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Arithmetic Progressions Class 10 Notes

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Arithmetic Progression

An Arithmetic progression is a sequence, list or series of numbers where the difference between two terms is constant, and we can get the preceding and succeeding terms by performing the arithmetic operation.
An arithmetic progression is denoted by $AP$, and the first term is denoted by $a$₁ and the last term is denoted by $a_n$.
It implies that the number of terms in $AP$ is $n$.
Example: 5, 8, 11, 14, 17, 20, 23......

Common difference

In an arithmetic progression, the difference between two consecutive terms is called the common difference, and it is denoted by $d$.
Example: 7, 12, 17, 22, 27, 32
In this arithmetic progression, the common difference ($d$) = 5

Formula For Calculating Common Difference

$d=\left(a_n-a_{n-1}\right)$
Where,
$d$ = Common difference
$a_n$ = nth term of the sequence
$a_{n - 1}$ = (n - 1)th term of the sequence

There are three types of common difference $(d)$:

1. Positive Difference: The $AP$ is increasing in this case.
Example: 1, 2, 3, 4, 5

2. Negative Difference: In this case, the $AP$ decreases.
Example: 100, 90, 80, 70, 60, 50

3. Zero Difference: In this case, the $AP$ is constant.
Example: 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Types of Arithmetic Progression (AP)

There are two types of arithmetic progression (AP):

1. Finite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is finite is called a finite arithmetic progression (AP). The finite AP have an nth term.
Example: 1, 3, 5, 7, 9, 11

2. Infinite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is infinite is called an infinite arithmetic progression (AP). The infinite AP does not have the nth term.
Example: 2, 4, 6, 8, 10, ........

The nth Term Of An Arithmetic Progression (AP)

The nth term in the last term in AP and the formula for determining the nth term is as follows:
$a_n$ = $a + (n − 1)d$
Here,
a = First term
d = Common difference
n = number of terms

Example: Determine the 10th term of the arithmetic progression 4, 8, 12, 16, .......
Given:
AP = 4, 8, 12, 16, .......
Here, a = 4
(8 - 4) = 4 and (12 - 8) = 4
So, d = 4
And now, we have to find the 10th term of the given AP.
Therefore, n = 10
$a$_n = $a + (n − 1)d$
After substituting the values in the formula, we get
a₁₀ = 4 + (10 – 1)4
a₁₀ = 4 + 9 × 4
a₁₀ = 4 + 36
a₁₀ = 40
So, the 10th term of the given arithmetic sequence 4, 8, 12, 16,... is 40.

The General Form Of An Arithmetic Progression (AP)

The general form of an AP is as follows:
a, a + d, a + 2d, a + 3d, a + 4d ....
Where a is the first term and d is the common difference, the value of d = 0, d > 0 or d < 0.

The Sum Of The Terms In Arithmetic Progression (AP)

The formula for calculating the sum of the terms in AP is as follows:

$S$ = $\frac{n}{2}$ $[2a + (n - 1)d]$
Where,
S = Sum of the terms
n = Number of terms
a = First term
d = Common difference
Or we can also write this as,
$S$ = $\frac{n}{2}$ $(a + a_n$)

Example: Determine the sum of the first 18 terms of AP 17, 11, 5, -1, .......
Given:
AP = 17, 11, 5, -1, .......
First term a = 17
Common difference d = 11 - 17 = -6, and 5 - 11 = -6
Therefore, the common difference d = -6
Number of terms n = 18

$S$ = $\frac{n}{2}$ $[2a + (n - 1)d]$
After substituting all the values, we get

$S$ = $\frac{18}{2}$ $[2 × 17 + (18 - 1) × (-6)]$

$S$ = $9$ × $[34 + (17) × (-6)]$

$S$ = $9$ × $[34 + (-102)]$

$S$ = $9$ × $(-68)$

$S$ = $-612$
Therefore, the sum of the 18th term is -612.

Arithmetic Mean (AM)

The average of the terms in an arithmetic progression is called the Arithmetic Mean (AM). Therefore, the arithmetic mean formula is as follows:
$AM$ = $\frac{\text{Sum of the terms}}{\text{Number of the terms}}$

The Sum of the First n Natural Numbers

The sum of the first n natural numbers can be determined by:

$S_n$ = $\frac{n(n + 1)}{2}$
Here,
n = Number of terms

Example: Determine the sum of the first 15 natural numbers.
Given:
n = 15

$S_n$ = $\frac{n(n + 1)}{2}$

$S_n$ = $\frac{15(15 + 1)}{2}$

$S_n$ = $\frac{15 ×16}{2}$

$S_n$ = $120$
Therefore, the sum of the first 15 natural numbers is 120.

Arithmetic Progressions: Previous Year Question and Answer

Question 1:
What is the sum of all three digits numbers which are divisible by 20?

Solution:
The first three-digit number divisible by 20 is 100, and the last is 980
This forms an arithmetic series with first term ($a$) = 100, common difference ($d$) = 20 and last term ($l$) = 980
⇒ The sum of an arithmetic series = $\frac{n}{2} × (a + l)$, where $n$ is the number of terms
We know, $n = \frac{l - a}{d} + 1$
⇒ $n=\frac{980-100}{20}+1$
⇒ $n=\frac{880}{20}+1$
⇒ $n=44+1=45$
So, the sum $=\frac{45}{2} × (100 + 980)= 24300$
Hence, the correct answer is 24300.

Question 2:
Ramu had to select a list of numbers between 1 and 1000 (including both), which are divisible by both 2 and 7. How many such numbers are there?

Solution:
The numbers that are divisible by both 2 and 7 = 2 × 7 = 14
The smallest number divisible by 14 = 14
The largest number divisible by 14 = 994 (nearest to 1000)
Let the required number be $n$.
Now, $a_n=a+(n-1)d$, where $a_n$ = last term, $a$ = first term $d$ is the common difference.
⇒ $994=14+(n-1)14$
⇒ $984=14+14n-14$
⇒ $984=14n$
$\therefore n=71$
Hence, the correct answer is 71.

Question 3:
What is the value of $99 \frac{11}{99}+99 \frac{13}{99}+99 \frac{15}{99}+.....+99 \frac{67}{99}$?

Solution:
Given: The series is $99 \frac{11}{99}+99 \frac{13}{99}+99 \frac{15}{99}+....+99 \frac{67}{99}$.
⇒ $(99+99+99+......)+(\frac{11}{99}+\frac{13}{99}+...+\frac{67}{99})$
The number of terms in the series is given by $t_n=a+(n–1)d$ where $t_n$ is the last term, $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.
⇒ $\frac{67}{99}=\frac{11}{99}+(n–1)\times\frac{2}{99}$
⇒ $67=11+(n–1)\times 2$
⇒ $56=2\times (n–1)$
⇒ $ (n–1)=28$
⇒ $n=29$
So, the number of terms is 29.
The given series can be written as $99\times 29+(\frac{11}{99}+\frac{13}{99}+...+\frac{67}{99})$.
$S_n=\frac{n}{2}[2a+(n–1)d]$
= $99\times 29+\frac{29}{2}\times[2\times\frac{11}{99}+28\times\frac{2}{99}]$
= $99\times 29+\frac{29\times11}{99}+29\times\frac{28}{99}=99\times 29+29(\frac{11}{99}+\frac{28}{99})$
= $99\times 29+29\times\frac{11+28}{99}=99\times 29+29\times\frac{39}{99}$
= $29[99+\frac{13}{33}]=29\times\frac{3267+13}{33}$
= $29\times\frac{3280}{33}$
= $\frac{95120}{33}$
Hence, the correct answer is $\frac{95120}{33}$.

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