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Find three consecutive numbers such that twice the first, three times the second and four times the third together make 191.

Text Solution

Verified by Experts

According to Question
`2n+3(n+1)+4(n+2)=191`
`2n+3n+3+4n+8=191`
`9n+11=191`
`n=20`
numbers are 20,21,22
Option 3 is correct.
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