Question:

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

Solution:

It is given that the numbers (2n − 1), (3n + 2) and (6− 1) are in AP.

$\therefore(3 n+2)-(2 n-1)=(6 n-1)-(3 n+2)$

$\Rightarrow 3 n+2-2 n+1=6 n-1-3 n-2$

$\Rightarrow n+3=3 n-3$

$\Rightarrow 2 n=6$

$\Rightarrow n=3$

When n = 3,

$2 n-1=2 \times 3-1=6-1=5$

$3 n+2=3 \times 3+2=9+2=11$

$6 n-1=6 \times 3-1=18-1=17$

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