Question

Solve the following question in detail:
How do you find the next term of the arithmetic sequence shown below?
\[\dfrac{1}{6},\dfrac{1}{3},\dfrac{1}{2},\dfrac{2}{3}\]

Answer 1

Hint: Consider the progression and find the type of the progression it is. Find the common difference by subtracting two consecutive terms. Then identify the first term and the term which we have to find out. Substitute it in the formula and simplify the terms to get the final answer.

Complete step-by-step solution:
Given sequence;
\[\dfrac{1}{6},\dfrac{1}{3},\dfrac{1}{2},\dfrac{2}{3}\]
The type of progression is the arithmetic progression since there is a common difference between each term of the sequence.
Let us find the common difference;
Common difference \[ = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}\]
Now the first term of the sequence is represented as \[{a_1}\]
\[{a_1} = \dfrac{1}{6}\]
The \[{n^{th}}\] term of the equation is represented as \[{a_n}\]
Let us take the common difference of the progression to be considered as \[d\].
\[d = \dfrac{1}{6}\]
We have the formula to find the \[{n^{th}}\] term of the progression.
\[ \Rightarrow {a_n} = {a_1} + d(n - 1)\]
Here in the progression, we have four terms given. We have to find the fifth term of the progression.
\[ \Rightarrow n = 5\]
Substituting the terms in the above formula, we get;
\[ \Rightarrow {a_5} = \dfrac{1}{6} + \dfrac{1}{6}\left( {5 - 1} \right)\]
Simplifying the parenthesis, we get;
\[ \Rightarrow {a_5} = \dfrac{1}{6} + \dfrac{1}{6} \times 4\]
Simplifying the multiplication of the terms, we get;
\[ \Rightarrow {a_5} = \dfrac{1}{6} + \dfrac{2}{3}\]
Adding both the terms after taking the L.C.M. and multiplying them to the opposite denominators, we get;
\[ \Rightarrow {a_5} = \dfrac{{1 + 4}}{6}\]
Now, adding the numerator, we get;
\[ \Rightarrow {a_5} = \dfrac{5}{6}\]

Therefore, the fifth term of the sequence \[ = \dfrac{5}{6}\].

Note: The arithmetic sequence is a list of numbers with some definite pattern and a common difference between them. Taking a number in the sequence and subtracting the previous number from this number, we get the common difference which remains constant throughout the sequence.