Arithmetic Progression: Definition, Formulas, Sum and Examples

Prior to learn the Arithmetic Progression, let’s revise the concept of sequence. A sequence is formed when terms are written in order such that they follow a particular pattern. A sequence can have any number of terms which can be finite or infinite. In real life, we can use Arithmetic Progression to find roll numbers of students in a class, days in a week, or months in a year.

In this article, we will cover the concept of the Arithmetic Progression. This category falls under the broader category of Sequence and series, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Questions based on this topic have been asked frequently in JEE Mains.

Arithmetic Progression

An arithmetic progression is a sequence in which each term increases or decreases by a constant term or fixed number. This fixed number is called the common difference of an AP and is generally denoted by ' $d$ '.

If $a_1, a_2, a_3, a_4 \ldots \ldots . a_{n-1}, a_n$ are in AP
Then
$
d=a_2-a_1=a_3-a_2=\ldots \ldots \ldots \ldots=a_n-a_{n-1}
$

Eg, 1, 4, 7, 10,.... is an AP with a common difference 3

Also, 2, 1, 0, -1,.... is an AP with a common difference of -1

In AP, the first term is generally denoted by ‘a’

General Term of an AP

We found a formula for the general term of a sequence, we can also find a formula for the general term of an arithmetic sequence. First, write the first few terms of a sequence where the first term is ‘a’ and the common difference is ' $d$ '. We will then look for a pattern.

i.e. $a, a + d, a + 2d, a + 3d, ………$

The $\mathrm{n}^{\text {th }}$ term (general term) of the A.P. is $a_n=a+(n-1) d$.
$
\begin{aligned}
& a_1=a+(1-1) d=a \\
& a_2=a+(2-1) d=a+d \\
& a_3=a+(3-1) d=a+2 d \\
& a_4=a+(4-1) d=a+3 d \\
& \cdots \\
& \cdots \\
& a_n=a+(n-1) d=l=\text { last term }
\end{aligned}
$

On simplification of a general term, we can see that the general term of an AP is always linear in $n$
$
T_n=a n+b
$

The sum of Arithmetic progressions

The sum, Sn of n terms of an AP with the first term ‘a’ and common difference ‘d’ is given by

$\begin{aligned} & S_n=\frac{n}{2}[2 a+(n-1) d] \\ & \text { OR } \\ & S_n=\frac{n}{2}[a+l] \\ & a \rightarrow \text { first term } \\ & d \rightarrow \text { common difference } \\ & n \rightarrow \text { number of terms }\end{aligned}$

Types of Arithmetic Progression

We have two types Of Arithmetic progressions based on numbers,

1. Finite AP
2. Infinite AP

Finite AP: If the AP has only a finite number of terms, then the AP is called a finite AP.

Eg- 2, 4, 6, 8

This AP has the first term as ‘2’ and the Common difference as ‘2’

Number of terms =4

Infinite AP: If an AP has three dots at the end, then it indicates the list never ends. It has an infinite number of terms. Then, the AP is said to be an infinite AP.

E.g., 2, 4, 6, 8, 10, …

This AP has the first term as ‘2’ and the common difference as ‘2’ but we don’t know the number of terms this AP contains.

Important Properties of an AP

1. If a fixed number is added to or subtracted from each term of a given A.P., then the resulting series is also an A.P. and its common difference remains the same.

2. If each term of an A.P. is multiplied by a fixed constant or divided by a non-zero fixed constant then the resulting series is also in A.P.

3. If $a_1, a_2, a_3 \ldots$ and $b_1, b_2, b_3 \ldots$ are two A.P’s, then $a_1 \pm b_1, a_2 \pm b_2, a_3 \pm b_3 \ldots$ are also in A.P.

4. If terms of an A.P. are taken at equal intervals, then the new sequence formed as also an A.P.

Eg, If from A.P., 1, 3, 5, 7, 9, 11, 13,.... we take terms at equal intervals, let's say first, third, fifth, seventh,....terms, then the resultant sequence will be

1, 5, 9, 13, ....which is also an A.P.

