If the numbers 2n-1 , 3n+2 and 6n-1 are in A.P, find n and hence find the numbers
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5
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10
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Given: (2n−1),(3n+2) and (6n−1) are in AP.
So, difference between two consecutive terms is same.
So, (3n+2)−(2n−1)=(6n−1)−(3n+2)
(3n+2)+(3n+2)=6n−1+2n−1
6n+4=8n−2
8n−6n=4+2
or n=3
Numbers are:
2×3−1=5
3×3+2=11
6×3−1=17
Answer: (5,11,17) are required numbers.
Step-by-step explanation:
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