A.M. and Useful Results in Arithmetic Progression: Here, the full form of A.M. means the arithmetic mean. The mean is the measure of central tendency, which is most commonly used for averaging. Arithmetic mean is generally signified as mean. The mean of the values of certain items is the average of the values of those items. 

They may be related to the average run taken by a cricket player in the last five years, or average profit and loss of a company, temperature of cities, etc. We can find the sum of the first and last terms of an A.P. series using the arithmetic mean, and eventually, we can find the sum of the A.P. series using A.M. Let us learn about A.M. and its usefulness in A.P.

Arithmetic Progression

Consider the following examples:

1. \(1, 3, 5, 7,9, 11, 13\) 
2. \(3, 6, 9, 12, 15, 18, 21\) 
3. \(10, 7, 4, 1, -2, -5\) 

These are sequences, and each of the numbers in each sequence is called a term.

In case (i), \(1\) is the first term, and each term is \(2\) more than its previous term.

In case (ii), \(3\) is the first term, and each term is \(3\) more than its previous term.

In case (iii), \(10\) is the first term, and each term is \(3\) less than its previous term.

Learn About Arithmetic Progression Here

We can observe that successive terms are found by adding or subtracting a fixed number to the preceding terms in all the above cases. Hence, such a sequence is called an arithmetic progression (A.P.).

Therefore, an arithmetic progression is a set of numbers in which each term is found by adding a fixed number to the preceding term, excluding the first term.

This fixed number is known as the common difference (CD) of the A.P.  The common difference can be positive, zero or negative.

Let us represent the first term of an A.P. by \(a_{1}\), the second term by \(a_{2}\), the third term by \(a_{3}\), ………, \(n^{\text {th }}\) term by \(a_{n}\) and the common difference by \(d\). Then A.P. becomes \(a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots, a_{n}\)

So, \(a_{2}-a_{1}=a_{3}-a_{2}=\ldots \ldots \ldots=a_{n}-a_{n-1}=d\)

Using the above cases, we can see \(a, a+d, a+2 d, a+3 d\)…… represents an arithmetic progression where \(a\) is the first term and \(d\) the common difference (CD). 

Last Term or nth of an A.P.

Let us think about an example: A sum of \(₹2000\) is spent at \(8\%\) simple interest per annum. Compute the interest at the end of each year. Do these interests form an A.P. series? If so, determine the interest at the end of \(20\) years applying this fact.

We know that simple interest \(=\frac{P \times R \times T}{100}\)

So, the interest at the end \(1^{s t}\) year \(= \frac{{2000 \times 8 \times 1}}{{100}} = ₹160\)

The interest at the end of \(2^{n d}\) year \(= \frac{{2000 \times 8 \times 2}}{{100}} =₹320\)

The interest at the end of \(3^{r d}\) year \( = \frac{{2000 \times 8 \times 3}}{{100}} =₹480\)

Now, looking at the arrangement formed above, we can find the interest deposited yearly for the \(10^{\text {th }}\) year.

Interest at the end of \(10^{\text {th }}\) year \(=\) Interest for the \(9^{t h}\) year \(+ ₹160\)

\(= ₹[160+(9-1)(160)]+ ₹160\)

\(= ₹[160+(8)(160)]+ ₹160\) 

\(=₹[160(1+8)]+ ₹160\) 

\(=₹[(160)(9)]+ ₹160\) 

\(=₹[160(10-1)+160]=₹1600\)

 \(\Rightarrow\) Simple interest of first-year \(+(10-1)×\) raise of simple interest each year.

Similarly, the simple interest at the end of \(20^{\text {th }}\) year would be: 

Simple interest of first-year \( + \left( {20 – 1} \right) \times \) raise of simple interest each year.

\(=₹160+(19)₹160\) 

\(=₹3200\) 

The above problem would have given you some clue about how to write the \(10^{t h}\) term, or the \(15^{\text {th }}\) term, and more generally, the \(n^{t h}\) term of the A.P.

