Question

Find the $\mathop{n}^{th}$ term of $3,6,9,12,15............$

Answer 1

Hint: First, students have to identify the series that the given series is in arithmetic progression, geometric progression or in harmonic progression. Then, apply the formula accordingly. If the common difference (second term-first term) is the same between successive numbers, then the series is in arithmetic progression.

Complete step by step answer:
The given series is $3,6,9,12,15............$
The given series is in arithmetic progression.
The $\mathop{n}^{th}$ term of an A.P. is given by the formula ${{a}_{n}}=a+\left( n-1 \right)d$
Where, ${{a}_{n}}$ = $\mathop{n}^{th}$ term of the sequence
$a$ = First term of the sequence
$d$ = common difference of successive numbers
$n$ = number of terms
Now, we have to find the $\mathop{n}^{th}$ term of the given sequence.
So, first we calculate the value of $d$ i.e. common difference.
$d$ = Second term – first term
 $d=6-3=3$
First term $=3$
Now put all these values in the formula, we get
${{a}_{n}}=a+\left( n-1 \right)d$
$\begin{align}
  & =3+(n-1)3 \\
 & =3+3n-3 \\
 & =3n \\
\end{align}$
So, the $\mathop{n}^{th}$ term of the given sequence is $3n$

Note: Before solving these types of questions, students must remember that first they should analyze the pattern of series and then solve accordingly. An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. If the given sequence is an arithmetic progression then first calculate the common difference and then, put the values in the formula. Now, suppose we had a common ratio between the terms of the series, then we would have gone for geometric progression.