Question

How many arithmetic progression with 10 terms are there whose first term is in the set $\left\{ 1,2,3 \right\}$ and whose common difference is in the set $\left\{ 2,3,4 \right\}$ ?

Answer 1

Hint: For solving this question, first we will understand what we mean by arithmetic progression and ${{n}^{th}}$ term of the A.P. can be written as ${{a}_{n}}={{a}_{1}}+d\left( n-1 \right)$ . After that, we will find the number of ways in which we can select the first term and common difference for the A.P. from the given sets separately. Then, we will use the concept of the fundamental principle of multiplication and multiply both the results and solve for the final answer.

Complete step-by-step answer:
We have to find the total number of arithmetic progression with 10 terms are there whose first term is in the set $\left\{ 1,2,3 \right\}$ and whose common difference is in the set $\left\{ 2,3,4 \right\}$ .
Now, first, we will understand when a sequence is called an A.P. and what are important conditions for a sequence to be in arithmetic progression.
Arithmetic Progression:
In a sequence when the difference between any two consecutive terms is equal throughout the series then, such sequence will be called to be in arithmetic progression and the difference between consecutive terms is called as the common difference of the arithmetic progression. If ${{a}_{1}}$ is the first term of an A.P. and common difference of the A.P. is $d$ then, ${{n}^{th}}$ term of the A.P. can be written as ${{a}_{n}}={{a}_{1}}+d\left( n-1 \right)$.Now, from the above discussion, we conclude that for an arithmetic progression if, first term and the common difference is fixed. Then, the series will be fixed or in other words, we need two inputs for fixing any arithmetic progression.
Now, before we proceed we should know the following important concept and formulas:
Fundamental Principle of Multiplication:
If there are two jobs such that one of them can be completed in $m$ ways, and when it has been completed in any of these $m$ ways, the second job can be completed in $n$ ways. Then, two jobs in succession can be completed in $m\times n$ ways.
Now, we come back to our problem in which we have to find the total number of arithmetic progression with 10 terms are there whose first term is in the set $\left\{ 1,2,3 \right\}$ and whose common difference is in the set $\left\{ 2,3,4 \right\}$ .
Now, as in each of the given sets, there are 3 elements only. Then,
The number of ways in which we can select the first term of the A.P. from the set $\left\{ 1,2,3 \right\}=m=3$ .
The number of ways in which we can select a common difference for the A.P. from the set $\left\{ 2,3,4 \right\}=n=3$ .
Now, from the fundamental principle of multiplication, we can say that the total number of arithmetic progression with 10 terms are there whose first term is in the set $\left\{ 1,2,3 \right\}$ and whose common difference is in the set $\left\{ 2,3,4 \right\}$ will be equal to $m\times n=3\times 3=9$ ways.

Note: Here, the student should know the concept of A.P. and how to express the general expression of ${{n}^{th}}$ the term of an A.P. and important points should be remembered. After that, we should apply the fundamental principle of multiplication without any mistake directly to get the correct answer quickly.