Question

How do you find three consecutive integers such that the sum of twice the smallest and three times the largest is $126$ ?

Answer 1

Hint:Consecutive numbers are numbers that follow an order from smallest to largest. For example, $2$ and \[3\] are consecutive integers. Also, $2$, \[3\] and $4$ are three consecutive integers. Here, we let the smallest integer as a variable and the greater integer as $1$ more than that variable and the greatest integer as two more than the variable. We will then form a linear equation in one variable using the data given in the question.

Complete step by step answer:
Here, we are required to find the value of the three consecutive integers. Clearly, out of the three integers one will be the smallest integer, second will be $1$ more than the smallest integer and third will be $2$ more than the smallest integer. Let, the smallest integer be called as ‘$a$’.Since the three numbers are consecutive in nature, the second integer will be $1$ more than the smaller number.

Thus, the second integer $ = a + 1$ and the third integer \[ = a + 2\].Now we see that twice the smallest integer will be two times $a$, that is, $ = 2a$.Now three times the largest number will be
$3 \times (a + 2) \\
\Rightarrow 3a + 3 \times 2 \\
\Rightarrow 3a + 6 \\ $
Thus, according to the question, we have:
$2a + 3a + 6 = 126 \\
\Rightarrow 5a + 6 = 126 \\
\Rightarrow 5a = 126 - 6 \\
\Rightarrow 5a = 120 \\ $
Dividing both the LHS and the RHS, we have:
$\dfrac{{5a}}{5} = \dfrac{{120}}{5} \\
\therefore a = 24 \\ $
Hence, the first integer is $a = 24$,the second integer is $a + 1 = 24 + 1 = 25$ AND the third integer is $a + 2 = 24 + 2 = 26$.

Note:We have taken here the three consecutive integers to be $a$, $a + 1$, \[a + 2\]. We can also take the three numbers to be $a$, $a - 1$ and $a - 2$, where $a$ would be the greatest integer, $a - 1$ will be the second integer and $a - 2$ will be the smallest and the third integer.