Question

friends help me plz



plz​

Answer 1

Answer:

$\boxed{ \pink{ \huge{7, \: 8, \: 9}}}$

Step-by-step explanation:

$\bold{\large{ Given}} \longrightarrow \\ three \: consecutive \: integers \: when \: taken \: in \: increasing \: order \\ and \: multiplied \: by \: 2 ,\: 3, \: 4 \: they \: add \: up \: to \: 74$

$\bold{\large{To \: find} }\longrightarrow \\ consecutive \: integers \\ \large{Concept \: used} \\ differenc \: between \: consecutive \: integers \: is \: one \: \\ example \\ 2 ,\: 3 \: 4 \: and \: 11 ,\: 12 ,\: 13 \: etc \\ \bold{\large{Solution}} \longrightarrow \\ let \: first \: integer \: be \: x \\ secod \: integer = x + 1 \\ third \: integer = x + 2 \\ now \: according \: to \: question \\2( first \: integer) + 3(second \: integer) + 4(third \: integer) = 74 \\ = > 2x + 3(x + 1) + 4(x + 2) = 74 \\ = > 2x + 3x + 3 + 4x + 8 = 71 \\rearranging \: like \: terms \\ = > (2x + 3x + 4x) + (3 + 8) = 74 \\ = > 9x + 11 = 74 \\ = > 9x = 74 - 11 \\ = > 9x = 63 \\ = > x = \dfrac{63}{9} \\ = > x = 7 \\ first \: integer \: = x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7 \\ second \: integer = x + 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7 + 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 8 \\ third \: integer = x + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 \\ so \: integers \: are \: 7 ,\: 8, \: 9$

Answer 2

Solution:

Let the no. be x , x+ 1 and x + 2

they are taken in increasing order and multiplied by 2, 3 and 4.

so,

x becomes 2 x

(x + 1) becomes 3(x+2)

(x+2) becomes 4(x + 2)

2x + 3(x+1) + 4 (x + 2) = 74

2x + 3(x + 1) + 4(x +2 ) = 74

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 74

9x = 74 - 11

9x = 63

x = 63/ 9

x = 7

$\sf{consecutive \: integers \: are \: 7 \: and \: 8 \: and \: 9}$