Sum of N Terms of an AP

Arithmetic Sequence is defined as the sequence of numbers such that the difference between any two consecutive numbers is always constant. That is we can say that the next number in the arithmetic sequence is always the previous sequence plus the constant term, where the constant term can be both positive or negative. We han have infinite terms in the arithmetic sequence, so the natural question arises what is the sum of these terms of the arithmetic sequence? We can easily find the sum of the first “n” terms of the arithmetic sequence using the formula discussed in the article.

Johann Carl Friedrich Gauss in the 19th century was the first to find the sum of the first “n” terms of the arithmetic sequence where n is any natural number. Let us learn all about the sum of n terms of an AP along with examples in this article.

What is Sum of n Terms of an AP?

The sum of n terms of an AP is considered to be the sum of n consecutive terms of any arithmetic sequence. We can understand this by the example discussed below. Suppose we have to find the sum of all the terms from 1 to 100. The first way is to find the sum by adding all the terms individually but this process is very lengthy. Another method is to consider the sequence of 1, 2, 3,…, 98, 99, 100 as the group of 50 pairs taking then one from the beginning and from the end as,

(1, 100), (2, 99), (3, 98),…

Sum of each pair is 101 and we have a total of 50 pairs, thus, the total sum will be

1+2+3+4+5+…+99+100 = 101×50 = 5050

Sum of n Terms of AP Formula

The formula which gives the sum of n terms of the AP is discussed in the article below. Suppose we have an AP such that its n terms are,

a, a+d, a+2d, a+3d, a+4d, ………. a+ (n-1)d

Then the sum of its first n-terms is given using the formula,

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

where 
n represents the number of terms of AP
a is the first term of the AP
d is the common difference of AP

Sum of n Terms in AP if Last Term is Given

If we are given the first and the last term of the AP then its sum can be easily calculated as,

Sum of n Terms = n/2(First Term + Last Term)

Suppose the given AP is a, a+d, a+2d, a+3d, a+4d, ………. a+ (n-1)d where the first term is a and the last term is l [a+ (n-1)d]. Then the sum of n terms of the AP is,

S_n = n/2(a + l)

Substituting the value of the first and the last term,

S_n = n/2(a + a + (n-1)d)

S_n = n/2(2a + (n-1)d)

which is the original formula for finding the sum of n-terms of the AP.

Sum of n Terms of AP Proof

Let’s consider the generalized representation of Arithmetic Progression, the sum of all the terms in the above sequence is given as

a, a+d, a+2d, a+3d, a+4d, ………. a+ (n-1)d

S_n = (a+ a+ d+ a+ 2d+ a+ 3d+ a+ 4d+….. a+ (n-1)d) ⇢ (i)

Rewriting the above equation in reverse order we get,

S_n = (a+ (n-1)d + a+ (n-2)d+ a+ (n-3)d+ ….. +a) ⇢ (ii)

Adding eq (i) and eq (ii)

2S_n = (2a+ (n-1)d + 2a+ (n-1)d+…….. + 2a+ (n-1)d) (n terms)

2S_n = [2a + (n-1)d] × n

Thus, the sum of n terms of the AP formula is Proved.

Sum of AP Formula for an Infinite AP

As we already know that AP are sequences that go up to infinity but finding the sum of AP up to an infinite term is a tedious task. If the AP is increasing AP the sum of the infinite terms of the AP is positive infinite but if the AP is decreasing AP then its sum of infinite terms is negative infinity. Before learning about them in detail let’s first learn what is Increasing AP, and what is decreasing AP.

Increasing AP

An AP where the next consecutive term is greater than the previous term is called an increasing AP. The common difference is always positive in increasing AP. (i.e. d > 0). Some examples of increasing AP are,

• S(n) = 2, 6, 10, 14,…
• S(n) = 5, 8, 11, 14,…
• S(n) = 6, 11, 16, 21,…

Sum of Infinite Terms of increasing AP =  ∞

This can be proved as, suppose we have to find the sum of S(n) = 2, 6, 10, 14,… up to infinite terms, the formula for the sum of n-terms is,

S_n = n/2(2a + (n-1)d)

Putting a = 2 and n = ∞ we get,

S_n = ∞/2[2(2) + (∞-1)d]

S_n = ∞

Decreasing AP

An AP where the next consecutive term is lesser than the previous term is called a decreasing AP. The common difference is always negative in decreasing AP. (i.e. d < 0). Some examples of decreasing AP are,

• S(n) = 8, 4, 0, -4,…
• S(n) = -2, -4, -6, -8,…
• S(n) = 10, 7, 4, 1,…

Sum of Infinite Terms of decreasing AP =  -∞

This can be proved as, suppose we have to find the sum of S(n) = 8, 4, 0, -4,… up to infinite terms, the formula for the sum of n-terms is,

S_n = n/2(2a + (n-1)d)

Putting a = 8 and n = ∞ we get,

S_n = ∞/2[2(8) + (∞-1)d]

As we know that if d is negative here.

