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An arithmetic progression (AP) is a sequence of numbers in which the difference between any two successive members is always the same. This fixed difference is called the common difference.
If the first term is a, the last term is l, and the common difference is d, then the number of terms n satisfies the relation
l = a + (n - 1)d

Thus,
n = (l - a)/d + 1
The sum of the first n terms, Sn, can be obtained in two common forms.
Sn = n/2 [2a + (n - 1)d]
or, if the last term l = Tn is known,
Sn = n/2 (a + l)

where a = first term, d = common difference, and Tn = l = a + (n - 1)d.
If three numbers a, b, c are in AP, then b is the arithmetic mean of a and c. In symbols,
b = (a + c)/2

The sum of the first n natural numbers is given by S = n(n + 1)/2. A standard derivation is shown below.
Write the sum forwards and backwards and add corresponding terms.
Let S = 1 + 2 + 3 + ... + (n - 1) + n
Write S in reverse order: n + (n - 1) + (n - 2) + ... + 2 + 1
Add the two expressions termwise to obtain 2S = (n + 1) + (n + 1) + ... + (n + 1) (n terms)
Therefore 2S = n(n + 1)
S = n(n + 1)/2

The sum of squares of the first n natural numbers is
S = 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)/6

The sum of the first n odd numbers is
S = n2
That is, 1 + 3 + 5 + ... (n terms) = n2.
The sum of the first n even numbers is
S = n(n + 1)
That is, 2 + 4 + 6 + ... (n terms) = n(n + 1).
A geometric progression (GP) is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio.
If successive terms are a1, a2, a3, ..., then the common ratio r is
r = a2 / a1 = a3 / a2 = ...

For r ≠ 1, the sum of the first n terms is
Sn = a1 (1 - rn)/(1 - r)


Note: the algebraic form is the same for r > 1 and for |r| < 1; the displayed placeholders correspond to the original presentation. When r = 1, every term equals a1 and Sn = n·a1.
If |r| < 1, an infinite GP has a finite sum. The sum to infinity is
S∞ = a1 / (1 - r)

If |r| ≥ 1, the sum to infinity does not exist (it diverges).
Solution:
First term a = 4
Common difference d = 4
Use Tn = a + (n - 1)d
T10 = 4 + (10 - 1) × 4
T10 = 4 + 9 × 4
T10 = 4 + 36
T10 = 40
Solution:
First term a = 3
Common difference d = 4
Number of terms n = 20
Use Sn = n/2 [2a + (n - 1)d]
S20 = 20/2 [2 × 3 + (20 - 1) × 4]
S20 = 10 [6 + 19 × 4]
S20 = 10 [6 + 76]
S20 = 10 × 82
S20 = 820
Solution:
First term a1 = 5
Common ratio r = (5/2)/5 = 1/2
Since |r| < 1, S∞ = a1/(1 - r)
S∞ = 5/(1 - 1/2)
S∞ = 5/(1/2)
S∞ = 10
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| 1. What is an arithmetic progression? | ![]() |
| 2. What is the formula to find the nth term of an arithmetic progression? | ![]() |
| 3. How can we find the sum of the first n terms of an arithmetic progression? | ![]() |
| 4. Can an arithmetic progression have a negative common difference? | ![]() |
| 5. How can we determine the number of terms in an arithmetic progression? | ![]() |