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Question

Question: In the sequence \[1,2,2,3,3,3,4,4,4,4,...\] where n consecutive terms have the value \(n\), find the...

In the sequence 1,2,2,3,3,3,4,4,4,4,...1,2,2,3,3,3,4,4,4,4,... where n consecutive terms have the value nn, find the 150150 th term of the sequence.

Explanation

Solution

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,...1,2,2,3,3,3,4,4,4,4,... where n consecutive terms have the value n. We have to find the 150150 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 150150 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,...1,2,2,3,3,3,4,4,4,4,...
Let the sequence be an{a_n} , then
an=1,2,2,3,3,3,4,4,4,4,...\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 11. Here each term repeats itself equal to the number of times of its own value, i.e. the nn consecutive terms have the value nn. We have to find the 150150 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 11 is 11 .
The last position of the natural number 22 is 33 .
The last position of the natural number 33 is 66 .
The last position of the natural number 44 is 1010 .
Therefore, there exists a unique pattern for each term’s last position.

In general, the last position for a natural number term nn is given by the expression,
n(n+1)2\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 150150 , let us take the natural number 1616 i.e. the term n=16n = 16 , that gives us the last position of 1616 as,
16(17)2=2722\Rightarrow \dfrac{{16\left( {17} \right)}}{2} = \dfrac{{272}}{2}
That is,
136\Rightarrow 136
Which is less than 150150 .
Now, we will take the next term, i.e. for n=17n = 17 .
Hence, the last position of the natural number 1717 is given by,
17(18)2=3062\Rightarrow \dfrac{{17\left( {18} \right)}}{2} = \dfrac{{306}}{2}
That gives us,
153\Rightarrow 153
Which is greater than 150150 .
That means the 150150 th term is also 1717 .

Therefore, the 150150 th term of the sequence 1,2,2,3,3,3,4,4,4,4,...1,2,2,3,3,3,4,4,4,4,... is 1717.

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 an{a_n} can be obtained by applying that unique rule.