# Geometric Mean IIT JEE Study Material

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) = $\mathbf{\;\sqrt{a\;c}}$ [Where, a and c > 0]

The Geometric Mean of as sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, . . . . .$ up to n terms can be defined as:

$\mathbf{GM\;=\;\left ( \;a_{1}\;\times \;a_{2}\;\times\; a_{3}\;\times \;a_{4}\;\times \;a_{n} \;\right )^{\frac{1}{n}}}$

If ‘n’ number of Geometric Means $G_{1}, G_{2}, G_{3}, G_{4}, . . . , G_{n}$ are inserted between two numbers a and b such that the resulting sequence a, G1, G2, G3, G4, . . . , Gn, b forms the geometric progression, then,

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

i.e. $b = a r ^{[ (n + 2) – 1 ]} = a r ^{(n + 1)}$

Or, $\mathbf{\frac{b}{a}\;=\;r^{n\;+\;1}}$

Or, $\mathbf{r\;=\;\left ( \;\frac{b}{a} \;\right )^{\frac{1}{n\;+\;1}}}$

Therefore, $G_{1}$ = $\mathbf{\;a\;\left ( \;\frac{b}{a}\; \right )^{\frac{1}{n\;+\;1}}}$

Similarly, $G_{2}$ = $\mathbf{\;a\;\left ( \;\frac{b}{a}\; \right )^{\frac{2}{n\;+\;1}}}$

Therefore, $G_{n}$ = $\mathbf{\;a\;\left ( \;\frac{b}{a}\; \right )^{\frac{n}{n\;+\;1}}}$

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.

i.e. $\mathbf{\sum_{r\;=\;1}^{n}\;\;G_{r}\;=\;G^{n}}$ [where G = single geometric mean between a and b]

Or, $G_{1} \times G_{2} \times G_{3} \times G_{4} \times . . . . . . . . . \times G_{n} = G_{n}$

## Relationship between Arithmetic Mean and Geometric Mean:

1: Arithmetic Mean ≥ Geometric Mean

AM = A = $\mathbf{\frac{a\;+\;b}{2}}$ . . . . . . . (1)

And, GM = G = $\mathbf{\sqrt{a\;b}}$ . . . . . . . . . . (2)

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

A – G = $\mathbf{\frac{a\;+\;b}{2}\;-\;\sqrt{ab}}$

Or, $\mathbf{\frac{a\;+\;b\;-\;2\;\sqrt{ab}}{2}\;=\;\frac{\left ( \;\sqrt{a}\;-\;\sqrt{b}\; \right )^{2}}{2}\;\geq\; 0}$

i.e. A – G ≥ 0

This implies that A ≥ G.

2: The quadratic equation whose roots are a and b can be written as $x^{2} – 2Ax + G^{2} = 0$, [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, $x^{2} – (a+b)x + ab = 0$

Since, a + b = 2A and ab = $G^{2}$

Therefore, the above equation becomes $x^{2} – 2A x + G_{2} = 0$

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

$\mathbf{a\;=\;A\;+\;\sqrt{A^{2}\;-\;G^{2}}\;\;and\;\;b\;=\;A\;-\;\sqrt{A^{2}\;-\;G^{2}}}$

Since, $x^{2} – 2A x + G_{2} = 0$,

Therefore, by using quadratic formula we get

(a, b) = $\mathbf{\frac{2\;A\;\pm \;\sqrt{4\;A^{2}\;-\;4\;G^{2}}}{2}}$

Or, (a, b) = $\mathbf{A\;\pm \;\sqrt{A^{2}\;-\;G^{2}}}$

## 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, $\mathbf{r\;=\;\left ( \frac{b}{a} \right )^{\frac{1}{n\;+\;1}}\;=\;\left ( \frac{162}{2} \right )^{\frac{1}{4}}\;=\;\left ( 81 \right )^{\frac{1}{4}}\;=\;3}$

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

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

Solution:

Given a = 4 and b = 36

Therefore, the Geometric Mean = $\mathbf{\sqrt{ab}\;=\;\sqrt{4\times 36}\;=\;12}$

Note: the sequence 4, 12, 36 also forms a geometric progression.

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, $\mathbf{\left ( \sqrt{a} \right )^{2}\;-\;\left ( \sqrt{b} \right )^{2}\;=\;8}$

Or, $\mathbf{\left ( \sqrt{a}\;-\;\sqrt{b} \right )\;\;\left ( \sqrt{a}\;+\;\sqrt{b} \right )\;=\;8}$ . . . . . . (1)

And, AM – GM = 1 (given)

i.e. $\mathbf{\frac{a\;+\;b}{2}\;-\;\sqrt{ab}\;=\;1}$

Or, $\mathbf{a\;+\;b\;-\;2\;\sqrt{ab}\;=\;2}$

Or, $\mathbf{\left ( \;\sqrt{a}\;-\;\sqrt{b}\; \right )^{2}\;=\;2}$

Or, $\mathbf{\;\sqrt{a}\;-\;\sqrt{b}\;=\;\pm \;2}$ . . . . . . . . (2)

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

$\mathbf{\left ( \pm\; 2 \right ) \;\;\left ( \sqrt{a}\;+\;\sqrt{b} \right )\;=\;8}$

$\mathbf{\sqrt{a}\;+\;\sqrt{b}\;=\;\pm \;4}$ . . . . . . . . . . (3)

Now on squaring both LHS and RHS of equation (2) and (3) we get

$\mathbf{a\;+\;b\;-\;2\;\sqrt{ab}\;=\;4}$ . . . . . . . . . . . (4)

And, $\mathbf{a\;+\;b\;+\;2\;\sqrt{ab}\;=\;16}$ . . . . . . . . . . (5)

On Solving equation (4) and equation (5), we get

ab = 9 . . . . . . (6)

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

On substituting the values of equation (7) in equation (6), we get

$a^{2} – 10a + 9 = 0$

$\mathbf{a\;=\;\frac{10\;\pm \;\sqrt{100\;-\;36}}{2}}$

Or, $\mathbf{a\;=\;\frac{10\;\pm \;8}{2}\;=\;5\;\pm \;4}$

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 respectively then:

$\mathbf{a\;=\;A\;+\;\sqrt{A^{2}\;-\;G^{2}}\;\;and\;\;b\;=\;A\;-\;\sqrt{A^{2}\;-\;G^{2}}}$

Therefore,

$\mathbf{a\;=\;13\;+\;\sqrt{13^{2}\;-\;12^{2}}\;\;and\;\;b\;=\;13\;-\;\sqrt{13^{2}\;-\;12^{2}}}$

Or, $\mathbf{a\;=\;13\;+\;\sqrt{169\;-\;144}\;\;and\;\;b\;=\;13\;-\;\sqrt{169\;-\;144}}$

Therefore, a = 18, and b = 8