = \frac{20n + 80}{n + 1}\)

\(x_n = 20 + nd = \frac{20 + 80n}{n + 1}\)

Given, \(x_1:x_n = 1:3\) \(\Rightarrow \frac{20n + 80}{80n + 20} = \frac{1}{3}\)

\(\therefore n = 11\)

Let the \(n\) arithmetic means are \(x_1, x_2, ..., x_n\)

Then \(a, x_1, x_2, ..., x_n, b\) will be in A. P.

\(\therefore x_1 = a + d\) and \(x_n = b - d\) where \(d\) is the common difference.

Thus, \(x_1 + x_2 + ... + x_n = \frac{n}{2}(a + b)\) (\(\because x_1 + x_2 + ... + x_n = \frac{n}{2}(\text{first term} + \text{last term})\))

Let \(2n\) be the number of means between two numbers \(a\) and \(b\).

Now, sum of the \(2n\) means = \(\frac{a + b}{2}.2n = (a + b)n\)

Given, \((a + b)n = 2n + 1 \Rightarrow \frac{13}{6} = 2n + 1 \Rightarrow 2n = 12\)

between \(a\) and \(b = \frac{a + b}{2}\)

Given, \(\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{a + b}{2}\)

\(\Rightarrow (a - b)(a^n - b^n) = 0\)

\(\because a\ne b \therefore a^n = b^n \Rightarrow n = 0\)

Let \(x_1, x_2, x_3, x_4\) be the four G. M. between \(5\) and \(160\).

Thus, \(5, x_1, x_2, x_3, x_4, 160\) will be in G. P.

\(160 = 5r^5\) where \(r\) is the common ratio. \(\Rightarrow r = 2\)

Thus, means are \(10, 20, 40, 80\)

Let \(x_1, x_2, ..., x_n\) be \(n\) G. M. between two quantities \(a\) and \(b\)

\(x_1 = a \left(\frac{b}{a}\right)^{\frac{1}{n + 1}}, x_2 = a \left(\frac{b}{a}\right)^{\frac{2}{n + 1}}, ..., x_n = a \left(\frac{b}{a}\right)^{\frac{n}{n + 1}}\)

Thus, \(x_1x_2 ... x_n = a^n\left(\frac{b}{a}\right)^{\frac{1 + 2 + ... + n}{n + 1}} = a^n\left(\frac{b}{a}\right)^{\frac{n}{2}} = (\sqrt{ab})^n\)

Let \(x_1, x_2, ..., x_6\) be the six H. M. between \(3\) and \(\frac{6}{23}\)

Thus, \(\frac{1}{3}, \frac{1}{x_1}, \frac{1}{x-2}, ..., \frac{1}{x_6}, \frac{23}{6}\) are in A. P.

\(\frac{23}{6} = \frac{1}{3} + 7d\) where \(d\) is the common difference. \(\Rightarrow d = \frac{1}{2}\)

Now it is trivial to calculate the values of harmonic means.

Let \(a\) and \(b\) be two numbers. Thus, \(\frac{a + b}{2} = 5 \Rightarrow a + b = 10\)

Since \(3\) is the G. M. between \(a\) and \(b\), \(a, 3, b\) are in G. P. Let \(r\) be the common ratio then \(ar = 3\) and \(b = ar^2\).

Now, we have \(a + ar^2 = 10\) and \(ar = 3\) and thus, \(\frac{1 + r^2}{r} = \frac{10}{3}\)

Solving this we get \(r = 3, \frac{1}{3}\) and thus the two numbers are \(9, \frac{1}{9}\).

Let \(A\) and \(G\) be the A. M. and G. M. between two numbers \(a\) and \(b\) then \(A = \frac{a + b}{2}\) and \(G = \sqrt{ab}\).

Given, \(\frac{a + b}{2} = 2\sqrt{ab} \Rightarrow \frac{a + b}{\sqrt{ab}} = \frac{2}{1}\)

By componendo and dividendo, we get

\(\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{3}{1} \Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{3}}{1}\)

Again by componendo and dividendo, we get

\(\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \Rightarrow \frac{a}{b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)

Given, \(a = \frac{b + c}{2}\). Let \(r\) be the common ration then \(g_1 = br\) and \(g_2 = br^2\)

\(g_1^3 + g_2^3 = (br)^3 + (br^2)^3 = b^3r^3(1 + r^3) = b^3.\frac{c}{b}\left(1 + \frac{c}{b}\right)\)

\(= bc(b + c) = 2abc\)

Given \(a, b, c\) are in G. P.

