Question

Three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectively , they add up to 74. find these numbers.......


Mates give answer quickly please.....​

Answer 1

Let the three consecutive integers be x, (x + 1) and (x + 2) respectively

$\bf{\underline{\underline{According\:to\:the\:question\: :}}}$

$\mathtt{\implies\: 2(x) + 3(x+1) + 4(x+2) = 74}$

$\mathtt{\implies\: 2x + 3x + 3 + 4x + 8 = 74}$

$\mathtt{\implies\: 9x + 11 = 74}$

$\mathtt{\implies\: 9x = 74 - 11}$

$\mathtt{\implies\: 9x = 63}$

$\mathtt{\implies\: x = \Large{\frac{63}{9}}}$

$\mathtt{\implies\: x = 7}$

$\bf{\underline{\underline{Answer\: :}}}$

The three consecutive no's are:

$\mathtt{\bigstar\: \green{ x = 7}}$

$\mathtt{\bigstar\: \blue{ (x + 1) = 8}}$

$\mathtt{\bigstar\: \red{ (x + 2) = 9}}$

$\bf{\underline{\underline{Verification\: :}}}$

$\mathtt{\diamond\:\: 2 × 7 + 3 × 8 + 4 × 9}$

$\mathtt{\implies\: 14 + 24 + 36}$

$\mathtt{\implies\: 74}$

Answer 2

$\huge{ \underline{ \underline{ \boxed{ \sf{ \blue{Answer :}}}}}}$

Let us assume those three consecutive numbers to be x, ( x + 1 ) & ( x + 2 ).

As given in the question,

$2(x) + 3(x + 1) + 4(x + 2) = 74 \\ \\2x + 3x + 3 + 4x + 8 = 74\\ \\ 9x + 11 = 74 \\ \\ 9x = 74 - 11 \\ \\ x = \frac{63}{9} \\ \\ x = 7$

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Now,

$\\ (x + 1) = 8 \\ \\ (x + 2) = 9$

Thank You! :)

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