Examples of arithmetic sequences

Examples of arithmetic sequences. Here it’s all about the arithmetic sequence. Have a look!

Arithmetic sequence

Well, it’s an old topic from high school. So I’ll not go into much detail.

Suppose I have a sequence like $F(n) = A + (n-1) D$ .

Here $A$ is the first term and $D$ is the common difference in the sequence.

Then this sequence is an arithmetic sequence.

There are also other sequences like geometric sequence, harmonic sequence and so on.

Now I’ll give some examples of arithmetic sequences.

Examples of arithmetic sequences

Here are some examples of arithmetic sequences.

Disclaimer: None of these examples is mine. I have chosen these from some book or books. The references are at the end of the post.

Example 1

According to Stroud and Booth (2013)* “The seventh and eleventh terms of an arithmetic sequence are $7b + 5c$ and $11b + 9c$ . Find the first term and the common difference.”

Solution

The standard form of an arithmetic sequence is

(1) $\begin{equation*} f (n) = A + (n-1) D. \end{equation*}$

Here $A$ is the first term and $D$ is the common difference.

Also, I already know the seventh and eleventh terms of the sequence as $7b + 5c$ and $11b + 9c$ respectively.

So, first of all, I’ll put $n = 7$ in equation (1). Thus I get

$\begin{equation*} f (7) = A + 7 D. \end{equation*}$

But in the example, the seventh term is $7b + 5c$ .

Thus it will be

(2) $\begin{equation*} A + 6 D = 7b + 5c. \end{equation*}$

Next, I’ll put $n = 11$ in equation (1). Thus I get

$\begin{equation*} f (11) = A + 10 D. \end{equation*}$

But in the example, the eleventh term is $11b + 9c$ .

Thus it will be

(3) $\begin{equation*} A + 10 D = 11b + 9c. \end{equation*}$

Now I’ll solve equations (2) and (3) to get the values of $A$ and $D$ .

So, I’ll subtract equation (2) from equation (3).

Therefore I get

$\begin{equation*} 4D = 4b + 4c. \end{equation*}$

So it means $D = b + c$ .

Now I’ll substitute this value of $D$ in equation (2).

Therefore I get

$\begin{equation*} A + 6 (b + c) = 7b + 5c. \end{equation*}$

Next, I’ll simplify it to get the value of $A$ .

Thus the value of $A$ is

$\begin{eqnarray*} A + 6 b + 6c &=& 7b + 5c\\ A &=& 7b + 5c -6b -6c\\ A &=& b - c. \end{eqnarray*}$

So I can conclude that the first term and the common difference are $(b - c)$ and $(b + c)$ respectively.

This is the answer to this example.

Now I’ll go to my next example.

Example 2

According to Stroud and Booth (2013)* “If $x, 5x, 6x + 9$ form three successive terms of an arithmetic sequence, find the next four terms.”

Solution

First of all, I’ll choose the general term of the arithmetic sequence, say $f(m) = A + (m - 1)D$ .

Here $A$ is the first term and $D$ is the common difference.

Let me choose three successive terms in this sequence.

So let them be $f(m), f(m + 1)$ and $f(m + 2)$ .

So that means,

$\begin{eqnarray*}f(m) &=& A + (m - 1)D \\ f (m+ 1) &=& A + m D\\ f(m+ 2) &=& A + (m + 1)D. \end{eqnarray*}$

But from the example, I already know that the three successive terms are $x, 5x, 6x + 9$ .

Therefore I can say

(4) $\begin{equation*} A + (m - 1) D = x, \end{equation*}$

(5) $\begin{equation*} A + m D = 5x \end{equation*}$

and

(6) $\begin{equation*} A + (m + 1)D = 6x + 9. \end{equation*}$

Now my job is to find out the values of $A$ , $D$ and $x$ .

Step 1

First of all, I’ll add equations (4), (5) and (6).

Therefore I get

$\begin{equation*}3A + 3 m D = 12x + 9. \end{equation*}$

This gives

$A + m D = 4x + 3.$

But from equation (5), I already knpw that

$A + m D =5x.$

So this means

$\begin{equation*}4x + 3 = 5x. \end{equation*}$

Thus it gives $x = 3$ .

Therefore the three successive terms of the arithmetic sequence are $3, 5 (3), 6(3) + 9$ . This gives the terms as $3, 15, 27.$

Thus I can say the first term $A$ is 3. the common difference $D$ is $(15 - 3) = 12$ .

Hence I can say that the general form of this arithmetic sequence is $f(m) = 3 + (m-1)12.$

Now I already know the first three terms, that is $f(1), f(2), f(3)$ .

Next, I’ll find out the values of $f(4), f(5), f(6), f(7).$

Thus it will be

$\begin{eqnarray*}f (4) &=& 3 + (4-1) 12 = 3 + (3)12 = 39\\ f(5) &=& 3 + (5-1) 12 = 3 + (4)12 = 51\\f (6) &=& 3 + (6 - 1)12 = 3 + (5)12 = 63\\ f(7) &=& 3 + (7-1)12 = 3 +(6) 12 = 75.\end{eqnarray*}$

So the next four terms in this sequence are 39, 51, 63 and 75.

This is the answer to this example.

Now I’ll go to the next example.

Example 3

According to Stroud and Booth (2013)* “If $\cfrac{1}{b+c}, \cfrac{1}{c+a}, \cfrac{1}{a+b}$ form three successive terms of an arithmetic sequence, show that $a^2, b^2, c^2$ also form three successive terms of another arithmetic sequence.”

Solution

Here $\cfrac{1}{b+c}, \cfrac{1}{c+a}, \cfrac{1}{a+b}$ are three successive terms of an arithmetic sequence. So I can say

$\frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}.$

This gives

(7) $\begin{equation*}\frac{2}{c + a} = \frac{1}{a + b} + \frac{1}{b + c}.\end{equation*}$

Now I have to show that $a^2, b^2, c^2$ also form three successive terms of another arithmetic sequence. This means that I have to prove

$2 b^2 = a^2 +c^2.$

So, first of all, I’ll simplify equation (7).

Thus it will be

$\begin{eqnarray*}\frac{2}{c + a} &=& \frac{a+ b+b+c}{(a+ b)(b + c)}\\ &=& \frac{a+ 2b+c}{(a+ b)(b + c)}\\ 2(a+b)(b+c) &=&(c+a)(a+2b+c) \\2(ab + b^2 + ac +bc) &=& ca + a^2 + 2bc + 2ab + c^2 + ac\\2ab + 2b^2 + 2ac +2bc &=& a^2 +c^2 +2ab+2bc+2ca\\ 2b^2 &=& a^2 + c^2. \end{eqnarray*}$

So this means $a^2, b^2, c^2$ also form three successive terms of another arithmetic sequence.

Hence I have proved this statement.

This is the answer to this last example.

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering mathematics, Industrial Press, Inc.; 7th Edition (March 8, 2013), Chapter: Sequences, Further problems 10, p. 637, Q. No. 12 (Example 1); Q. No. 13 (Example 2); Q. No. 17 (Example 3) .

Arithmetic sequence

Examples of arithmetic sequences

Example 1

Solution

Example 2

Solution

Step 1

Example 3

Solution

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering mathematics, Industrial Press, Inc.; 7th Edition (March 8, 2013), Chapter: Sequences, Further problems 10, p. 637, Q. No. 12 (Example 1); Q. No. 13 (Example 2); Q. No. 17 (Example 3) .

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