Let \(r\) be the common ratio and thus \(b = ar, c = ar^2\)

Given, \(x = \frac{a + b}{2}\) and \(y = \frac{b + c}{2}\)

Now, \(\frac{a}{x} + \frac{b}{y} = \frac{2a}{a + b} + \frac{2c}{b + c} = \frac{2a}{a(1 + r)} + \frac{2ar^2}{ar(1 + r)} = 2\)

Again, \(\frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2}{a(1+r)}\left(1 + \frac{1}{r}\right) = \frac{2}{b}\)

Given, \(A = \frac{a + b}{2}\) and \(H = \frac{2ab}{a + b}\)

Now, \(\frac{a - A}{a - H}.\frac{b - A}{b - H} = \frac{(a - b)(b - a)(a + b)^2}{4(a^ + ab - 2ab)(ab + b^2 - 2ab)}\)

\(= \frac{(a + b)^2}{4ab} = \frac{A}{H}\)

Let the two numbers be \(a\) and \(b\).

\(\therefore A_1 + A_2 = a + b, G_1G_2 = ab\)

Also, \(\frac{1}{H_1} + \frac{1}{H_2} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}\)

Thus, \(\frac{G_1G_2}{H_1H_2} = \frac{A_1 + A_2}{H_1 + H_2}\)

Let the two numbers be \(a\) and \(b\) and \(A, G, H\) be the respective A. M., G. M., H. M. between them.

Given, \(A = G + \frac{3}{2}\) and \(G = H + \frac{6}{5}\)

\(A.H = G^2 \Rightarrow \left(G + \frac{3}{2}\right)\left(G - \frac{6}{5}\right) = G^2 \Rightarrow G = 6\)

Thus, \(a + b = 15\) and \(ab = 36\). Solving we get numbers as \(12\) and \(3\)

Since \(a, b, c, d\) are in H. P. thus \(b\) is H. M. of \(a\) and \(c\) and \(c\) is H. M. of \(b\) and \(d\)

Since A. M. > H. M. \(\therefore \frac{a + c}{2} > b\) or \(a + c > 2b\)

Similarly \(b + d > 2c\). Adding these two \(a + b + c + d > 2b + 2c \Rightarrow a + d > b + c\)

Also, since G. M. > H. M. \(\sqrt{ac} > b\) and \(\sqrt{bd} > c\) and multiplying these two \(ad > bc\)

Since \(a, b, c\) are in H. P. \(b\) is H. M. of \(a\) and \(c\).

Since, G. M. > H. M. \(\therefore \sqrt{ac} > b\)

Now, A. M. of \(a^n\) and \(c^n = \frac{a^n + c^n}{2}\) and G. M. \(= (\sqrt{ac})^n\)

But A. M. > G. M. \(\therefore \frac{a^n + b^n}{2} > (\sqrt{ac})^n > b^n\)

\(a + b = x + y + z = \frac{a + b}{2}.3 \Rightarrow a + b = 10\) when they are in A. P.

When they are in H. P. \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\left(\frac{1}{a} + \frac{1}{b}\right)}{2} \Rightarrow ab = 9\)

Thus the numbers are \(a = 9, b = 1\)

Since A. M \(\geq\) G. M. \(\therefore \frac{x + \frac{1}{x}}{2} > \sqrt{x. \frac{1}{x}} \Rightarrow x + \frac{1}{x} > 2\)

Let \(x_1, x_2, ..., x_8\) are \(8\) arithmetic means between \(5\) and \(32\).

Let \(d\) be the common difference. Then, \(32 = 5 + 9d \Rightarrow d = 3\)

Thus, the means are \(8, 11, 14, 17, 20, 23, 26, 29\)

This is similar to previous problem and has been left as an exercise.

This is similar to previous problem and has been left as an exercise.

Let the means are \(x_x, x_2, ..., x_{31}\) between \(1\) and \(31\) then \(d = \frac{30}{n + 1}\) where \(d\) is the common difference.

