If the terms a, b, and c are in geometric progression (GP), then the middle term (b) is called the geometric mean (GM) of the other two terms (a and c). Therefore, the geometric mean between two terms, a and c, is given by

GM (b) = √(ac) (Where, a and c > 0)

The geometric mean of as sequence a₁, a₂, a₃, a₄, a₅,…. up to n terms can be defined as

\begin{array}{l} GM = {(a_{1} \times a_{2} \times a_{3} \times a_{4} \times a_{n})}^{\frac{1}{n}} \end{array}

If ‘n’ number of geometric means G₁, G₂, G₃, G₄, . . . , G_n are inserted between two numbers a and b, such that the resulting sequence a, G₁, G₂, G₃, G₄, . . ., G_n, b forms the geometric progression, then,

The first term = a and the (n + 2)th term = b

$\begin{array}{l} i . e . b = a r^{[(n + 2) - 1]} = a r^{(n + 1)} \end{array}$

Or,

\begin{array}{l} \frac{b}{a} = r^{n + 1} \end{array}

Or,

\begin{array}{l} r = {(\frac{b}{a})}^{\frac{1}{n + 1}} \end{array}

Therefore,

\begin{array}{l} G_{1} = a {(\frac{b}{a})}^{\frac{1}{n + 1}} \end{array}

Similarly,

\begin{array}{l} G_{2} = a {(\frac{b}{a})}^{\frac{2}{n + 1}} \end{array}

Therefore,

\begin{array}{l} G_{n} = a {(\frac{b}{a})}^{\frac{n}{n + 1}} \end{array}

Also, the product of n geometric means between a and b is equal to the nth power of the single geometric mean between a and b.

\begin{array}{l} i . e . \sum_{r = 1}^{n} G_{r} = G^{n} \end{array}

[where G = single geometric mean between a and b]

Or,

$\begin{array}{l} G_{1} \times G_{2} \times G_{3} \times G_{4} \times . . . . . . . . . \times G_{n} = G^{n} \end{array}$

Relationship between Arithmetic Mean and Geometric Mean

1: Arithmetic Mean ≥ Geometric Mean

AM = A = (a + b)/2 . . . . . . . (1)

And, GM = G = √(ab) . . . . . . . . . . (2)

Now, Equation (1) – Equation (2), we get

\begin{array}{l} A - G = \frac{a + b}{2} - \sqrt{ab} \end{array}

Or,

\begin{array}{l} \frac{a + b - 2 \sqrt{ab}}{2} = \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{2} \geq 0 \end{array}

i.e. A – G ≥ 0

This implies that A ≥ G.

2: The quadratic equation whose roots are a and b can be written as x² – 2Ax + G² = 0. (Where A and G are arithmetic and geometric mean between a and b, respectively.)

If a and b are the roots of the quadratic equation, then x² – (a + b)x + ab = 0

Since, a + b = 2A and ab = G²

Therefore, the above equation becomes x² – 2Ax + G² = 0

3: If A and G are arithmetic and geometric mean between two numbers a and b, respectively, then

$\begin{array}{l} a = A + \sqrt{A^{2} - G^{2}} and b = A - \sqrt{A^{2} - G^{2}} \end{array}$

Since, x² – 2Ax + G² = 0,

Therefore, by using the quadratic formula, we get

\begin{array}{l} (a, b) = \frac{2 A \pm \sqrt{4 A^{2} - 4 G^{2}}}{2} \end{array}

Or,

\begin{array}{l} (a, b) = A \pm \sqrt{A^{2} - G^{2}} \end{array}

Geometric Mean IIT JEE Problems

Example 1: Insert three geometric means between 2 and 162.

Solution:

Given a = 2, b = 162, and n = 3

Therefore,

\begin{array}{l} r = {(\frac{b}{a})}^{\frac{1}{n + 1}} = {(\frac{162}{2})}^{\frac{1}{4}} = {(81)}^{\frac{1}{4}} = 3 \end{array}

Hence, the 3 geometric means between 2 and 162 are 6, 18, and 54.

Example 2: Find the geometric mean of 4 and 36.

Solution:

Given a = 4 and b = 36

Therefore, the geometric mean =

\begin{array}{l} \sqrt{ab} = \sqrt{4 \times 36} = 12 \end{array}

Example 3: The arithmetic mean of two positive numbers, a and b, is 1 more than its geometric mean. Find the value of a and b, if their difference is 8.

Solution:

According to the given condition,

a – b = 8

Or,

\begin{array}{l} {(\sqrt{a})}^{2} - {(\sqrt{b})}^{2} = 8 \end{array}

Or,

\begin{array}{l} (\sqrt{a} - \sqrt{b}) (\sqrt{a} + \sqrt{b}) = 8 \dots . . (1) \end{array}

And, AM – GM = 1 (given)

i.e.

\begin{array}{l} \frac{a + b}{2} - \sqrt{ab} = 1 \end{array}

Or,

\begin{array}{l} a + b - 2 \sqrt{ab} = 2 \end{array}

Or,

\begin{array}{l} {(\sqrt{a} - \sqrt{b})}^{2} = 2 \end{array}

Or,

\begin{array}{l} \sqrt{a} - \sqrt{b} = \pm 2 \dots . (2) \end{array}

Now, on substituting the values of equation (2) to equation (1), we get

\begin{array}{l} (\pm 2) (\sqrt{a} + \sqrt{b}) = 8 \end{array}

\begin{array}{l} \sqrt{a} + \sqrt{b} = \pm 4 \dots . (3) \end{array}

Now, on squaring both the LHS and RHS of equations (2) and (3), we get

\begin{array}{l} a + b - 2 \sqrt{ab} = 4 \dots . . (4) \end{array}

And,

\begin{array}{l} a + b + 2 \sqrt{ab} = 16 \dots . . (5) \end{array}

On solving the equation (4) and equation (5), we get

ab = 9

so b = 9/a……(6)

And a + b = 10 . . . . . (7)

On substituting the values of b in equation (7), we get

a + 9/a = 10

\begin{array}{l} a^{2} - 10 a + 9 = 0 \end{array}

Now, using the quadratic formula

\begin{array}{l} a = \frac{10 \pm \sqrt{100 - 36}}{2} \end{array}

Or,

\begin{array}{l} a = \frac{10 \pm 8}{2} = 5 \pm 4 \end{array}

= 9 or 1

Hence, when a = 9, b = 1 and when a = 1, b = 9

Therefore, the required numbers are 9 and 1.

Example 4: If the arithmetic and geometric mean of two positive real numbers a and b are 13 and 12, respectively. Find the value of a and b.

Solution:

Given: A = 13 and G = 12

Note: If A and G are arithmetic and geometric mean between two numbers a and b, then

\begin{array}{l} a = A + \sqrt{A^{2} - G^{2}} and b = A - \sqrt{A^{2} - G^{2}} \end{array}

Therefore,

\begin{array}{l} a = 13 + \sqrt{13^{2} - 12^{2}} and b = 13 - \sqrt{13^{2} - 12^{2}} \end{array}

Or,

\begin{array}{l} a = 13 + \sqrt{169 - 144} and b = 13 - \sqrt{169 - 144} \end{array}

Therefore, a = 18, and b = 8

Relationship between Arithmetic Mean and Geometric Mean

Geometric Mean IIT JEE Problems

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