Question

Find three consecutive numbers such that the sum of twice the first, three times the second and four times the third togetherbmakes 191.​

Answer 1

find 3 consecutive numbers such that twice the first , 3 times the second and 4 times the third together make 191 ?

Let n-1,n and n+1 be three consecutive numbers.

Given

2(n-1)+3n+4(n+1)=191

2n-2+3n+4n+4=191

9n+2=191

9n = 189

n = 21

So, n-1=20 and n+1=22

So the answer is 20,21,22

Answer 2

`\huge \bold{\color{red}{ANSWER}}`

`\pmb { \orange{Given }:} \sf{ Three \: Consecutive \: numbers}`

`\pmb{\orange{To find}: } \sf{Least \: of \: the \: numbers}`

`\pmb{\orange{Solution}: }`

`\sf{Let \: the \: three \: consecutive \: numbers \: be }`

`\sf\red x, \red {x+1}, \red{x+2}`

`\sf{Twice \: the \: number = 2\red x}`

`\sf{3 \: times \: the \: Second \: number = 3(\red{ x+1})}`

`\sf{4 \: times \: the \: third \: number= 4(\red {x+2})}`

☆ according to question

`\pmb{2\red x + 3(\red{x+1})+4(\red{x+2})= 191}`

`\pmb{2\red{x} + 3\red{x}+3+4\red{x}+8= 191}`

`\pmb{9\red{x}=191-11 = 180}`

`\pmb{\red{x}=20}`

`\sf{∴the \: three \: Consecutive \: numbers \: are}`

`\pmb{\pink{20}, \pink{21}, \pink{22}}`

_________________......

`\purple{ \pmb{LearningLifeII}}`