Grade 10 → Algebra → Arithmetic Progressions ↓
Sum of First n Terms of an AP
An Arithmetic Progression (AP) is a sequence of numbers where the difference of any two successive members is a constant. This constant difference, usually represented by the letter d
, is called the common difference.
For example, the sequence 2, 4, 6, 8, 10
is an arithmetic progression because the difference between successive terms is always 2. Another sequence is 5, 10, 15, 20
, where the common difference is 5.
Elements of arithmetic progression
Arithmetic progression is represented as:
A, A+D, A+2D, A+3D, ..., A+(n-1)D
a
is the first term.d
is the common difference.n
is the number of terms.
The sum of the first n terms of an arithmetic progression
To find the sum of the first n
terms of an AP, we use the following formula:
S n = n/2 × (2a + (n − 1) × d)
or equivalent
S n = n/2 × (a + l)
Where:
S n
is the sum of the firstn
terms.l
is the last term of the sequence. Thus,l = a + (n-1) × d
.
Derivation of the formula
Let us derive this formula with a simple logical interpretation. Consider a sequence of n
terms:
a, (a + d), (a + 2d), ..., [a + (n - 1)d]
Now, if we write the sequence in reverse order:
[a + (n - 1)d], [a + (n - 2)d], ..., (a + d), a
Adding these sequences term by term, we get the pairs:
(a + [a + (n - 1)d]), ((a + d) + [a + (n - 2)d]), ..., ([a + (n - 1)d] + a)
Each pair has the same sum:
2a + (n - 1)d
There are n
such pairs, so the total becomes:
n × [(2a + (n – 1)d) / 2] = n/2 × (2a + (n – 1)d)
Example 1
Consider the sequence 2, 5, 8, 11, 14
Find the sum of these first 5 terms.
Here, a = 2
, d = 3
, and n = 5
.
S 5 = 5/2 × (2 × 2 + (5 - 1) × 3)
S 5 = 5/2 × (4 + 12)
S 5 = 5/2 × 16 = 5 × 8 = 40
Hence, the sum of the first 5 terms is 40.
Visual example
Example 2
Let us find the sum of the first 10 terms of the arithmetic series 3, 7, 11, 15...
Here, a = 3
and d = 4
.
First, find the 10th term:
L = a + (n − 1) × d = 3 + (10 − 1) × 4 = 3 + 36 = 39
Now, use the sum formula:
S 10 = 10/2 × (3 + 39)
S 10 = 5 × 42 = 210
Therefore, the sum of the first 10 terms is 210.
Example 3: AP with negative difference
Consider an arithmetic progression where a = 20
, d = -3
, and n = 6
.
Sequence: 20, 17, 14, 11, 8, 5
.
Calculate the totals:
S 6 = 6/2 × (2 × 20 + (6 - 1) × (-3))
S 6 = 3 × (40 – 15)
S 6 = 3 × 25 = 75
Hence, the sum of the first 6 terms is 75.
Generalization
The formula S n = n/2 × (2a + (n - 1) × d)
is universally applicable to all arithmetic progressions - whether the common difference is positive, negative, or zero.
Mathematical explorations
This method of adding terms in an arithmetic progression illustrates an important concept: the balancing of increasing and decreasing numbers to get a total. In arithmetic progressions, this occurs because the progressive addition of the common difference creates a symmetric operation from both ends of the progression.
Practical applications
Understanding the sum of an arithmetic progression is useful in real life where situations proceed in a linear manner, such as calculating the total amount of equally spaced installments, or estimating the amount of savings over a period of time.