Arithmetic Mean and Special Series

# Arithmetic Mean and Special Series | Mathematics (Maths) Class 11 - Commerce PDF Download

Arithmetic Mean

The sum of all of the numbers in a list divided by the number of terms in that list gives the arithmetic mean of that list. Let us first understand the concept of arithmetic progression.
In the arithmetic progression, we know that if the three numbers are in AP, that means if a, b and c are in AP, then basically the first two terms a and b will have the difference which will be equal to the next two terms b and c.

So we can say, b – a = c – b. Rearranging the terms,
2b = a + c
b =   a+c/2
So we can say that this term b is the average of the other two terms a and c. This average in arithmetic progression is called the arithmetic mean.

Arithmetic mean = A = S/N
where A =  arithmetic mean, N = the number of terms and S = the sum of the numbers in the list.

Multiple Arithmetics Means between Two Given Numbers

Let a and b be the two given numbers and A1, A2……An be the arithmetic means between them. Then a, A1, A2….An, b will be in AP.  If a is the first term, then b will be the (n+2)th term.

Hence, b = a + (n + 2 – 1 )d

Geometric Mean

Geometric mean b of two terms a and c is given by √(ac). If a, b and c are in geometric progression, then the ratio of the two consecutive terms should be equal.

This means that b is the geometric mean of a and c
Multiple Geometric Means between Two Given Numbers
Let a and b be the two given numbers. Let, G1, G2, G3….Gn  be n geometric mean between them.

Example
Question: Insert 3 numbers between 1 and 256 so that the resulting sequence is GP
Solution:
So we are supposed to insert three numbers, say
1   G1   G2   G3   256

Let’s insert these numbers in such a way that all these numbers are in G.P. We know that the general form of a GP is :a1,  a1r,  a1r², a1r³,  a1rn-1,  arn … So, let us first write a = 1,  Gas ar, G2 as ar², and G3 = ar³. So,
a = 1 and ar= 256
or, r= (4)4
or, r = 4, -4
GP: 1  ar  ar²  ar³  256

Case 1: r = 4
ar = 1 × 4
ar² =  1 × 4²
ar³ = 1 × 4³
GP: 1, 4,  16,  64, 256
Hence, 4, 16 and 64 are the three terms which we can add between 1 and 256.

Case 2: r = -4
ar = 1 × -4
ar² =  1 × -4²
ar³ = 1 ×-4³

GP: 1, -4,  16,  -64, 256
Hence, -4, 16 and -64 are the three terms which we can add between 1 and 256.

Relationship between A.M and G.M
For terms a and b, let A be A.M and B be G.M, then A.M >= G.M

Let us take a = 5 and b = 5. In this case,

Here we see that A.M > G.M
So either A.M = G.M or A.M > G.M

Solved Examples

Question: Find the arithmetic mean of the series 1, 3, 5,……(2n-1)
A. n
B. 2n
C. n/2
D. n-1
Solution: A is the correct option. First term, a =1 and the common difference, d = 3 – 1 = 2. Let 2n – 1 b ethe kth term. Then, from the general term formula,
ak = a + (k-1)d
2n-1 = 1 + (k-1) 2
k-1 = n-1
or k =n
Now, Sk = [(k/2)*(2a+ (k-1)d]
And A.M = Sk/k = [2a + (k-1)d]/2

A.M = (2+(n-1)2)/2 = n

What is a Series?
We can define the series as the sum of all the numbers of the given sequence. The sequences are finite as well as infinite. In the same way, the series can also be I finite or infinite. For example, consider a sequence as 1, 3, 5, 7, … Then the series of these terms will be 1 + 3 + 5 + 7 + …

What are the Special Series?
The series special in some way or the other is called a special series. The following are the three types of special series.
• 1 + 2 + 3 +… + n (sum of first n natural numbers)

• 12 + 22 + 32 +… + n2(sum of squares of the first n natural numbers)

• 13 + 23 + 33 +… + n3(sum of cubes of the first n natural numbers)

Let’s now look at the sum to n terms of special series:
Sum of First n Natural Numbers (1 + 2 + 3 +… + n)
The above series is an AP. Here a = 1, d =1 because the difference between these terms is 1 and also 1st term is 1.

So above is the formula of the seies.