Let \(a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots, a_{n}\) be an A.P. whose first term \(a_{1}\) is a and the common difference is \(d\).

As a result, adding a constant quantity to each term of an arithmetic progression produces terms that are also in an arithmetic progression with the same common difference. 

Then,

The second term \(a_{2}=a+d=a+(2-1) d\)

The third term \(a_{3}=a+2 d=a+(3-1) d\)

…………………………………………………………..

…………………………………………………………..

Observing the pattern, we can say that the \(n^{t h}\) term 

The arithmetic progression, therefore, becomes \(a, a+d, a+2 d, a+3 d, a+4 d, \ldots \ldots . a+n d\) etc. By adding a constant quantity:

\(a_{n}=a+(n-1) d\)

\(a_{n}\) is also named as the general term of the A.P. If there are m terms in the A.P., then \(a_{m}\) signifies the last term, which is sometimes also denoted by \(l\).

Sum of n Terms of an Arithmetic Progression

Let us consider an example where Rimi was asked to find the sum of the positive integers from \(1\) to \(50\). She immediately replied that the sum is \(1275\). Can you guess how did she do? She wrote: \(S=1+2+3+…….+49+50\)

And then, reversed the numbers to write \(S=50+49+48+……+2+1\)

Adding these two, he got:

\(2 S=(50+1)+(49+2)+(48+3)+\ldots \ldots . .+(3+48)+(2+49)+(1+50)\)

\(=51+51+\ldots \ldots .+51+51\) (\(50\) times)

So, \(S=\frac{50 \times 51}{2}=1275\)

Therefore, the sum from \(1\) to \(50 =1275\)

We will now apply the same technique to find the sum of the first \(n\) terms of an A.P. 
\(a, a+d, a+2 d, a+3 d, \ldots \ldots\)

The \(n^{t h}\) term of this A.P. is \(a+(n-1) d\). Let S denote the sum of first \(n\) terms of the A.P.

So, \(S=a+(a+d)+(a+2 d)+\ldots \ldots \ldots .+[a+(n-1) d] \ldots . .(\mathrm{i})\)

Revising the sum in reverse order, we get:

\(S=[a+(n-1) d]+[a+(n-2) d]+\ldots \ldots \ldots .+(a+2 d)+(a+d) \ldots \ldots(ii)\)

Adding (i) and (ii), we get:

\(2 S=[2 a+(n-1) d]+[2 a+(n-1) d]+[2 a+(n-1) d]+\ldots \ldots \ldots\) (\(n\) times)

or, \(2 S=n[2 a+(n-1) d]\) (Since there are \(n\) terms)

or, \(S=\frac{n}{2}[2 a+(n-1) d]\)

So, the sum of the first \(n\) terms of an A.P. is given by:

\(S=\frac{n}{2}[2 a+(n-1) d]\) 

We can also write it as:

\(S=\frac{n}{2}\left[a+a_{n}\right]\)

or,

\(S=\frac{n}{2}[a+l]\)

Where \(a\) is the first term and \(a_{n}\) or \(l\) is the last term.

Arithmetic Mean (A.M.)

Arithmetic mean is the most commonly applied measure of a mean or an average. Mean, often referred to as the arithmetic average or arithmetic mean, is calculated by adding all the numbers in each set and dividing by the total number of items within that set. 

In another way, we can find the arithmetic mean of an A.P. series if the first term and the last term and the total number of terms are known. And if the arithmetic mean of an A.P. series is known, we can easily find out the summation of the A.P. series.

