Question

In the sequence \[1,2,2,3,3,3,4,4,4,4,...\] where n consecutive terms have the value $n$, find the \[150\] th term of the sequence.

Answer 1

Hint: In the above given problem, we are given an infinite sequence of natural numbers which is given as \[1,2,2,3,3,3,4,4,4,4,...\] where n consecutive terms have the value n. We have to find the \[150\] th term of this given sequence. In order to approach the solution, first we have to find the pattern of each term’s last position in this sequence.Then the term which has its value on the position of \[150\] will be the required term of the sequence.

Complete step by step answer:
Given that, an infinite sequence of natural numbers that is written as \[1,2,2,3,3,3,4,4,4,4,...\]
Let the sequence be \[{a_n}\] , then
\[ \Rightarrow {a_n} = 1,2,2,3,3,3,4,4,4,4,...\]
The above given sequence has an infinite number of terms which are all natural numbers, starting from \[1\]. Here each term repeats itself equal to the number of times of its own value, i.e. the \[n\] consecutive terms have the value \[n\]. We have to find the \[150\] th term of this sequence.Now, notice each of the natural number term’s positions, in fact their last positions.
Here, the last position of the natural number \[1\] is \[1\] .
The last position of the natural number \[2\] is \[3\] .
The last position of the natural number \[3\] is \[6\] .
The last position of the natural number \[4\] is \[10\] .
Therefore, there exists a unique pattern for each term’s last position.

In general, the last position for a natural number term \[n\] is given by the expression,
\[ \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\]
Now we can find the last position of each natural number.
To be near the position of \[150\] , let us take the natural number \[16\] i.e. the term \[n = 16\] , that gives us the last position of \[16\] as,
\[ \Rightarrow \dfrac{{16\left( {17} \right)}}{2} = \dfrac{{272}}{2}\]
That is,
\[ \Rightarrow 136\]
Which is less than \[150\] .
Now, we will take the next term, i.e. for \[n = 17\] .
Hence, the last position of the natural number \[17\] is given by,
\[ \Rightarrow \dfrac{{17\left( {18} \right)}}{2} = \dfrac{{306}}{2}\]
That gives us,
\[ \Rightarrow 153\]
Which is greater than \[150\] .
That means the \[150\] th term is also \[17\] .

Therefore, the \[150\] th term of the sequence \[1,2,2,3,3,3,4,4,4,4,...\] is \[17\].

Note: A sequence is an arrangement of a number of finite or infinite objects or elements, or a set of numbers in a particular order which is followed by some unique rule. Here, the order i.e. rank or position of each element is unique and special. The nth term of the sequence \[{a_n}\] can be obtained by applying that unique rule.