Arithmetic Progression Class 10 Maths Formulas

The List of Important Formulas for Class 10 Arithmetic Progression is provided on this page. We have everything covered right from basic to advanced concepts in Arithmetic Progression. Make the most out of the Maths Formulas for Class 10 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Arithmetic Progression Class 10 covering numerous concepts and use them to solve your Problems effortlessly.

We have observed many things in our daily life, follow a certain pattern.
(a) 1, 4, 7, 10, 13, 16, …….
(b) 15, 10, 5, 0, -5, -10,………….
(c) 1,\(\frac{1}{2}\),0,\(-\frac{1}{2}\)………………
These patterns are generally known as sequence. Two such sequences are arithmetic and geometric sequences. Let us investigate the Arithmetic sequence.

1. Sequence: A sequence is a ordered list of numbers.
Terms: The various numbers occurring in a sequence are called its terms. Terms of sequence are denoted by a₁ a₂, a₃, …………… a_n.

2. Arithmetic Progression: An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms are equal.

3. Common difference: The difference between two consecutive terms of an arithmetic progression is called common difference.
d = a₂ – a₁
d = a₃ – a₂
d = a₄ – a₃
……………..
……………..
d = a_n – a_n-1

4. Finite Arithmetic Progression: A sequence which has finite or definite number of terms is called finite sequence.
Example, (1, 3, 5, 7, 9)… which has 5 terms.

5. Infinite Arithmetic Progression: A sequence which has indefinite or infinite number of terms is called infinite arithmetic progression.
Example, 1, 2, 3, 4, 5, …

In general, arithmetic progression can be written as a, a + d, a + 2d, where a is the first term and d is called the common difference i.e. difference between two consecutive terms.

6. General form of an AP: Let a be the first term and d is the common difference then the AP is

Here
a₁ = a (we take) (a is first term of AP)
a₂ = a₁ + d = a + d
a₃ = a₂ + d = a + d + d = a + 2d
a₄ = a₃+ d = a + 2d + d = a + 3d
…………….
……………
a_n = a + (n – 1) d
i.e. AP is a, a + d, a + 2d, a + 3d,………… , a + (n – 1)d.
n^th term of AP = a + (n -1)d
Note: Common difference of AP can be positive, negative or zero.

1. n^th term or General term of an AP
n^th term of an AP = a + (n – 1) d where
a → first term of the AP
n → number of terms
d→common difference of an AP.

2. n^th term of an AP from the end: Let us consider an AP where first term a and common difference is If m is number of terms in the AP. then
n^th term from the end = [m – n + 1]^th term from the beginning.
n^th term from the end = a + (m-n +1 – 1)d – a + (m – n) d
It l is the last term of the AP, then n^th term from the end is the n^th term of an AP where first term is l and common difference is – d
n^th term from the end – 1 + (n – 1) (-d)
= 1 – (n – 1) d

Sum of first n terms of an AP
Let S_n denote the sum of first n terms of an AP
S_n = a + a + d + a + 2d + a + 3d …. + a + (n – 1)d ……….. (1)
Rewriting the terms in reverse order.
S_n = a + (n – 1) + a + (n – 2)d + a + (n-3)d + ………….+a ……… (2)
Adding equations (1) and (2)
2S_n = [2a + (n – 1)d] + [2a + (n-1)d] + … + [2a + (n – 1)d]
2S_n = n[2a + (n – 1)d]
S_n=\(\frac{n}{2}\)[2a+(n-1)d]
We can Write
S_n=\(\frac{n}{2}\)[a+a+(n-1)d] [l=a+(n-1)d]
S_n=\(\frac{n}{2}\)[a+l]

Note:
(i) The Tith term of an AP = S_n – S_n-1 or a_n = S_n+1 – S_n
Sum of first n positive integer
\(S_{n}=\frac{n(n+1)}{2}\)
(iii) Sum of n odd positive integer = n²
(iv) Sum of n even positive integer = n(n + 1)