S_n = -∞

We can summarize these conditions as,

• Sum of infinite AP = ∞, if d > 0
• Sum of infinite AP = -∞, if d > 0

Sum of AP Formulas

Various formulas are used for finding the sum of n-terms of AP are discussed in the table below,

Conditions	Formulas
Sum of n terms of AP when the first term is a	Sn = n/2[a+(n-1)d]
Sum of n terms of AP when the first term is a and the last term is l	Sn = n/2[a+l]
Sum of first n Natural Number	Sn = n(n+1)/2
Sum of square of first n Natural Number	Sn = n(n+1)(2n+1)/6
Sum of cube of first n Natural Number	Sn = [n(n+1)/2]2

Sum of Natural Numbers

The sum of the first 10,100, 1000, 10000, and 100000 natural numbers is discussed below.

Numbers	Sum
1-10	55
1-100	5050
1-1000	500500
1-10000	50005000
1-100000	500005000

Sum of Square Series

A square series is represented as,

1² + 2² + 3² + 4² + ………. + n²

The sum of n terms of the square series is given using the formula,

S_n = n(n+1)(2n+1)/6

Sum of Cubic Series

A square series is represented as,

1³ + 2³ + 3³ + 4³ + ………. + n³

The sum of n terms of the square series is given using the formula,

S_n = [n(n+1)/2]²

Read More,

• Geometric Progression
• Common Difference

Solved Examples on Sum of n Terms

Example 1: Consider the AP = 2, 4, 6, 8, 10,… Find the sum of the first 20 terms of this A.P.

Solution:

Given AP,

2, 4, 6, 8, 10,…

First Term (a) = 2

Common Difference (d) = 4 – 2 = 6 – 4 = 2

Sum of n terms is,

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

n = 20

S_n = 20/2 [2×2 + (20-1)×2]

S_n = 10(42) = 420

Thus, the sum of the of first 20 terms in the sequence are, 420

Example 2: Calculate the sum of the first 16 terms of the AP: S = 98 + 95 + 92 +…

Solution:

Given AP,

S = 98 + 95 + 92 +…

First Term (a) = 98

Common Difference (d) = 95 – 98 = 92 – 95 = -3

Sum of n terms is,

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

n = 16

S_n = 16/2 [2×98 + (16-1)×-3]

S_n = 8(151) = 1208

Thus, the sum of the of first 20 terms in the sequence are, 1208

Example 3: Calculate the sum of the first 24 terms of the AP: S = 8 + 13 + 18 +…

Solution:

Given AP,

S = 8 + 13 + 18 +…

First Term (a) = 8

Common Difference (d) = 13 – 8 = 18 – 13 = 5

Sum of n terms is,

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

n = 24

S_n = 24/2 [2×8 + (24-1)×5]

S_n = 8(131) = 1572

Thus, the sum of the of first 24 terms in the sequence are, 1572

Example 4: Find the sum of squares of the first 10 natural numbers.

Solution:

We know that formula for the sum of n terms of the square of the natural number is,

S_n = n(n+1)(2n+1)/6

For finding the square of the sum of the first 10 natural numbers

S_n = 10(10+1)(20+1)6

= 10(11)(21)/6

= 385

Thus, the sum of the first 10 squares of the natural number is 385.

FAQs on Sum of n Terms of an AP

Q1: What is an AP?

Answer:

AP or also called the Arithmetic Progression is a sequence of numbers which have a common difference, i.e. the difference between any two consecutive terms is always equal.

Q2: What is the Sum of n Terms of an AP Formula?

Answer:

The formula for finding the sum of n terms of the AP is,

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

where,
S_n represents the sum of the first n terms
a is the first term of the sequence
d is the common difference of the sequence

Q3: What is the Sum of n Terms of an AP when the last term is given?

Answer:

Suppose we have an AP whose first term is a and the last term is l then the sum of its first n term is given by,

S_n = n/2 [a + l]

Q4: What is the formula for the sum of the first n natural number?

Answer:

The formula for finding the sum of the first n natural number is,

S_n = n(n+1)/2