As averages is a very crucial topic for the CAT exam, here is a briefing of the classification of AVerages. Averages can be broadly classified into the following 3 types of Mean: – Arithmetic Mean, Geometric Mean and Harmonic Mean :

**A) Arithmetic Mean (AM)**:

Arithmetic mean of n numbers a1,a2,……an is given by (a_1+a_2+a_3+?…a_n)/n

It is nothing but the normal average.

If a,b,c are in AP , then Arithmetic Mean, b = (a+c)/2

In an AP the middle term will be the arithmetic mean of the previous term and the next term

If you have three numbers in AP the middle term will be the arithmetic mean of the first term and the last term.

And if you have 5 numbers in AP then the middle term i.e. the third term will be the arithmetic mean.

Also if you have an AP then the sum of the first term and last term divided by 2 will give the arithmetic mean of that AP which will also be the middle term.

**Point to Note**: This is why while calculating the sum of ‘n’ terms we take the sum of first term and last term divided by 2 to get the average and then multiply by the number of terms to get the sum.

Inserting n Arithmetic Means (AMs) between two numbers A and B Before explaining the concept of inserting AMs try to solve this question.

**Example: **

**Find the sum of 500 arithmetic means between 2 and 3?**

**Solution:**

When you insert “n” AMs between A and B, we will get an AP with n+2 terms, and the common difference is given by (B-A)/(n+1)

Taking a similar question-

For eg: Inserting 4 AMs between 10 and 20 implies that we will get an AP with 6 terms.

The AP can be formed by calculating the common difference as (20-10)/5= 2.

Thus the 4 Arithmetic Means will be 12,14,16 and 18.

If you observe, AM ( average) of 12,14,16 and 18 will be the AM of 10 and 20 = 15.

i.e (12+14+16+18)/4=(10 +20)/2= 15

So if the question is to find the sum of 4 AMs between 10 and 20 , answer is easy, it will be nothing but

= 4 x (average of the 4AMs between 10 and 20)

= 4 x [(12+14+16+18)/4]

= 4 x 15 = 60.

Extending the same concept to a larger number of terms, say “N” AMs, we can find out

The Sum of n AM’s between a and b = n×(a+b)/2

a= first term, b = last term ? (a + b)/2 = average of terms. So for ‘n’ terms inserted in between ‘a’ and ‘b’ sum = (n×(a + b))/2

Thus the solution for question, Find the sum of 500 arithmetic means between 2 and 3 is (500×(2+3))/2 = 500 × 2.5 = 1250.

**Example:**

** The first and the last term of an AP are 107 and 253. If there are five terms in this sequence, find the sum of the sequence.**

**Solution:**

First term = 107 last term = 253

Average or mean = (107 + 253)/2 = 180 Number of terms in the sequence = 5.

So, sum of sequence = 180 * 5 = 900.

**B) Geometric Mean (GM):**

Geometric mean of n numbers a1, a2,…..an is given by (a_1×a_2×a_3×……a_n )^{(1/n)}

e.g. geometric mean four numbers a,b,c,d = (abcd)^{(1/4)}

e.g. geometric mean of 2, 4, 8 = (2 ×4 × 8)^{(1/3)} = (64)1/3 = 4

If a,b,c are in GP, geometric mean, b = vac

i.e In a GP if you take three consecutive terms middle term will be geometric mean of first term and the third term. And if you take a GP with 5 terms, them the middle term or 3rd term will be the GM of the series.

**Application of GM :**

The Concept is GM is used in finding out the compounded annual growth rate (CAGR)

CAGR = [((final value)/(initial value))^{(1/n)}-1]×100 where n is the number of years.

**Example:**

**The population of a village is 10000 in the year 2000 and is 14400 in 2002. Find Compounded Annual Growth Rate?**

**Solution:**

CAGR = [(14400/10000)^{(1/2)} -1]× 100 = 20 %

This means an annual increase of 20 %.

To verify 10000 × 1.2 = 12000 12000× 1.2 = 14400

A factor of multiplication 1.2 is used to increase a number by 20% and to get the final value,

One needs multiply by 120/100= 1.2 CAGR is similar to rate percent, R in compound Interest calculations.

Final amount = P(1+R/100)^{n} where P = initial amount, R = rate percent, n = number of years.

So, CAGR or R = [((final amount)/(initial amount))^{(1/n)}– 1]× 100

CAGR implies the annual increase in growth.

