Question

Very-Short and Short-Answer Questions

If the numbers (2n − 1), (3n + 2) and (6n − 1) are in AP, find the value of n and the numbers. [CBSE 2013C]

Answer 1

It is given that the numbers (2n − 1), (3n + 2) and (6n − 1) are in AP.
$$\begin{gathered} \therefore \left(3n+2\right)-\left(2n-1\right)=\left(6n-1\right)-\left(3n+2\right) \\ \Rightarrow 3n+2-2n+1=6n-1-3n-2 \\ \Rightarrow n+3=3n-3 \\ \Rightarrow 2n=6 \end{gathered}$$
$\Rightarrow n=3$

When n = 3,
$$\begin{gathered} 2n-1=2\times 3-1=6-1=5 \\ 3n+2=3\times 3+2=9+2=11 \\ 6n-1=6\times 3-1=18-1=17 \end{gathered}$$

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.