Question

How do you write the first five terms of the sequence ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$?

Answer 1

Here, we will find the terms in a sequence by using the given ${n^{th}}$ term of an AP i.e. the given equation. Then we will substitute different values of $n$, to find the required consecutive terms. An arithmetic sequence is a sequence of numbers such that the common difference between any two consecutive numbers is a constant.

Complete Step by Step Solution:
The equation is the ${n^{th}}$ term of an AP.
First, we will find the first term of the sequence by substituting $n = 1$ in ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$. Therefore, we get
${a_1} = \dfrac{{3{{\left( 1 \right)}^2} - \left( 1 \right) + 4}}{{2{{\left( 1 \right)}^2} + 1}}$
Simplifying the equation, we get
$\Rightarrow {a_1} = \dfrac{{3 - 1 + 4}}{{2 + 1}}$
Adding and subtracting the terms, we get
$\Rightarrow {a_1} = \dfrac{6}{3}$
$\Rightarrow {a_1} = 2$
Now, we will find the second term of the sequence by substituting $n = 2$ in ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$, so we get
${a_2} = \dfrac{{3{{\left( 2 \right)}^2} - \left( 2 \right) + 4}}{{2{{\left( 2 \right)}^2} + 1}}$
Simplifying the equation, we get
$\Rightarrow {a_2} = \dfrac{{12 - 2 + 4}}{{8 + 1}}$
Adding and subtracting the terms, we get
$\Rightarrow {a_2} = \dfrac{{14}}{9}$
Now, we will find the third term of the sequence by substituting $n = 3$ in ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$. Therefore, we get
${a_3} = \dfrac{{3{{\left( 3 \right)}^2} - \left( 3 \right) + 4}}{{2{{\left( 3 \right)}^2} + 1}}$
Simplifying the equation, we get
$\Rightarrow {a_3} = \dfrac{{27 - 3 + 4}}{{18 + 1}}$
Adding and subtracting the terms, we get
$\Rightarrow {a_3} = \dfrac{{28}}{{19}}$
Now, we will find the fourth term of the sequence by substituting $n = 4$ in ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$. So, we get
${a_4} = \dfrac{{3{{\left( 4 \right)}^2} - \left( 4 \right) + 4}}{{2{{\left( 4 \right)}^2} + 1}}$
Simplifying the equation, we get
$\Rightarrow {a_4} = \dfrac{{48 - 4 + 4}}{{32 + 1}}$
Adding and subtracting the terms, we get
$\Rightarrow {a_4} = \dfrac{{48}}{{33}}$
Now, we will find the fifth term of the sequence by substituting $n = 5$ in ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$. Therefore, we get
$\Rightarrow {a_5} = \dfrac{{3{{\left( 5 \right)}^2} - \left( 5 \right) + 4}}{{2{{\left( 5 \right)}^2} + 1}}$
Simplifying the equation, we get
${a_5} = \dfrac{{75 - 5 + 4}}{{50 + 1}}$
Adding and subtracting the terms, we get
$\Rightarrow {a_5} = \dfrac{{74}}{{51}}$

Therefore, the first five term of the sequence ${a_n} = \dfrac{{3{n^2} - n + 4}}{{2{n^2} + 1}}$ are $2,\dfrac{{14}}{9},\dfrac{{28}}{{19}},\dfrac{{48}}{{33}},\dfrac{{74}}{{51}}$.

Note:
We know that a sequence of real numbers is defined as an arrangement or a list of real numbers in a specific order. We should know that if a sequence has only a finite number of terms then it is called a finite sequence and if a sequence has infinitely many terms, then it is called an infinite sequence. If we are given a general term of a sequence and then we will be able to find any particular term of the sequence directly.