Arithmetic Progressions Class 10- NCERT Notes
Last Updated :
30 Aug, 2024
Arithmetic Progressions (AP) are fundamental sequences in mathematics where each term after the first is obtained by adding a constant difference to the previous term. Understanding APs is crucial for solving problems related to sequences and series in Class 10 Mathematics. These notes cover the essential concepts and formulas related to APs, including how to find specific terms and the sum of terms in a progression.
Arithmetic Progressions Class 10- NCERT Notes
Arithmetic Progressions (AP) are fundamental sequences in mathematics where each term after the first is obtained by adding a constant difference to the previous term. Understanding APs is crucial for solving problems related to sequences and series in Class 10 Mathematics. These notes cover the essential concepts and formulas related to APs, including how to find specific terms and the sum of terms in a progression.
A progression is a sequence or series of numbers in which they are arranged in a particular order such that the relation between the consecutive terms of a series or sequence is always constant. In a progression, it is possible to obtain the nth term of the series.
In mathematics, there are 3 types of progressions:
- Arithmetic Progression (AP)
- Geometric Progression (GP)
- Harmonic Progression (HP)
let's learn about AP in this article.
Arithmetic Progressions
Arithmetic Progression (AP) also known as Arithmetic Sequence is a sequence or series of numbers such that the common difference between two consecutive numbers in the series is constant.
For example:
In this series, the common difference between any two consecutive numbers is always 2.
- Series 2: 28,25,22,19,16,13….
In this series, the common difference between any two consecutive numbers is strictly -3.
Terminology and Representation in Arithmetic Progressions
- Common difference, d = a2 - a1 = a3 - a2 = ……. = an - an - 1
- an = nth term of Arithmetic Progression
- Sn = Sum of first n elements in the series
Nth Term of an AP
If a is taken as the first term and d is taken as the common difference, then the Nth term of the AP will be given by the formula:
By computing the n terms of an AP with the upgiven formula, the general form of the AP is as follows:
Example: Find the 35th term of the series 5,11,17,23.....
Solution:
In the given series,
a = 5, d = a2 - a1 = 11 - 5 = 6, n = 35
We have to find out the 35th term, hence, apply the formulae,
an = a + (n - 1)d
an = 5 + (35 - 1) x 6
an = 5 + 34 x 6
an = 209
Hence 209 is the 35th term.
Sum of n Terms of Arithmetic Progression
The formula for the arithmetic progression sum is,
Sn = (n/2)[2a + (n - 1) × d]
Sn = (n/2)[a + l]
where,
- a is the First Term of Series
- l is the Last Term of Series
- n is the Number of Terms in Series
Let 'l' denote the nth term of the series and Sn be the sum of first n terms of AP a, (a+d), (a+2d), ...., a+(n-1)d then,
Sn = a1 + a2 + a3 + ....an-1 + an
Sn = a + (a + d) + (a + 2d) + ........ + (l - 2d) + (l - d) + l ...(1)
Writing the series in reverse order, we get,
Sn = l + (l - d) + (l - 2d) + ........ + (a + 2d) + (a + d) + a ...(2)
Adding equation (1) and (2),
2Sn = (a + l) + (a + l) + (a + l) + ........ + (a + l) + (a + l) + (a + l)
2Sn = n(a + l)
Sn = (n/2)(a + l) ...(3)
Hence, the formulae for finding the sum of a series is,
Sn = (n/2)(a + l)
Replacing the last term l by the nth term in equation 3 we get,
nth term = a + (n - 1)d
Sn = (n/2)(a + a + (n - 1)d)
Sn = (n/2)(2a + (n - 1) x d)
Note: Consecutive terms in an Arithmetic Progression can also be represented as,
........, a-3d , a-2d, a-d, a, a+d, a+2d, a+3d, ........
Related Article:
Sample Problems on Arithmetic Progressions
Problem 1. Find the sum of the first 35 terms of series 5,11,17,23.....
Solution:
In the given series,
a = 5, d = a2 - a1 = 11 - 5 = 6, n = 35
Sn = (n/2)(2a + (n - 1) x d)
Sn = (35/2)(2 x 5 + (35 - 1) x 6)
Sn = (35/2)(10 + 34 x 6)
Sn = (35/2)(10 + 204)
Sn = 35 x 214/2
Sn = 3745
Problem 2. Find the sum of the series when the first term of the series is 5 and the last of the series is 209 and the number of terms in the series is 35.
Solution:
In the given series,
a = 5, l = 209, n = 35
Sn = (n/2)(a + l)
Sn = (35/2)(5 + 209)
Sn = 35 x 214/2
Sn = 3745
Problem 3. 21 Rupees is divided among three brothers where the three parts of money are in AP and the sum of their squares is 155. Find the largest amount.
Solution:
Let the money is deivided in three parts as (a-d), a, (a+d)
Given,
(a - d) + a + (a + d) = 21
Therefore,
3a = 21
a = 7
Again, (a - d)2 + a2 + (a + d)2 = 155
a2 + d2 – 2ad + a2 + a2 + d2 + 2ad = 155
3a2 + 2d2 = 155
Putting the value of 'a' we get,
3(7)2 + 2d2 = 155
2d2 = 155 – 147
d2 = 4
d = ±2
Three parts of distributed money are:
a + d = 7 + 2 = 9
a = 7
a - d = 7 - 2 = 5
Hence, the largest part is Rupees 9
Practice Questions on Arithmetic Progression
Q1. Sum of the first 12 terms of an arithmetic progression is equal to the sum of the next 12 terms. Prove that the 13th term is zero.
Q2. In an arithmetic progression, if the sum of the first p terms is equal to the sum of the first q terms, prove that the (p+q)th term is zero.
Q3. The 3rd term of an arithmetic progression is 4, and the 8th term is -11. Find the 20th term.
Q4. If the sum of the first n terms of an arithmetic progression is Sn = n/2.(3n+1), find the 15th term.
Q5. Sum of the first n terms of an arithmetic progression is Sn = 5n2 - n. Find the common difference and the 10th term.
Conclusion
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference. Understanding APs is crucial in Class 10 Mathematics as it forms the basis for solving problems related to sequences and series
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