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# Arithmetic Progression Formula

Arithmetic progression is defined as the series of numbers having the difference between two consecutive term is constant every time. (A.P) is also known as Arithmetic Sequence. It’s series is obtained by adding or subtracting the finite number to its successive term. A.P generally deals as a part of mathematics which involves series in defining the position of the given number, addition or subtraction of numbers in series and much more. Formula are listed here.

Let us take an example

2, 4, 6, 8, 10………. is a series of number which is in A.P.

We can say that this series is in A.P because the difference in each consecutive term is fixed i.e. 2. If 2 is being added to each successive term gives the next term.

Every series in A.P is in **Infinite series** so is said **Infinite A.P **and the finite part of Infinite A.P series is called **Finite Arithmetic Progression.**

The Sum of Finite A.P is known to be **Arithmetic Series.**

**Terms Related to A.P:**

**a –**First term in A.P**d**– Common difference**n –**Number of Terms in A.P**a**n_{n}–^{th}term of A.P**S**Sum of n terms of A.P_{n}–

**Formula for n**^{th} Term of A.P:

^{th}Term of A.P:

**a _{n} = a + (n – 1)d**

**For Example:**

*If 2, 4, 6, 8, …. are in A.P then find its 20*^{th}term from left.

Solution:

Given,

a = 2, d = (4 – 2) = 2, n = 20

**a _{n} = a + (n – 1)d**

a_{n} = 2 + (20 – 1)x 2 = 40

Thus, 40 is the 20^{th} term of given A.P.

**Formula for Sum of n Terms of an A.P:**

**S _{n} = \(\frac{n}{2}\){2a + (n-1)d}**

Also,

** S _{n} = \(\frac{n}{2}\)(first term + last term)**

**For Example:**

*If 4, 8, 12, 16, …… are in A.P then Find the sum of its first 20 terms.*

**Solution:**

Given,

a = 4, d = 4, n = 20

**S _{n} = \(\frac{n}{2}\){2a + (n-1)d}**

S_{n} = \(\frac{20}{2}\){2×4 + (20 – 1)x4}

Thus, S_{n} = 840

*If 8, 10, 12, 14, ……. , 42 are the terms of an A.P than Find the sum of terms.*

**Solution:**

a = 8, d = 2, n = ?

a_{n} = 42

First let’s find n

**a _{n} = a + (n – 1)d**

42 = 8 + (n – 10)x2

n = 27

Now, **S _{n} = \(\frac{n}{2}\)(first term + last term)**

S_{n} = \(\frac{27}{2}\)(8 + 42)

S_{n} = 675

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