Question

How do you write the first six terms in the following sequence ${a_n} = 3n + 1$?

Answer 1

The sequence of each term gives us many different values. The sequence ${a_n}$ has a n number of values, when we put different values to the sequence it gives a different number.
Now we are going to substitute positive integers for n of order $n = 1,2,3,4,5,6$.

Complete step-by-step solution:
Now given sequence is ${a_n} = 3n + 1$,
Take $n$ values for 6 as we are asked to find the first 6 terms, so, $n = 1,2,3,4,5,6$,
Now substituting each values we get,
Now we take $n = 1$,
${a_n} = 3n + 1$,
Substituting $n = 1$ in the sequence we get,
$\Rightarrow {a_1} = 3\left( 1 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_1} = 3 + 1$,
Now adding we get,
$\Rightarrow {a_1} = 4$
So the first term is 4.
Now we take$n = 2$,
${a_n} = 3n + 1$,
Substituting $n = 2$ in the sequence we get,
$\Rightarrow {a_{}} = 3\left( 2 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_2} = 6 + 1$,
Now adding we get,
$\Rightarrow {a_2} = 7$,
So the second term is 7.
Now we take $n = 3$,
${a_n} = 3n + 1$,
Substituting $n = 3$ in the sequence we get,
$\Rightarrow {a_3} = 3\left( 3 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_3} = 9 + 1$,
Now adding we get,
$\Rightarrow {a_3} = 10$,
So the third term is 10.
Now we take $n = 4$,
${a_n} = 3n + 1$,
Substituting $n = 4$ in the sequence we get,
$\Rightarrow {a_4} = 3\left( 4 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_4} = 12 + 1$,
Now adding we get,
$\Rightarrow {a_4} = 13$,
So the third term is 13.
Now we take $n = 5$,
${a_n} = 3n + 1$,
Substituting $n = 5$ in the sequence we get,
$\Rightarrow {a_5} = 3\left( 5 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_5} = 15 + 1$,
Now adding we get,
$\Rightarrow {a_5} = 16$,
So the fifth term is 16.
Now we take $n = 6$,
${a_n} = 3n + 1$,
Substituting $n = 6$ in the sequence we get,
$\Rightarrow {a_6} = 3\left( 6 \right) + 1$,
Now simplifying we get,
$\Rightarrow {a_6} = 18 + 1$,
Now adding we get,
$\Rightarrow {a_6} = 19$,
So the sixth term is 19.
The first 6 terms for the sequence are 4, 7, 10, 13, 16, and 19.

$\therefore$The six terms of the sequence ${a_n} = 3n + 1$ are 4, 7, 10, 13, 16, and 19.

Note: A sequence is an ordered list of numbers.
Example: 1, 3, 5, 7, 9…….
The three dots mean to continue forward in the pattern established. Each number in the sequence is called a term. In the sequence 1, 3, 5, 7, 9, …, 1 is the first term, 3 is the second term, 5 is the third term, and so on. The notation ${a_1},{a_2},{a_3}........{a_n}$ is used to denote the different terms in a sequence.
The expression ${a_n}$ is referred to as the general or $n^{th}$ term of the sequence.