Arithmetic Sequence

If the numbers in a sequence follows arithmetic addition or subtraction, we call it arithmetic sequence. For example, 1, 3, 5, 7, \ldots ,33 is an arithmetic sequence.

Its first number, a = 1

Interval, d = 2 [ (second term -first term) or (third term – second term) and so on]

Last number, n = 33

The number of elements in this sequence is not equal to 33 since there are some missing numbers. We can calculate the number of elements in this series by using the following equation:

n^{th} \text{element} = a + (n-1)d \\

or, 33 = 1 + (n-1)2 \\

or, 33 = 1 + 2n - 2 \\

or, n = 17

Now, we will calculate the sum of all elements in the above series

1 + 3 + 5 + \ldots + 33 \\

Sum, S = \frac{n}{2}\left\{2a + (n-1)d\right\} \\

= \frac{17}{2}\left\{2\times1 + (17-1)\times2 \right\} \\ \\

= 17\times17 \\ \\

= 289

Interval can be negative number. For example, for the following sequence calculate number of elements and their summation

23 + 20 + 17 + \ldots + 5 \\ \\

n^{th} \: \text{element} = a + (n-1)d \\

or, 5 = 23 + (n-1)(-3) \\

or, 5 = 23 - 3n + 3 \\

or, 3n = 21 \\

or, n = 7

Sum, S = \frac{n}{2} \left\{2a + (n-1)d \right\} \\

= \frac{7}{2} \left\{2\times23 + (7-1)\times(-3)\right\} \\ \\

= 7\times14 \\ \\

= 98

Geometric Sequence:

If the numbers in a sequence follow some multiplication (greater or less than 1) factor, then it is known as geometric sequence. For example, 1, 2, 4, 8, \ldots , 256

Here first number, a = 1

Ratio, r = 2/1 = 4/2 = 2 (ratio of second term to first term or third term to second term)

n^{th} \text{term} = ar^{n-1} \\

or, 256 = 1\times2^{n-1} \\

or, 2^{n-1} = 2^8 \\

or, n-1 = 8 \\

or, n = 9

Sum, S = a \frac{1-r^n}{1-r} \\

= 1 \frac{1-2^9}{1-2} \\

= 511

Calculate the sum of first 10 elements of the following geometric series

1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots , \ldots

Here, a= 1, r= 1/2

Sum, S = a \frac{1-r^n}{1-r} \\

= 1. \frac{1- \frac{1}{2^{10}}}{1-\frac{1}{2}} \\

= 1.998

If instead of 10 elements, we want to determine the summation of the infinite sequence, then the above equation reduces to

Sum, S = a. \frac{1}{1-r} since \frac{1}{2^n} \text{becomes zero as} n \Rightarrow \infty

= 1. \frac{1}{1-\frac{1}{2}} \\

= 2