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