Question

How do you write the expression for the nth term of the given sequence 3, 7, 11, 15, 19, …?

Answer 1

We start solving the problem by writing each term of the sequence as the sum of two numbers. We then write the second term as a multiple of two numbers. We then make the necessary calculations and then check the form that each term resembles. We then substitute n in this resembled form and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the nth term of the given sequence 3, 7, 11, 15, 19, ….
Now, we have the sequence 3, 7, 11, 15, 19, ….
$\Rightarrow$ 3, 7, 11, 15, 19, … = $\left( 3+0 \right)$, $\left( 3+4 \right)$, $\left( 3+8 \right)$, $\left( 3+12 \right)$, $\left( 3+16 \right)$, ….
$\Rightarrow$ 3, 7, 11, 15, 19, … = $\left( 3+0\left( 4 \right) \right)$, $\left( 3+1\left( 4 \right) \right)$, $\left( 3+2\left( 4 \right) \right)$, $\left( 3+3\left( 4 \right) \right)$, $\left( 3+4\left( 4 \right) \right)$, ….
$\Rightarrow$ 3, 7, 11, 15, 19, … = $\left( 3+\left( 1-1 \right)\left( 4 \right) \right)$, $\left( 3+\left( 2-1 \right)\left( 4 \right) \right)$, $\left( 3+\left( 3-1 \right)\left( 4 \right) \right)$, $\left( 3+\left( 4-1 \right)\left( 4 \right) \right)$, $\left( 3+\left( 5-1 \right)\left( 4 \right) \right)$, ….
We can see that every term of the series resembles the form $3+\left( r-1 \right)4$, $1\le r<\infty$.
Now, let us substitute n in place of r to get the nth term of the given sequence.
So, the nth term of the given sequence is ${{T}_{n}}=3+\left( n-1 \right)4$.
$\Rightarrow {{T}_{n}}=3+4n-4$.
$\Rightarrow {{T}_{n}}=4n-1$.
$\therefore$ We have found the nth term of the given sequence 3, 7, 11, 15, 19, … as $4n-1$.

Note:
We can also solve the given problem as shown below:
We have given the sequence 3, 7, 11, 15, 19, ….
$\Rightarrow$ 3, 7, 11, 15, 19, … = $\left( 4-1 \right)$, $\left( 8-1 \right)$, $\left( 12-1 \right)$, $\left( 16-1 \right)$, $\left( 20-1 \right)$, ….
$\Rightarrow$ 3, 7, 11, 15, 19, … = $\left( 4\left( 1 \right)-1 \right)$, $\left( 4\left( 2 \right)-1 \right)$, $\left( 4\left( 3 \right)-1 \right)$, $\left( 4\left( 4 \right)-1 \right)$, $\left( 4\left( 5 \right)-1 \right)$, ….
We can see that every term of the series resembles the form $4r-1$, $1\le r<\infty$.
Now, let us substitute n in place of r to get the nth term of the given sequence.
So, the nth term of the given sequence is ${{T}_{n}}=4n-1$.