5. If $a_1, a_2, a_3, \ldots, a_n$ are in A.P., then $
a_r=\frac{a_{r-k}+a_{r+k}}{2}, \forall k, 0 \leq k \leq n-r
$

6. Choosing terms in A.P.

If some of a few terms (like 3, 4, or 5 terms) in an A.P. is given in the problem, then selecting the following terms reduces the calculation
- If we need to choose three terms in an A.P., then choose $(a-d), a,(a+d)$
[Note: Here the first term is $a-d$, and the common difference is $d$]
- If we need to choose four terms in an A.P., then choose $(a-3 d),(a-d),(a+d),(a+3 d)$
[Note: Here first term is $a-3d$, and the common difference is $2d$]
- If we need to choose five terms in an A.P., then choose $(a-2 d),(a-d), a,(a+d)$, $(\mathrm{a}+2 \mathrm{~d})$
[Note: Here the first term is $a-2d$, and the common difference is $d$]

7. The sum of terms equidistant from the beginning and end of an AP is constant and it equals the sum of the first and the last terms.

$a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\ldots \ldots=a_r+a_{n-r+1}$

8. If $a, b, c$ are in A.P., then $2 b=a+c$

Solved Examples Based on Arithmetic Progression

Example 1: The $8^{\text {th }}$ common term of the series $
\begin{aligned}
& \mathrm{S}_1=3+7+11+15+19+\cdots \\
& \mathrm{S}_2=1+6+11+16+21+\cdots
\end{aligned}
$
is $\qquad$ [JEE MAINS 2023]
Solution
$8^{\text {th }}$ common term of the series
$
\begin{aligned}
& \mathrm{S}_1=3+7+11+15+19+\ldots \ldots \\
& \mathrm{S}_2=1+6+11+16+21+\ldots \ldots
\end{aligned}
$

First common term $=11$
common diff of the AP of common terms
$
\begin{aligned}
& =\mathrm{L} . \mathrm{C} . \mathrm{M} \text { of }\{4,5\} \\
& =20
\end{aligned}
$
$
\begin{aligned}
& \therefore \mathrm{AP} \\
& 11,31,51, \quad \ldots \ldots \\
& \mathrm{T}_8=11+(8-1) 20 \\
& =11+140 \\
& \mathrm{~T}_8=151
\end{aligned}
$

Hence, the answer is (151).

Example 2: Let $a, b, c>1, a^3, b^3$ and $c^3$ be in A.P., and $\log _a b, \log _c$ a and $\log _b c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is -444, then abc is equal to:
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \text { If } \log _{\mathrm{a}} \mathrm{b}, \log _{\mathrm{c}} a, \log _b c \rightarrow \text { G.P. } \\
& \left(\log _c a\right)^2=\log _a b \times \log _b c \\
& \left(\log _{\mathrm{c}} \mathrm{a}\right)^2=\log _{\mathrm{a}} \mathrm{c} \\
& \Rightarrow\left(\log _c a\right)^2=\frac{1}{\log _c a} \\
& \Rightarrow\left(\log _{\mathrm{c}} \mathrm{a}\right)^3=1 \\
& \Rightarrow \log _{\mathrm{c}} \mathrm{a}=1 \\
& \mathrm{a}=\mathrm{c} \\
& \text { If } a^3 b^3 c^3 \rightarrow \mathrm{A} . \mathrm{P} \\
& 2 b^3=a^3+c^3 \\
& \text { If } \mathrm{a}=\mathrm{c} \\
& \Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c} \\
\end{aligned}
$

For AP
$
\begin{aligned}
& A=\frac{a+4 a+a}{3} \quad D=\frac{a-8 a+a}{10} \\
& \mathrm{~A}=2 \mathrm{a} \quad \mathrm{D}=\frac{-3 \mathrm{a}}{5} \\
& S_{20}=\frac{20}{2}\left[2 \times 2 a+(20-1)\left(\frac{-3 a}{5}\right)\right] \\
& =10\left[4 a-\frac{57 a}{5}\right] \\
& =10\left[-\frac{37 a}{5}\right]=-444 \\
& \Rightarrow \mathrm{a}=\frac{444 \times 5}{37 \times 10} \\
& a=6 \\
& \Rightarrow \mathrm{abc}=6 \times 6 \times 6=216 \\
&
\end{aligned}
$

