Question

If a, b, c, and d are consecutive integers such that a<b<c<d, then in terms of a, the sum a + b + d =

• A. a + 4
• B. 2a + 3
• C. 3a + 2
• D. 3a + 3
• E. 3a + 4

Hint:

For problems with abstract variables, you can check your answer by picking a simple number. Let \(a=1\). Then \(b=2\), \(c=3\), \(d=4\). The sum is \(a+b+d = 1+2+4=7\). Now, plug \(a=1\) into the options. Only option (E) gives \(3(1)+4=7\). This confirms the answer.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This is an algebraic manipulation problem involving consecutive integers. We need to express the other integers in terms of the first integer, \(a\), and then simplify the given sum.
Step 2: Detailed Explanation:
1. Express b, c, and d in terms of a.
Since the integers are consecutive and \(a\) is the smallest, we have:
\(b = a + 1\)
\(c = a + 2\)
\(d = a + 3\)
2. Formulate the sum.
We need to find the sum \(S = a + b + d\).
3. Substitute the expressions for b and d into the sum.
\[ S = a + (a + 1) + (a + 3) \]
4. Simplify the expression.
Combine the 'a' terms and the constant terms:
\[ S = (a + a + a) + (1 + 3) \]
\[ S = 3a + 4 \]
Step 3: Final Answer:
The sum \(a + b + d\) in terms of \(a\) is \(3a + 4\).