Now \(\frac{x_6}{x_{n -1}} = \frac{5}{9} \Rightarrow \frac{1 + 5d}{1 + (n - 1)d} = \frac{5}{9} \Rightarrow n = 14\)

In first case \(x_r = x + \frac{2y - x}{n + 1}r\) and in second case \(y_r = 2x + \frac{y - 2x}{n + 1}r\) where \(x_r\) and \(y_r\) are \(r\) th means respectively.

Equating them we get \(ry = (n + 1 -r)x\)

Let \(g_1, g_2, ..., g_7\) be the \(7\) geometric means between \(2\) and \(162\) and let \(r\) be the common ratio. Then, \(162 = 2.r^8 \Rightarrow r = \sqrt{3}\)

Thus, geometric means are \(2\sqrt{3}, 6, 6\sqrt{3}, 18, 18\sqrt{3}, 54, 54\sqrt{3}\)

This is similar to previous problem and has been left as an exercise.

Let \(2n + 1\) geometric means are inserted between \(a\) and \(b\) and that \(r\) is the common ratio. Then \(b = a.r^{2n + 2} \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{2n + 2}}\)

Thus middle geometric mean \(g_{n + 1} = a.r^{n + 1} = \sqrt{ab}\)

Let \(x_1, x_2, x_3, x_4\) be 4 harmonic means between \(1\) and \(\frac{1}{11}\). Now it is as simple as treating them in A. P. and finding out the means which has been shown previously.

Let \(h_1, h_2, ..., h_n\) be \(n\) harmonic means between \(1\) and \(4\) then \(1, \frac{1}{h_1}, \frac{1}{h_2}, ..., \frac{1}{h_n}, \frac{1}{4}\) are in A. P. Let \(d\) be the common difference then \(d = \frac{\frac{1}{4} - 1}{n + 1}\)

Solving like 22 we have \(n = 11\)

between \(a\) and \(b = \frac{2ab}{a + b}\)

Given, \(\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{2ab}{a + b}\)

\(a^{n + 2} + ab^{n + 1} + ba^{n + 1} + b^{n + 2} = 2a^{n+1}b + 2ab^{n + 1}\)

\(a^{n + 2} - ab^{n + 1} - ba^{n + 1} + b^{n + 2} = 0\)

\(\Rightarrow a^{n + 1} - b^{n + 1} = 0 \Rightarrow n = -1\)

Clearly \(H_1 = \frac{ab(n + 1)}{a + nb}\) and \(H_n = \frac{ab(n + 1)}{na + b}\)

Now it is trivial to substitute and solve for equality.

Let \(A\) be the A. M. and \(G\) be the G. M between \(a\) and \(b\) then \(A = \frac{a + b}{2}\) and \(G = \sqrt{ab}\).

Now, \(A + \sqrt{A^2 - G^2} = \frac{a + b}{2} + \sqrt{\frac{(a + b)^2}{4} - ab} = a\)

and \(A - \sqrt{A^2 - G^2} = \frac{a + b}{2} - \sqrt{\frac{(a + b)^2}{4} - ab} = b\)

Following from previous question \(\frac{a + b}{2} : \sqrt{ab} = m : n\).

Thus, \(a : b = m + \sqrt{m^2 - n^2} : m - \sqrt{m^2 - n^2}\)

Let \(a\) and \(b\) be two numbers. \(G = \sqrt{ab}\). Let \(d\) be the common difference. Then, \(d = \frac{b - a}{3}\).

Thus, we have \(p = a + d = \frac{b - 2a}{3}\) and \(q = \frac{2b - a}{3}\)

Now, \((2p - q)(2q - p) = \frac{(2b - 4a - 2b - a)}{3}.\frac{(4b - 2a - b - 2a)}{3}\)

\(= ab = G^2\)

We have, \(A = \frac{(a + b)}{2}\). Let \(r\) be the common ratio, then \(r = \sqrt[3]{\left(\frac{b}{a}\right)}\)

\(p = a\sqrt[3]{\left(\frac{b}{a}\right)}\) and \(q = a\left(\frac{b}{a}\right)^{\frac{2}{3}}\)