Example 3: If $a_1, a_2, a_3 \ldots \ldots$ and $b_1, b_2, b_3 \ldots \ldots$ are A.P., and $a_1=2, a_{10}=3, a_1 b_1=1=a_{10} b_{10}$ then $a_4 b_4$ is equal to
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \mathrm{a}_1=2, \quad \mathrm{a}_{10}=2+(10-1) \mathrm{d}=3 \Rightarrow \mathrm{d}=\frac{1}{9} \\
& \mathrm{a}_4=2+(4-1) \times \frac{1}{9}=\frac{7}{3} \\
& \mathrm{a}_1 \mathrm{~b}_1=1 \Rightarrow \mathrm{b}_1=\frac{1}{2}, \quad \mathrm{a}_{10} \mathrm{~b}_{10}=1 \Rightarrow \mathrm{b}_{10}=\frac{1}{3} \\
& \mathrm{~b}_{10}=\frac{1}{2}+(10-1) \times \mathrm{d}=\frac{1}{3} \Rightarrow 9 \mathrm{~d}=-\frac{1}{6} \Rightarrow \mathrm{d}=-\frac{1}{5^4} \\
& \mathrm{~b}_4=\frac{1}{2}+(4-1) \times\left(-\frac{1}{54}\right)=\frac{1}{2}-\frac{1}{18}=\frac{4}{9} \\
& \mathrm{a}_4 \mathrm{~b}_4=\frac{7}{3} \times \frac{4}{9}=\frac{28}{27}
\end{aligned}
$

Example 4: If the 10th term of an A.P. is $\frac{1}{20}$ and its 20th term is $\frac{1}{10}$, then the sum of its first 200 terms is
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& a_{10}=a+9 d=1 / 20 \\
& a_{20}=a+19 d=1 / 10
\end{aligned}
$

On subtracting them, we get $d=1 / 200$
$
\begin{aligned}
& a+9 / 10=1 / 20 \Rightarrow a=1 / 200 \\
& S_{200}=\frac{200}{2}[2 / 200+(199) \times 1 / 200]=\frac{201}{2}
\end{aligned}
$

Example 5: Let $f: R \rightarrow R$ be such that for all $x \in R\left(2^{1+x}+2^{1-x}\right), f(x)$ and $\left(3^x+3^{-x}\right)$ are in A.P., then the minimum value of $f(x)$ is:
[JEE MAINS 2023]

Solution
If $a, b, c$ are in $A P$, then $2 b=a+c$
$
2 f(x)=\left(2 \frac{1}{2^x}+2.2^x\right)+\left(3^x+\frac{1}{3^x}\right)
$

Now, $A M \geq G M$
$
\begin{aligned}
& \frac{a+b}{2} \geq \sqrt{a b} \\
& \frac{3^x+\frac{1}{3^x}}{2} \geq \sqrt{1} \\
& \left(3^x+\frac{1}{3^x}\right) \geqslant 2
\end{aligned}
$

Value equals 2 at $3^x=\frac{1}{3^x} \Rightarrow x=0$
Similarly, $\frac{1}{2^x}+2^x \geq 2$, and this holds at $x=0$
So, $2($ minimum of $f(x))=2(2)+(2)$ $\min f(x)=3$

Summary

Arithmetic Progression is a sequence in which there is a constant increase or decrease between successive terms. It is used to generalize the pattern we see in day-to-day life. Its direct formula to find the number of terms, the sum of terms, and most importantly, the nth term makes it suitable for use in many fields. Arithmetic progressions are essential in mathematics for their ability to model and predict sequential patterns with precision and efficiency.