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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. |
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Answer» 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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