Question

The nth term of a pattern of number is $ \left( {2n + 1} \right) $ . Is the pattern of numbers, so formed, in A.P? If so, find its 12th term.

Answer 1

Hint: A series or a sequence is said to be in Arithmetic progression only if every term of the series starting from the second term is obtained by adding a fixed number to its previous term. This fixed number is called the common difference. So here we have the nth term as $ \left( {2n + 1} \right) $ . Using this, find 1st term, 2nd term, 3rd term, so on. And notice whether there is a common difference between every two consecutive numbers of this series or not. If yes, then this pattern formed is in A.P. Find the 12th term using the nth term of an A.P formula.
Formula used:
The nth term of an A.P is $ t = a + \left( {n - 1} \right)d $ , where a is the first term and d is the common difference.

Complete step by step solution:
We are given that the nth term of a pattern of numbers is $ \left( {2n + 1} \right) $ .
We have to find whether the pattern formed is in A.P or not. If yes we have to find its 12th term.
We have nth term as $ \left( {2n + 1} \right) $
This means that the value of the 1st term is $ 2\left( 1 \right) + 1 = 3 $ as n is equal to 1.
 The value of the 2nd term is $ 2\left( 2 \right) + 1 = 5 $ as n is equal to 2.
The value of the 3rd term is $ 2\left( 3 \right) + 1 = 7 $ as n is equal to 3.
The value of the 4th term is $ 2\left( 4 \right) + 1 = 9 $ as n is equal to 4 and so on.
The series formed is 3, 5, 7, 9, ….
As we can see every two consecutive terms have a difference 2, $ \left( {5 - 3} \right) = \left( {7 - 5} \right) = \left( {9 - 7} \right) = 2 $
So the given pattern of a number is in Arithmetic progression with first term 3 and common difference 2.
The 12th term of this A.P is $ a + \left( {n - 1} \right)d = 3 + \left( {12 - 1} \right) \times 2 = 3 + \left( {11 \times 2} \right) = 25 $ as n=12, a=3 and d=2.
So, the correct answer is “25”.

Note: Do not confuse an A.P with a G.P. A.P is obtained by adding a fixed number to the previous terms whereas G.P is obtained by multiplying a fixed number to the previous terms. Here as we can see the nth term formula, $ t = a + \left( {n - 1} \right)d $ , has $ \left( {n - 1} \right) $ in it. Many of us might confuse whether writing $ \left( {n - 1} \right) $ is correct or just writing n is correct. $ \left( {n - 1} \right) $ must be written because one term as ‘a’ is already added in the formula, so we have to just find the remaining $ \left( {n - 1} \right) $ terms’ summation. So we use $ \left( {n - 1} \right) $ instead of n.