Question

How do you write the first five terms of the arithmetic sequence given ${a_1} = 15,{a_{k + 1}} = {a_k} + 4$ and find the common difference and write the ${n^{th}}$ term of the sequence as a function of n?

Answer 1

Hint: As we know, the common difference is the difference between any two successive terms. So, move ${a_k}$ on the left side to get a common difference. After that, substitute $k = 1,2,3,4$ to get the first five terms of the arithmetic sequence. Then, substitute the value in the general term to find the ${n^{th}}$ term.

Complete step-by-step answer:
The given sequence is ${a_{k + 1}} = {a_k} + 4$.
Move ${a_k}$ on the left side to get a common difference.
$ \Rightarrow {a_{k + 1}} - {a_k} = 4$
So, the common difference is 4.
Let us find the second term of the series by substituting $k = 1$,
$ \Rightarrow {a_{1 + 1}} = {a_1} + 4$
As ${a_1} = 15$. Substitute the value,
$ \Rightarrow {a_2} = 15 + 4$
Add the terms,
$ \Rightarrow {a_2} = 19$
Let us find the third term of the series by substituting $k = 2$,
$ \Rightarrow {a_{2 + 1}} = {a_2} + 4$
As ${a_2} = 19$. Substitute the value,
$ \Rightarrow {a_3} = 19 + 4$
Add the terms,
$ \Rightarrow {a_3} = 23$
Let us find the fourth term of the series by substituting $k = 3$,
$ \Rightarrow {a_{3 + 1}} = {a_3} + 4$
As ${a_3} = 23$. Substitute the value,
$ \Rightarrow {a_4} = 23 + 4$
Add the terms,
$ \Rightarrow {a_4} = 27$
Let us find the second term of the series by substituting $k = 4$,
$ \Rightarrow {a_{4 + 1}} = {a_4} + 4$
As ${a_4} = 27$. Substitute the value,
$ \Rightarrow {a_5} = 27 + 4$
Add the terms,
$ \Rightarrow {a_5} = 31$
So, the first five terms of the sequence are $15,19,23,27,31$.
The general term of the sequence is,
${a_n} = {a_1} + \left( {n - 1} \right)d$
Now, substitute the value of the first term and the common difference in the general term,
$ \Rightarrow {a_n} = 15 + \left( {n - 1} \right) \times 4$
Open bracket and multiply the terms,
$ \Rightarrow {a_n} = 15 + 4n - 4$
Simplify the terms,
$\therefore {a_n} = 4n + 11$
So, the ${n^{th}}$ term is $4n + 11$.

Hence, the first five terms of the sequence are $15,19,23,27,31$ and the ${n^{th}}$ term is $4n + 11$.

Note:
A sequence is said to be A.P if and only if the common difference that is the difference between the consecutive terms remains constant throughout the series. It is always advisable to remember all the series-related formulas whether it is of nth term or the sum of nth as it helps to save a lot of time.