Question

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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given 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 56. Find the numbers.

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf\: Let \: -\begin{cases} &\sf{first \: integer \: be \: x} \\ \\ &\sf{second \: integer \: be \: x + 1}\\ \\ &\sf{third \: integer \: be \: x + 2} \end{cases}\end{gathered}\end{gathered}$

According to statement,

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 56.

So,

$\rm :\longmapsto\:2x + 3(x + 1) + 4(x + 2) = 56$

$\rm :\longmapsto\:2x + 3x + 3 + 4x + 8 = 56$

$\rm :\longmapsto\:9x + 11 = 56$

$\rm :\longmapsto\:9x = 56 - 11$

$\rm :\longmapsto\:9x = 45$

$\bf\implies \:x = 5$

$\begin{gathered}\begin{gathered}\bf\: Hence\: -\begin{cases} &\sf{first \: integer \: = \: x = 5} \\ \\ &\sf{second \: integer \: = \: x + 1 = 6}\\ \\ &\sf{third \: integer \: = \: x + 2 = 7} \end{cases}\end{gathered}\end{gathered}$

Verification :-

$\begin{gathered}\begin{gathered}\bf\: Given\: -\begin{cases} &\sf{first \: integer \: = \: x = 5} \\ \\ &\sf{second \: integer \: = \: x + 1 = 6}\\ \\ &\sf{third \: integer \: = \: x + 2 = 7} \end{cases}\end{gathered}\end{gathered}$

So,

These consecutive integers, when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, and added, we get

$\rm :\longmapsto\:2 \times 5 + 3 \times 6 + 4 \times 7$

$\rm \: = \:10 + 18 + 28$

$\rm \: = \:56$

Thus,

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 56.

Hence, Verified