 # Arithmetic Mean IIT JEE Study Material

If p, q, and r are in arithmetic progression (AP) then, the middle term (q) is called the arithmetic mean (AM) of the other two terms (p and r). Therefore, the Arithmetic Mean between two numbers p and r is given by:

AM (q) = $\mathbf{\frac{p\;+\;r}{2}}$

If p and q be any two given numbers such that $p, K_{1}, K_{2}, K_{3}, K_{4}, . . . . . , K_{n}, q$ are in Arithmetic progression then $K_{1}, K_{2}, K_{3}, K_{4}, . . . . . , K_{n}$ are ‘n’ arithmetic means between p and q such that:

The first term = p and the (n + 2)th term = q

i.e. q = p + [ (n + 2) – 1] d = p + (n +1)d

Or, d = $\mathbf{\frac{q\;-\;p}{n\;+\;1}}$

Therefore, $\mathbf{K_{1}\;=\;p\;+\;\frac{q\;-\;p}{n\;+\;1}}$

Similarly, $\mathbf{K_{2}\;=\;p\;+\;\frac{2\;(q\;-\;p)}{n\;+\;1}}$

Therefore, $\mathbf{K_{n}=p+\frac{n\;(q\;-\;p)}{n\;+\;1}}$

Also, the sum of ‘n’ arithmetic means inserted between p and q is equal to the ‘n’ times the single arithmetic mean between p and q.

i.e. $\mathbf{\sum_{r\;=\;1}^{n}\;K_{r}\;=\;n\;K}$ [where K = arithmetic mean between p and q]

## Arithmetic Mean of mth Power

Let $p_{1}, p_{2}, p_{3}, p_{4}, . . . . . . . . , p_{n}$ be n positive real numbers,

Case 1: $\mathbf{m\;\in \;(0,\;1)}$

$\mathbf{\frac{p_{1}^{m}\;+\;p_{2}^{m}\;+\;p_{3}^{m}\;+\;.\;.\;.\;+\;p_{1}^{m}}{n}\;<\;\left ( \frac{p_{1}\;+\;p_{2}\;+\;p_{3}\;+\;.\;.\;.\;+\;p_{n}}{n} \right )^{m}}$

Case 2: m $\mathbf{in}$ {0, 1}

$\mathbf{\frac{p_{1}^{m}\;+\;p_{2}^{m}\;+\;p_{3}^{m}\;+\;.\;.\;.\;+\;p_{1}^{m}}{n}\;=\;\left ( \frac{p_{1}\;+\;p_{2}\;+\;p_{3}\;+\;.\;.\;.\;+\;p_{n}}{n} \right )^{m}}$

Case 3: $\mathbf{m\;in \;[0,\;1]}$

$\mathbf{\frac{p_{1}^{m}\;+\;p_{2}^{m}\;+\;p_{3}^{m}\;+\;.\;.\;.\;+\;p_{1}^{m}}{n}\;>\;\left ( \frac{p_{1}\;+\;p_{2}\;+\;p_{3}\;+\;.\;.\;.\;+\;p_{n}}{n} \right )^{m}}$

## Relationship between Arithmetic, Geometric, and Harmonic Mean:

1. If A, H, & G be the arithmetic, harmonic and geometric mean of numbers p and q. Then, $G^{2} = AH$

2. If A and G are AM and GM between two numbers a and b then the quadratic equation whose roots are a and b can be written as $x^{2} – 2Ax + G_{2} = 0$

3. In terms of inequalities: Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean.

Arithmetic Mean IIT JEE Problems:

Example 1: If a, b, and c are in arithmetic progression then find the value of (a + 2b – c) (2b + c – a) (c + a – b).

Solution:

Since, a, b, and c are in arithmetic progression therefore, $\mathbf{b\;=\;\frac{a\;+\;c}{2}}$

From the above equation, 2b – c = a, 2b – a = c, and a + c = 2b

Therefore, (a + 2b – c) (2b + c – a) (c + a – b) = (2a) (2c) (b) = 4abc.

Example 2: An even number of arithmetic means are inserted between 2 numbers p and q whose sum is $\mathbf{\frac{13}{6}}$. Find the total number of arithmetic means if, the sum of these arithmetic means exceeds their number by 1.

Solution:

Let the total number of arithmetic means inserted between two number p and q be 2n.

Since, the sum of n AMs inserted between two numbers is equal to the n times the single AM between those two numbers and the sum of Arithmetic Means = 2n + 1 (Given)

Therefore, $\mathbf{2n\;\left [\; \frac{p\;+\;q}{2}\; \right ]\;=\;2n\;+\;1}$

i.e. $\mathbf{n\;\left [\; \frac{13}{6}\; \right ]\;=\;2n\;+\;1}$

i.e. n = 6

Therefore, there are total 6 Arithmetic Means between the two numbers p and q.

Example 3: Find the value of ‘n’, if a total of n arithmetic means are inserted between 20 and 80 such that the ratio if 1st mean to the last mean is 1 : 3.

Solution:

The given sequence can be written as $20, K_{1}, K_{2}, K_{3}, K_{4}, . . . . , a_{n}, 80$.

Here, 1st term (a) = 20 and (n + 2)th term = 80

i.e. 80 = 20 + [ (n + 2) – 1] d

Or, d = $\mathbf{\frac{60}{n\;+\;1}}$

Now, the first term of AM is given by,

a + d = 20 + $\mathbf{\frac{60}{n\;+\;1}}$ = $\mathbf{\frac{20n\;+\;80}{n\;+\;1}}$

And, the last term of AM is given by,

a + nd = $\mathbf{\frac{60\; \times \;n}{n\;+\;1}}$ = $\mathbf{\frac{80\;n\;+\;20}{n\;+\;1}}$

According to the given condition:

$\mathbf{\frac{20n\;+\;80}{n\;+\;1}\;:\;\frac{80\;n\;+\;20}{n\;+\;1}\;=\;1\;:\;3}$

Or, $\mathbf{\frac{20n\;+\;80}{n\;+\;1}\; \times \;\frac{n\;+\;1}{80\;n\;+\;20}\;=\;\frac{1}{3}}$

Or, $\mathbf{\frac{n\;+\;4}{4n\;+\;1}\;=\;\frac{1}{3}}$

Therefore, n = 11