Arithmetic Sequence and Series

An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.

In the arithemetic sequence
The constant difference in all pairs of consecutive or successive numbers is called the common difference. It is denoted by the letter d.
Please note that
Difference here means the second minus the first. We use the common difference to go from one term to another.
A sequence is said to be in a progression if its last term can be represented using a formula.

An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For example,
the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common difference of 2.

General term of arithmetic sequence

An arithmetic sequence with a starting value a and common difference d is a sequence of the form
A recursive definition for this sequence has two parts
a 1=a: initial condition
an=an-1+d for n>1:recursive formula
Therefore, an explicit definition for this sequence is a single formula as general terms given by
a n=a+(n-1)d for n > 1

Arithmetic mean

Let a, b, and c are three terms in an arithmetic progression (AP), then b is called Arithemetic mean of a and c.
In this case
, first two terms a and b will have the difference which will be equal to the next two terms b and c.
So we can say,
b-a = c-b.
Rearranging the terms, we get
or 2b = a + c
or b = \( \frac{a+c}{2} \)

Arithmetic Series

To find the explicit formula for the Sn: Sum of arithmetic sequence up to n th term, we write
Sn=a+(a+d)+(a+2d)+(a+3d)+---+[a+(n-2)d]+[a+(n-1)d] (1)
Writing in reverse order, we get
Sn=[a+(n-1)d]+[a+(n-2)d]+...+(a+3d)+(a+2d)+(a+d)+a (2)
Adding (1) and (2), we get
or Sn=[a+(n-1)d]+[a+(n-2)d]+[a+(n-3)d]+---+(a+d)+a
or 2Sn=2a+(n-1)d+2a+(n-1)d+2a+(n-1)d+---+2a+(n-1)d+2a+(n-1)d
or 2Sn=n[2a+(n-1)d]
or Sn=\(\frac{n}{2}[2a+(n-1)d] \)

Example 1

If the first term of an AP is 67 and the common difference is -13, find the sum of the first 20 terms.
Here, a = 67 and d= -13, thus
Sn = \(\frac{n}{2}[2a+(n-1)d]\)
or Sn = \(\frac{20}{2}[2 \times 67+(20-1)(-13)]=-1130\)
So, the sum of first 20 terms is -1130.

Example 2

[Modeling]:Look at the Figure below. The length of the side of each cube is 1cm. Copy the figure in isometric dot paper Based on the pattern

  1. Draw Figure no. 4 of this pattern on the dot paper
  2. Find the volume of the four Figures
  3. Write an equation to represent the volume of Figure n
  4. What would be the volume of Figure no. 12 of this pattern

Sum of Natural Numbers

Example 3

Show that sum of first n odd natural number is Sn=\(n^2\)
Here, a=1,d=2, thus the sum is
Sn=\( \frac{n}{2} [2a+(n-1)d]\)
or Sn=\( \frac{n}{2} [2.1+(n-1)2]\)
or Sn=\(\frac{n}{2} [2+2n-2]\)
or Sn=\(n^2\)

Example 4

Show that sum of first n even natural number is Sn=n(n+1)
Here, a=2,d=2, thus the sum is
Sn=\(\frac{n}{2} [2a+(n-1)d]\)
or Sn=\(\frac{n}{2} [2.2+(n-1)2]\)
or Sn=\(\frac{n}{2} [4+2n-2]\)
or Sn=n(n+1)

Example 5

Show that sum of first n natural number is Sn=\( \frac{n(n+1)}{2} \)
Suppose that 1+2+3+...+n=Sn, then we start from

\( r^2-(r-1)^2\) \(=2r-1\)
Taking r=1 \( 1^2-0^2\) \( =2 \times 1-1\)
Taking r=2 \( 2^2-1^2\) \( =2 \times 2-1\)
Taking r=3 \( 3^2-2^2\) \( =2 \times 3-1\)
------- ------- -------
Taking r=(n-1) \( (n-1)^2-(n-2)^2\) \( =2 \times (n-1)-1\)
Taking r=n \( n^2-(n-1)^2\) \( =2 \times n-1\)

Adding all we get
or \(n^2\)=2(Sn)-n
or 2(Sn)=\( n^2+n \)
or Sn=\( \frac{n(n+1)}{2}\)

Example 6

Show that sum of square of first n natural number is Sn=\(\frac{n(n+1)(2n+1)}{6} \)
Suppose that \( 1^2+2^2+3^2+...+n^2=\) Sn, then we start from

\( r^3-(r-1)^3\) \(=3 r^2- 3 r +1\)
Taking r=1 \( 1^3-0^3\) \(=3 \times 1^2- 3 \times 1 +1 \)
Taking r=2 \( 2^3-1^3\) \(=3 \times 2^2- 3 \times 2 +1 \)
Taking r=3 \( 3^3-2^3\) \(=3 \times 3^2- 3 \times 3 +1\)
------------- ------------- -------------
Taking r=(n-1) \( (n-1)^3-(n-2)^3\) \(=3 \times (n-1)^2- 3 \times (n-1) +1\)
Taking r=n \( n^3-(n-1)^3\) \(=3 \times n^2- 3 \times n +1\)

Adding all we get
\( n^3=3(1^2+2^2+3^2+...+n^2)-3(1+2+3+...+n)+n\)
or \(n^3=3(S_n)-3\left( \frac{n(n+1)}{2}\right )+n\)
or \(3(S_n)=n^3+3\left( \frac{n(n+1)}{2}\right )-n\)
or \(3(S_n)=n(n^2-1)+3\left( \frac{n(n+1)}{2}\right )\)
or \(3(S_n)=n(n-1)(n+1)+3\left( \frac{n(n+1)}{2}\right )\)
or \(3(S_n)=n(n+1)\left\{ (n-1)+\frac{3}{2}\right \}\)
or \(3(S_n)=n(n+1)\frac{(2n+1)}{2}\)
or \(S_n)=\frac{n(n+1)(2n+1)}{6}\)

Example 7

Show that sum of cube of first n natural number is Sn=\(\left (\frac{n(n+1)}{2} \right )^2 \)
Suppose that \(1^3+2^3+3^3+...+n^3\) =Sn, then we start from

\( r^4-(r-1)^4\) \(=4 r^3- 6 r^2+ 4 r-1\)
Taking r=1 \(1^4-0^4\) \(=4 \times 1^3- 6 \times 1^2+ 4\times 1-1 \)
Taking r=2 \(2^4-1^4\) \(=4 \times 2^3- 6 \times 2^2+ 4\times 2-1 \)
Taking r=3 \(3^4-2^4\) \(=4 \times 3^3- 6 \times 3^2+ 4\times 3-1\)
............................. ............................ ............................
Taking r=(n-1) \((n-1)^4-(n-2)^4\) \(=4 \times (n-1)^3- 6 \times (n-1)^2+ 4\times (n-1)-n \)
Taking r=n \(n^4-(n-1)^4\) \(=4 \times n^3- 6 \times n^2+ 4\times n-1\)

Adding, we get
\( n^4=4 (1^3+2^3...+ n^3)- 6 (1^2+2^2+...+ n^2)+ 4(1+2...+n)-n \)
or \(n^4\)=4 (Sn)- \(6 \frac{n(n+1)(2n+1)}{6}+ 4 \frac{n(n+1)}{2}-n \)
or 4 (Sn)=\( n^4+ n(n+1)(2n+1)- 2 n(n+1)+n \)
or 4Sn= \( n^4+ 2 n^3 +n^2 \)
or Sn=\( \left (\frac{n(n+1)}{2} \right )^2 \)

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