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.
Solution:
It is given that the numbers (2n − 1), (3n + 2) and (6n − 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.