Question

The sum of the first 50 positive integers is 1,275. What is the sum of the integers from 51 to 100?

• A. 2,525
• B. 2,550
• C. 3,250
• D. 3,775
• E. 5,050

Hint:

Recognizing the structure of arithmetic series can lead to clever shortcuts. The alternative method, which sees the second series as a simple transformation of the first, is very efficient and avoids calculating the sum to 100 from scratch.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question asks for the sum of an arithmetic series. We are given a piece of information (the sum of the first 50 integers) that we can use to find the answer.
Step 2: Key Formula or Approach:
The sum of the first \(n\) positive integers is given by the formula \( S_n = \frac{n(n+1)}{2} \). The sum of integers from 51 to 100 can be found by taking the sum of the first 100 integers and subtracting the sum of the first 50 integers. Sum(51 to 100) = Sum(1 to 100) - Sum(1 to 50).
Step 3: Detailed Explanation:
We are given Sum(1 to 50) = 1,275. We need to find Sum(1 to 100). Using the formula \( S_n = \frac{n(n+1)}{2} \) with \(n=100\): \[ S_{100} = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \] Now, we can find the sum of the integers from 51 to 100: \[ \text{Sum(51 to 100)} = \text{Sum(1 to 100)} - \text{Sum(1 to 50)} \] \[ \text{Sum(51 to 100)} = 5050 - 1275 \] \[ 5050 - 1275 = 3775 \] Alternative Method (Pairing): The sum we want is \(51 + 52 + \dots + 100\). Each term in this series is exactly 50 greater than the corresponding term in the series \(1 + 2 + \dots + 50\). \(51 = 1 + 50\) \(52 = 2 + 50\) ... \(100 = 50 + 50\) There are 50 terms in the series from 51 to 100. So we can write the sum as: \[ \text{Sum(51 to 100)} = (1+50) + (2+50) + \dots + (50+50) \] \[ = (1+2+\dots+50) + (50+50+\dots+50) \] \[ = \text{Sum(1 to 50)} + (50 \times 50) \] We are given that Sum(1 to 50) = 1,275. \[ = 1275 + 2500 = 3775 \] Step 4: Final Answer:
Both methods yield the same result: the sum of the integers from 51 to 100 is 3,775.