**Inserting “n” GM’s between A and B:**

When “n” GMs is inserted between A and B, we will get a GP with n+2 terms, the Common Ratio is given by

r = (b/a)^{(1/[n+1] )}.

GM1 = A.r

GM2 = A r2

…………

For example 2 GMs between 1 and 27 will be 3,9. We can find out the GMs by finding the common ratio = (27/1)^{(1/3)} = 3

We will get a GP with 4 terms 1,3,9,27

3 GMs between 1 and 16 will be 2,4 & 8, so we will get a GP with 5 terms 1,2,4,8,16.

**ILLUSTRATIONS:**

**1) Insert 8 geometric means between 3 and 1536.Find the 6th GM ?**

**Solution:**

Common ratio = (1536/3)^{(1/9)}= 2

GM1 = 3*2 = 6 GM2 = 3*22 =12 GM3 = 3*23 = 24…… GM6 = 3*26 = 192

Required answer = 192.

**2) Find the number of GMs in a series where the ratio between first and last terms is 16:1 and common ratio of GP=2**

**Solution:**

Common ratio=(b/a)^{(1/((n+1)))}

2=16^{(1/x)} 2x=16=>x=4 x=n+1=>n=4-1=3 Number of Geometric Means=3.

**C) Harmonic Mean (HM):**

Let a and b be two given quantities. It is required to insert n harmonic means h1, h2, h3,….hn between the quantities a and b

Harmonic mean of n numbers a_1,a_2,……an is given by n/[1/a_1 +1/a_2 +1/a_3 +?.1/a_n ]

If a,b,c are in Harmonic Progression, then HM, b =2/[1/a+1/c] =2ac/( a + c )

Harmonic mean of 4 numbers a,b,c,d = 4/(1/a+1/b+1/c+1/d)

i.e In an HP if you take three consecutive terms , middle term will be the harmonic mean of the first term and the third term.

**Application of harmonic mean:-**

Harmonic mean helps us find out the average speed of a journey when the distances covered are equal

Average speed when the distances covered are equal = HM of the speeds

Let’s say a person P, covers a distance A to B at a speed of “a” kmph and a distance B to A at a speed of “b” kmph.

As the distance is constant, the average speed can be found using the following formula Average speed=2ab/(a+b) kmph

When the two quantities are in direct proportion, take the Arithmetic Mean

When the two quantities are in inverse proportion, take the Harmonic Mean.

**ILLUSTRATION:**

**1) A person travels from A to B with a speed of 60 km/hr and returns from B to A at 40 km/hr. ****What is the average speed for the whole journey? **

**Solution:**

Here since the distances covered are equal, average speed for the whole journey = Harmonic mean of the speeds

= HM of 60,40 = 2/[(1/60)+ (1/40) ]

= (2×60×40)/(60+40)= 48 km/hr.

**2) Amit covers first 1 ^{rd}/3 of the distance walking at 2 km/hr, second 1^{rd}/3 of the Distance running at 3km/hr and the rest cycling at 6 km/hr. **

**Find the average speed for the whole journey?**

** a) 3 km/hr **

**b) 3.66 km/hr **

**c) 4 km/hr**

**d) 5 km/hr.**

**Solution: **

distances covered are equal for all 3 cases, each time 1/3rd of the total distance.

So average speed = Harmonic mean of 2, 3 and 6

= 3/[(1/2)+ (1/3)+ (1/6) ] = 3 km/hr.

**Median:-**

#### Median is the middle value of a group of numbers arranged in an ascending or descending order.

a) If the number of values (n) is odd, then median will be the ((n+1))/2 th value

Take for example,

**3) Find the median of the following numbers 51, 50, 49, 58, 48, 43 and 51**

**Solution:**

Arrange the group in the ascending order 43, 48, 49, 50, 51, 51, 58. Median will be the ((7+1))/2 th value = 4th value = 50

If the number of values (n) of a given set of data is even, then the median will be the mid value of the two middle numbers.

Take for example,

**4) Find the median of 54, 33, 23, 26, 25, 24**

**Solution:**

On arranging the values in ascending order, 23, 24, 25, 26, 33, 54. the two middle values are 25 and 26.

Median = (25+26)/2 = 52/2 = 25.5

Mode:- Mode is the number that occurs most frequently in a given set of numbers.

Take for example,

**5) Find the mode of 1,1,1,1,3,4,5,6,2,2,3,3,1,1,6,6,7,7,7,8,8,8,8**

**Solution:**

The mode will be 1 as it occurs the maximum number of times.

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