Question

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

Answer 1

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

∴ (3n + 2) - (2n-1) = (6n-1) - (3n+2)

⇒ 3n + 2-2n +1 = 6n-1-3n-2

⇒ n +3=3n-3

⇒ 2n = 6

 ⇒ n = 3

When , n = 3 

2n - 1 = 2×3 -1=6-1=5

3n + 2 = 3×3+2=9+2=11

6n -1 =  6 × 3-1=18-1=17

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