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.
For example, 3, 7, 11, 15,… is an arithmetic sequence where the difference between consecutive terms is 4.

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.

The sum of n Terms of AP Formula

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,

Sum of n Terms in AP if the 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,

S_n = n/2(a + l)

Where,
n represents the number of terms of AP
a is the first term of the AP
d is the common difference of AP
l is the last term

Example: Sum of Numbers from 1 to 100

The sum of n terms of an AP is considered to be the sum of n consecutive terms of any arithmetic sequence. 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

Using the AP Formula: Another method for larger or more general APs, where pairing is impractical, the formula for the sum of n terms is more efficient:

S_n = n/2(a + l)
For 1, 2, 3,…, 100:
n = 100,
a = 1,
l = 100.

S_n = 100/2(1 + 100) = 5050

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

S_n= \frac{n}{2}[2a+ (n-1)d]

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

Sum of AP Formula for an Infinite AP

As we know, an AP (Arithmetic Progression) is a sequence that can extend infinitely. However, finding the sum of an AP up to infinite terms is a challenging task.

• For an increasing AP, the sum of its infinite terms approaches positive infinity.
• For a decreasing AP, the sum of its infinite terms approaches negative infinity.

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 (the AP is increasing).
• Sum of infinite AP = -∞, if d < 0 (the AP is decreasing).

Sum of AP Formulas

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

Sum of Natural Numbers

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

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.