Sum of Squares of the First n Natural Numbers (12 + 22 + 32 +… + n2)
This series is neither in GP nor in AP. So let’s first convert this in GP or AP. Here we use the formula,
k– (k-1)3
= k– (k-1)3 = 3k– 3k +1
Let us take the value of k as k = 1
13 – 03 = 3 (1)2 – 3 (1) + 1
Now take k =2
23 – 13 = 3 (2)2 – 3 (2) + 1
Now k = 3
33 – 23 = 3(3)2 – 3 (3) + 1

Similarly if we take k =n, we get,
n3 – (n – 1)3 = 3 (n)2 – 3 (n) + 1

Now add both the sides, the equation we get is,
n3 – 03 = 3 (12 + 22 + 32 + … + n2) – 3 (1 + 2 + 3 + … + n) + n

Sum of Cubes of the First n Natural Numbers (13 + 23 + 33 +… + n3)
Here we use the formula,
(k + 1)4 – k4 = 4k3 + 6k2 + 4k + 1

Let us take the value of k as k = 1

24 – 14 = 4(1)3 + 6(1)2 + 4(1) + 1

Now take the value of k = 2

34 – 24 = 4(2)3 + 6(2)² + 4(2) + 1

Again take k = 3

44 – 34 = 4(3)3 + 6(3)+ 4(3) + 1

Similarly if we take k =n
(n + 1)4 – n4 = 4n3 + 6n+ 4n + 1
Now adding both  the sides, we get

(n + 1)4 – 14 = 4(13 + 23 + 33 +…+ n3) + 6(12 + 22 + 32 + …+ n2) + 4(1 + 2 +
3 +…+ n) + n
(n + 1)4 – 14 = 4 S+ 6  + n
=  n4+ 4n³ + 6n² + 4n- n (2n² + 3n +1 )-2n(n+1)-n
= n4+ 2n³ + n²
= n² (n+1)²

Hence we get the formula for all the three special cases.

Solved Examples
Question:
2, 3, 6, n, 42… Find the value of n in the above series.
A. 9
B. 12
C. 15
D. 18
E. 21
Solution: C is the correct option. Given series is 2, 3, 6, n, 42… Here we can decode the series by multiplying the previous number by 3 and then by subtracting 3 from it. As we can see 2 being the first numbers of the series.
2 × 3 = 6
6 – 3 = 3  (3 is the second number)
3 × 3 -3 = 6 ( 6 is the third number)
n will be 6 × 3 – 3 = 18-3 = 15
Hence, n is 15

Question: Consider the following statements:
(1) The sum of cubes of first 2o  natural numbers is 44400
(2) The sum of squares of first 20 natural numbers is 2870
Which of the above statements is/are correct?
Only 1st
Only 2nd

Solution: B is the correct option. The sum of the cubes of first n consecutive natural
numbers is given by
Putting n = 20, we have sum = = 44100
The sum of the squares of first n consecutive natural numbers is given by

Putting n = 20, we have the sum = = 2870

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## FAQs on Arithmetic Mean and Special Series - Mathematics (Maths) Class 11 - Commerce

 1. What is the arithmetic mean and how is it calculated?
Ans. The arithmetic mean, also known as the average, is a measure of central tendency. It is calculated by adding up all the values in a set of numbers and dividing the sum by the total number of values.
 2. How is the arithmetic mean useful in analyzing data?
Ans. The arithmetic mean is useful in analyzing data as it provides a representative value that summarizes the entire dataset. It helps in comparing different sets of data, identifying trends, and making predictions based on the average value.
 3. What are the limitations of using the arithmetic mean?
Ans. One limitation of using the arithmetic mean is that it can be influenced by extreme values, also known as outliers. These outliers can significantly skew the average and may not accurately represent the central tendency of the data. Additionally, the arithmetic mean does not provide information about the distribution and variability of the data.
 4. How is the arithmetic mean different from the median and mode?
Ans. The arithmetic mean, median, and mode are all measures of central tendency, but they are calculated differently. The median is the middle value when the data is arranged in ascending or descending order, while the mode is the most frequently occurring value. The arithmetic mean is calculated by summing up all the values and dividing by the total number of values.
 5. Can the arithmetic mean be used with non-numeric data?
Ans. No, the arithmetic mean is typically used with numerical data only. It involves adding and dividing numbers, which cannot be done with non-numeric data such as categories or qualitative information. For non-numeric data, other measures like mode or percentage may be more appropriate.

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