Question

If the numbers \[\left( {2n - 1} \right)\], \[\left( {3n + 2} \right)\] and \[\left( {6n - 1} \right)\] are in A.P., then find the value of n and the numbers.

Answer 1

Hint: This problem is from arithmetic progression. We are given three terms in the form of n and they are in A.P. so it is clear that there is a common difference d in two consecutive terms. So we will form two equations in n and d form. One equation from first and second term and the other is from first and third term. On solving them we will get the value of n and d. from that we will find the numbers. So let’s solve it!

Step by step solution:
Given that \[\left( {2n - 1} \right)\] , \[\left( {3n + 2} \right)\] and \[\left( {6n - 1} \right)\] are in A.P.
So let d be the common difference. Then we can write,
\[\left( {3n + 2} \right) - \left( {2n - 1} \right) = d\] ……..equation1
Now on solving the equation above we get,
\[ \Rightarrow 3n + 2 - 2n + 1 = d\]
Taking similar terms on one side we get,
\[ \Rightarrow 3n - 2n + 2 + 1 = d\]
\[ \Rightarrow n + 3 = d\] …….equation1.1
Now the difference in first and third term is 2d. then we can write,
\[\left( {6n - 1} \right) - \left( {2n - 1} \right) = 2d\] …….equation2
Now on solving the equation above we get,
\[ \Rightarrow 6n - 1 - 2n + 1 = 2d\]
Taking similar terms on one side we get,
\[ \Rightarrow 6n - 2n - 1 + 1 = 2d\]
\[ \Rightarrow 4n = 2d\]
So on further simplification,
\[ \Rightarrow n = \dfrac{d}{2}\]
\[d = 2n\]
This is the value of d.
From 1.1 we get \[ \Rightarrow d - n = 3\]
Putting the value of d in the equation above,
\[ \Rightarrow 2n - n = 3\]
\[ \Rightarrow n = 3\]
This is the value of n \[ \Rightarrow n = 3\] .
Now putting the value one by one in the numbers given we get the numbers also.
First number: \[\left( {2n - 1} \right) = 2 \times 3 - 1 = 6 - 1 = 5\]
Second number: \[\left( {3n + 2} \right) = 3 \times 3 + 2 = 9 + 2 = 11\]
Third number: \[\left( {6n - 1} \right) = 6 \times 3 - 1 = 18 - 1 = 17\]

Thus the numbers are \[5,11,17\].

Note:
Note that the numbers are given that they are already in A.P. so they have the relation in them. We can solve the problem by finding the value of common difference also. That is just need to find the first number and rest two can be found by adding the common difference. How? See below.
 \[
   \Rightarrow n + 3 = d \\
   \Rightarrow d - n = 3 \\
   \Rightarrow d - \dfrac{d}{2} = 3 \\
   \Rightarrow \dfrac{d}{2} = 3 \\
   \Rightarrow d = 3 \times 2 = 6 \\
 \]
Now First number: \[\left( {2n - 1} \right) = 2 \times 3 - 1 = 6 - 1 = 5\]
Then second number \[5 + d = 5 + 6 = 11\]
Then third number \[11 + d = 11 + 6 = 17\]