As we know, where \(a\) is the first term and \(a_{n}\) or \(l\) is the last term, \(n\) is the total number of terms and \(S\) is the summation of an A.P. series, then: 

\(S=\frac{n}{2}[a+l] \ldots \ldots(i)\)

\(\frac{S}{n}=\frac{a+l}{2}\)

And,

\(A . M .=\frac{a+l}{2} \ldots \ldots(ii)\)

Substituting the value of A.M. in equation (i), we have,

\(\frac{S}{n}=A M\)

Solved Examples: A.M. and Useful Results in Arithmetic Progression

Q.1. Write the first five terms of A.P., when \(a\,=\,23\) and \(d\,=\,5\)
Ans: The general form of A.P. is \(a, a+d, a+2d …\)
When \(a=23\) and \(d=5\),
The first five terms of the A.P. series are \(23, 23+(5), 23+2(5), 23+3(5), 23+4(5)\)
Hence, the first five terms of the A.P. series are \(23, 28, 33, 48\) and \(43.\)

Q.2. Find the \({50^{{\rm{th}}}}\) odd number.
Ans: Odd numbers are in A.P. In which, \(a=1\) and \(d=2\). So, using the formula \(a_{n}=(a+(n-1) d)\) we can find the \(50^{t h}\) odd number.
By substituting in the formula, we get,
\(a_{50}=1+(50-1) 2\)
\(=1+(49) 2=1+98=99\)
Thus, the \(50^{t h}\) odd number is \(99.\)

Q.3. Find the missing term in the following A.P. \(5,\,x,\,13\)
Ans: Three numbers \(x, y, z\) are in A.P. if \(2y=x+z\). Given \(x=5, z=13\) therefore \(y=\frac{x+z}{2}\)
By substituting values, we get,
\(y=\frac{5+13}{2}=\frac{18}{2}=9\)
Therefore, the missing term is \(9\).

Q.4. If the A.M. is \(25\) of an A.P. with \(10\) terms, then find the sum of all terms of the A.P.
Ans: By using the formula, 
\(A M=\frac{a+l}{2}=25\)
\(S=\left[\left(\frac{n}{2}\right) \times(a+l)\right]\)
We can find the sum by substituting \(AM=25\) in the above formula
\(S=[10 \times 25]=250\)
Therefore, the sum of the A.P. series is \(250\).

Q.5. Find the sum of the first \(15\) multiples of \(8.\)
Ans: The first multiple of \(8\) is \(8\), i.e. \(a=8, d=8\) and \(n=15\)
We know that, \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(\Rightarrow S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]\)
\(\Rightarrow S_{15}=\frac{15}{2}[16+(14) 8]\)
\(\Rightarrow S_{15}=\frac{15 \times 128}{2}\)
\(\Rightarrow S_{15}=960\)
Therefore, the sum of the first \(15\) multiples of \(8\) is \(960\).

Summary

This article explained that successive terms are found by adding or subtracting a fixed number to the preceding terms in an A.P. Hence, such a sequence is an arithmetic progression (A.P.). We can find the arithmetic mean of an A.P. series if the first and last terms and the total number of terms are known. And if the arithmetic mean of an A.P. series is known, we can easily find out the summation of the A.P. series. It shows some examples of arithmetic mean and its use in A.P.

Frequently Asked Questions (FAQs): A.M. and Useful Results in Arithmetic Progression

Q.1. What is arithmetic progression? Give an example.
Ans: An A.P. is a list of numbers in which each term is found by adding a fixed number to the preceding term, excluding the first term.
Example: \(1, 3, 5, 7, ……. \)are in A.P. with the first term as \(1\), common difference \(=3-1=2\)

Q.2. What are the types of mean?
Ans: The most common types of the mean are:
(i) Arithmetic mean 
(ii) Geometric mean 
(iii) Harmonic mean

Q.3. Write one real-life example of A.M.
Ans: The mean income of a country’s population is the country’s per capita income calculated using A.M.

Q.4. How to find the last term of an A.P. series?
Ans: Let us consider the last term as \(n^{t h}\) term.
We can find the last term in two ways:
Explicit formula: \(a_{n}=a_{1}+d(n-1)\) or
Recursive formula: \(a_{n}=a_{n-1}+d\)

Q.5. What is the arithmetic progression formula?
Ans: The general formula for the \(n^{t h}\) term of an arithmetic progression is \(a_{n}=a+(n-1) d\) and for the sum of \(n\) terms is \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)

NCERT Solutions for Arithmetic Progression

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