Sum of n terms in Arithmetic Sequence (AP)-Proof, Solved Examples, FAQs

📢 Master the Sum of n Terms of an Arithmetic Progression (AP) – Class 11 Mathematics 📚

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The Sum of n Terms of an Arithmetic Progression (AP) is a fundamental concept in mathematics, essential for solving sequence and series problems efficiently. Strengthen your understanding with clear explanations, formulas, and practice questions.

🔢 Key Topics Covered:

✅ Understanding Arithmetic Progression (AP)
✅ Sum of n Terms Formula:

✅ Step-by-Step Derivation & Practical Applications
✅ Shortcut Tricks for Quick Calculations in JEE & Exams

Sum of n terms in Arithmetic Sequence

The sum of terms of an Arithmetic Sequence is called an Arithmetic Series.

Suppose the first term of the arithmetic series is a and the common difference is d then the sum of the n term of this arithmetic series is given using the formula,

S_n= n/2 [2a + (n – 1)d]

If the common difference of the arithmetic series is not given but the last term of the series is given (say l). Then its sum is calculated as,

S_n= n/2 [a + l]

Proof : The sum of n terms of AP is the sum(addition) of first n terms of the arithmetic sequence. Let S_n denote the sum for the first n terms, we can write S_n in two ways:

Students can see both are the same but written in opposite directions, i.e. in the first elements are from the first term to the last term, and in the second the elements are from last to first. 

Now add the above two terms:-

If l is the last or nth term of an AP i.e. l = (a + (n – 1)d) , then

S_n= n/2 [2a + (n – 1)d]

S_n= n/2 [a + (a + (n – 1)d)]

We can write:

S_n= n/2 [a + l]

Solved Examples on Sum of n Terms of an AP

Example 1: If the first term of an AP is 67 and the common difference is -13, find the sum of the first 20 terms.

Solution: Here, a = 67 and d= -13

S_n = n/2[2a+(n-1)d]

S₂₀ =20/2[2×67+(20-1)(-13)]

S₂₀= 10[134 – 247]

S₂₀ = -1130

So, the sum of the first 20 terms is -1130.

Example 2: The sum of n and n-1 terms of an AP is 441 and 356, respectively. If the first term of the AP is 13 and the common difference is equal to the number of terms, find the common difference of the AP.

Solution: The sum of n terms S_n = 441

Similarly, S_n-1= 356

a = 13

d= n

For an AP, S_n = (n/2)[2a+(n-1)d]

Putting n = n-1 in above equation,

l is the last term. It is also denoted by a_n. The result obtained is:

S_n  -S_n-1 = a_n

So, 441-356 = a_n

a_n = 85 = 13+(n-1)d

Since d=n,

n(n-1) = 72

⇒n² – n – 72= 0

Solving by factorization method,

n²-9n+8n-72 = 0

(n-9)(n+8)=0

So, n= 9 or -8

Since number of terms can’t be negative,

n= d = 9

Example 3: Find the sum of the first 20 multiples of 5.

Solution:

First 20 multiples of 5 are 5, 10, 15, . . . , 100.

Here, it is clear that the sequence formed is an AP where,

a = 5, d = 5, an = 100, n = 20.

Thus, S_n = n/2 [2a + (n − 1) d]
⇒ S_n = 20/2 [2 × 5 + (20 − 1)5]
⇒ S_n = 10 [10 + 95]
⇒ S_n = 1050

Example 4:  Calculate the sum of natural numbers from 1 to 100. 

Solution:

We see the natural numbers (1, 2, 3, …) form an AP and hence we can calculate the sum of the first n terms as follows:
First-term, a = 1
Difference, d = 1
So, 

S_n = (n/2) × [2a + (n – 1) × d]
S_n = (n/2) × [2 × 1 + (n – 1) × 1]
S_n = (n/2) × (n + 1)

For n = 100, we have:

S₁₀₀ = (100/2) × (100 + 1)
S₁₀₀ = 5050

Sum of Square Series

• 1² + 2² + 3² + 4² + ………. + n²

This arithmetic series represents the sum of squares of n natural numbers. Let us try to calculate the sum of this arithmetic series.

To prove this let us consider the identity p³ – (p – 1)³ = 3p² – 3p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

1³ – (1 – 1)³ = 3(1)² – 3(1) + 1

2³ – (2 – 1)³ = 3(2)² – 3(2) + 1

3³ – (3 – 1)³ = 3(3)² – 3(3) + 1

………………………………………..

………………………………………..

………………………………………..

n³ – (n – 1)³ = 3(n)² – 3(n) + 1

Adding both the sides of the equation, we get

n³−0³=3(1²+2²+3²+…+n²)–3(1+2+3+…+n)+n

Therefore,

Sum of Square of n terms = [n(n+1)(2n+1)]/6

Sum of Cubic Series

• 1³ + 2³ + 3³ + 4³ + ………. + n³

This arithmetic series represents the sum of cubes of n natural numbers. Let us try to calculate the sum of this arithmetic series.

To prove this let us consider the identity (p + 1)⁴ – p⁴ =4p³ + 6p² + 4p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

2⁴ – 1⁴ =4(1)³ + 6(1)² + 4(1) + 1

3⁴ – 2⁴ =4(2)³ + 6(2)² + 4(2) + 1

4⁴ – 3⁴ =4(3)³ + 6(3)² + 4(3) + 1

……..………………………………………..

………..……………………………………..

…………..…………………………………..

(n + 1)⁴ – n⁴ =4n³ + 6n² + 4n + 1

Adding both the sides of the equation, we get

(n+1)⁴–1⁴=4(1³+2³+3³+…..+n³)+6(1²+2²+3²+…n²)+4(1+2+3+4+..+n)+n

Therefore, sum of cube of n terms is:

S = [n(n+1)/2]2

The basic sum of series are as follows :

Sum of n terms in AP	n/2[2a + (n – 1)d]
Sum of natural numbers	n(n+1)/2
Sum of square of ‘n’ natural numbers	[n(n+1)(2n+1)]/6
Sum of Cube of ‘n’ natural numbers	[n(n+1)/2]2

Practice Problems – Sum of First n Terms | Class 11 Maths

Problem 1: Find the sum of the first 10 terms of an arithmetic progression where the first term a = 3 and the common difference d = 2.

Problem 2: If the sum of the first 15 terms of an AP is 180 and the first term is 5, find the common difference.

Problem 3: The sum of the first 20 terms of an AP is 530, and the last term is 45. Find the first term.

Problem 4: Determine the number of terms in an AP if the first term is 2, the last term is 32, and the sum of all terms is 290.

Problem 5: The sum of the first n terms of an AP is given by S_n = 3n² – 2n. Find the first term and the common difference of this AP.

FAQs on Sum of First n Terms | Class 11 Maths

What is Arithmetic Progression (AP)?

Arithmetic Progression, often abbreviated as AP, is a sequence of numbers where the difference between consecutive terms remains constant.

How do you find the sum of the first n terms of an AP?

To find the sum of the first n terms of an AP, use the formula S_n=n/2[2a+(n–1)d], where a is the first term and d is the common difference.

What is the formula for the nth term of an AP?

The nth term of an AP can be found using the formula a_n = a + (n−1)d, where a is the first term and d is the common difference.

What are some real-life applications of Arithmetic Progression?

Arithmetic Progression is used in various fields such as finance for calculating interest payments, in physics for analyzing motion with constant acceleration, and in computer algorithms for generating sequences.

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