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Question: Identify the pattern with which each number in the list of numbers \[12,23,34,45,56,...\;\] is forme...

Identify the pattern with which each number in the list of numbers 12,23,34,45,56,...  12,23,34,45,56,...\; is formed.
A. New number == preceding number +11 + 11
B. New number == preceding number +12n−1 + 12n - 1
C. New number == preceding number 11n11n
D. There is no specific pattern between its numbers.

Explanation

Solution

In the above given question, we are given a list of numbers written as 12,23,34,45,56,...  12,23,34,45,56,...\; . Clearly given list of numbers is a sequence of natural numbers. As we can see the first term of this sequence is 1212 , the second term is then 2323 , and so on. We have to find the pattern involved in the given sequence. In order to approach the solution, we can start by looking for the difference between two consecutive terms of the given sequence to find the pattern that is involved in the above given sequence.

Complete step by step answer:
Given sequence is 12,23,34,45,56,...  12,23,34,45,56,...\; We need to find the pattern of the above sequence. First we should check the difference between two consecutive terms of this sequence and see if they have a similar pattern in each difference or not.Now, as it is given that the first term of the sequence is 1212 and the second term is 2323.
Hence we can see that the difference of these two terms is 23−12=1123 - 12 = 11 .
Also, the difference between the second and the third term is 34−23=1134 - 23 = 11 .

Similarly, the difference between the third and the fourth term is 45−34=1145 - 34 = 11 .
And the difference between the fourth and the fifth term is 56−45=1156 - 45 = 11 .
Therefore, we have found that the difference of any two consecutive terms of this sequence is equal to 1111 . Thus, that means each next term of this sequence is 1111 more than its preceding term. Hence the pattern involved in this sequence is given by,
New number == preceding number +11 + 11

So the correct option is A.

Note: Since in the given sequence 12,23,34,45,56,...  12,23,34,45,56,...\; , each next term is 1111 more than its preceding term i.e. the difference between any two consecutive terms of this sequence is constant and is equal to 1111 . Now, since the first term of this sequence is 1212 , therefore we can say that the given list of numbers is nothing else but an Arithmetic Progression i.e. AP where the first term is 1212 and the common difference is 1111 .
Hence, given AP can be written as,
⇒12,23,34,45,56,67,78,89,100...  \Rightarrow 12,23,34,45,56,67,78,89,100...\;
⇒an=a(n+1)d\Rightarrow {a_n} = a\left( {n + 1} \right)d
Where a=12a = 12 , d=11d = 11 and an{a_n} is the nthnth term of the AP.