Now, \(\frac{p^2}{q} + \frac{q^2}{p} = a + b = 2A\)

Let \(A\) be the A. M. then \(A = \frac{a + b}{2}\) and \(H\) be the H. M. then \(H = \frac{2ab}{a + b}\)

Given, \(A = mH \Rightarrow \frac{a + b}{2} = \frac{2abm}{a + b}\)

\(\frac{(a + b)^2}{4ab} = m\)

\(\sqrt{m} = \frac{a + b}{2\sqrt{ab}}\) and \(\sqrt{m - 1} = \frac{a - b}{2\sqrt{ab}}\)

Clearly, \(\sqrt{m} + \sqrt{m - 1}: \sqrt{m} - \sqrt{m - 1} = a : b\)

Let \(d\) and \(h\) be common differences for the A. P. and H. P. respectively.

Let \(a_1, a_2, ..., a_9\) be the arithmetic means and \(h_1, h_2, ..., h_9\) be the harmonic means between \(2\) and \(3\).

\(d = \frac{1}{10}\) and \(h = \frac{\frac{1}{3} - \frac{1}{2}}{10} = -\frac{1}{60}\)

\(A = 2 + rd = 2 + \frac{r}{10}\) and \(\frac{1}{H} = \frac{1}{2} - \frac{r}{60}\) where \(A\) and \(H\) are \(r\) th mean for A. P. and H. P. respectively.

Now it is trivial to show that \(A + \frac{6}{H} = 5\)

\(a = \frac{b + c}{2}\) and \(b = \sqrt{ac} \Rightarrow c = \frac{b^2}{a}\)

Substituting for \(c\) in A. M. we have

\(a = \frac{b(a + b)}{2a}\)

Substituting this value of \(a\) for H. M. between \(a\) and \(b\)

\(\frac{2ab}{a + b} = \frac{2b(a + b)b}{2a(a + b)} = \frac{b^2}{a} = c\) from the G. M.

Clearly, \(a_1 = \frac{2x - y}{3}\) and \(a_2 = \frac{x - 2y}{3}\)

\(g_1 = x\left(\frac{y}{x}\right)^{\frac{1}{3}}\) and \(g_1 = x\left(\frac{y}{x}\right)^{\frac{2}{3}}\)

\(h_1 = \frac{3xy}{x - 2y}\) and \(h_2 = \frac{3xy}{2x - y}\)

Substituting these we get, \(a_1h_2 = a_2h_1 = g_1g_2 = xy\)

Let the two quantities be \(a\) and \(b\). If \(2n - 1\) arithmetic means are inserted and if \(d\) is the common difference then \(d = \frac{b - a}{2n}\)

For \(2n - 1\) geometric means if \(r\) be the common ratio then \(r = \left(\frac{b}{a}\right)^{\frac{1}{2n}}\)

For harmonic means the common difference would be \(= \frac{a - b}{2nab}\)

\(n\) th arithmetic mean \(= a + \frac{b - a}{2n}.n = \frac{a + b}{2}\)

\(n\) th geometric mean \(= \sqrt{ab}\)

\(n\) th harmonic mean \(= \frac{2ab}{b + a}\)

Clearly, these three are in G. P.

This problem is same as 14 and has been left as an exercise.

Let \(a\) and \(b\) be the two numbers. Then, \(4 = \frac{2ab}{a + b}\)

\(A = \frac{a + b}{2}\) and \(G = \sqrt{ab}\)

\(A.H = G^2 \therefore G^2 = 4A\) Solving we get numbers as \(6\) and \(3\)

Following 16 both can be proved easily.

If \(a, b, c\) are in A. P. then \(b = \frac{a + c}{2}\)

\(b^2 = \frac{(a + c)^2}{4} > ac\) because \(\frac{(a + c^2)}{4} - ac > 0 \Leftrightarrow (a -c)^2 > 0\)

If \(a, b, c\) are in G. P. then \(b = \sqrt{ac} \Leftrightarrow b^2 = ac\)

If \(a, b, c\) are in H. P. then \(b = \frac{2ac}{a + c}\)

\(b^2 = \frac{4a^2c^2}{(a + c)^2} < ac\)

because \((a + c)^2 - 4ac > 0\)