Finding Consecutive Numbers Algebraically
Let the four consecutive numbers be represented as $x$, $x+1$, $x+2$, and $x+3$.
The problem states that twice the first, three times the second, four times the third, and five times the fourth sum up to 236. This can be written as an equation:
$2x + 3(x+1) + 4(x+2) + 5(x+3) = 236$
Solving the Algebraic Equation
- Expand the terms: $2x + 3x + 3 + 4x + 8 + 5x + 15 = 236$
- Combine like terms (the $x$ terms and the constant terms): $(2x + 3x + 4x + 5x) + (3 + 8 + 15) = 236$ $14x + 26 = 236$
- Isolate the variable term by subtracting 26 from both sides: $14x = 236 - 26$ $14x = 210$
- Solve for $x$ by dividing both sides by 14: $x = \frac{210}{14}$ $x = 15$
Determining the Consecutive Numbers
Since the first number ($x$) is 15, the four consecutive numbers are:
- First number: $x = 15$
- Second number: $x+1 = 15+1 = 16$
- Third number: $x+2 = 15+2 = 17$
- Fourth number: $x+3 = 15+3 = 18$
The four consecutive numbers are 15, 16, 17, and 18.
Verification: $2(15) + 3(16) + 4(17) + 5(18) = 30 + 48 + 68 + 90 = 236$. The condition is satisfied.
The correct option is 15, 16, 17, 18.
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