Question

In an increasing sequence of 6 consecutive odd integers, the sum of the second, third, fourth, and fifth integers is 192. What is the sum of the first and last integers?

• A. 88
• B. 92
• C. 96
• D. 100
• E. 104

Hint:

For any arithmetic sequence, the sum of any two terms equidistant from the center is the same. For example, in this sequence, 1st + 6th = 2nd + 5th = 3rd + 4th. Since we found the 2nd (45) and 5th (51) terms, their sum is 96, which must also be the sum of the 1st and 6th terms.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem deals with arithmetic sequences, specifically a sequence of consecutive odd integers. Consecutive odd integers differ by 2.
Step 2: Key Formula or Approach:
For a set of numbers in an arithmetic sequence, the mean is equal to the median. Also, the sum of the terms is the number of terms multiplied by the mean.
Step 3: Detailed Explanation:
Method 1: Using the Mean We are given that the sum of four consecutive odd integers (the 2nd, 3rd, 4th, and 5th terms of the sequence) is 192. Let's find the average of these four integers: \[ \text{Average} = \frac{\text{Sum}}{\text{Number of terms}} = \frac{192}{4} = 48 \] Since these are consecutive odd integers, their average (48) will lie exactly in the middle of the set. The four integers must be the two odd integers immediately below 48 and the two odd integers immediately above 48. The two odd integers around 48 are 47 and 49. So, the four integers are 45, 47, 49, and 51. These are the second, third, fourth, and fifth terms of the full sequence of six consecutive odd integers.

Second term = 45
Third term = 47
Fourth term = 49
Fifth term = 51
To find the first term, we subtract 2 from the second term: \(45 - 2 = 43\). To find the last (sixth) term, we add 2 to the fifth term: \(51 + 2 = 53\). The full sequence is: 43, 45, 47, 49, 51, 53. The question asks for the sum of the first and last integers. \[ \text{Sum} = 43 + 53 = 96 \] Method 2: Using Algebra Let the six consecutive odd integers be represented by: \(n, n+2, n+4, n+6, n+8, n+10\) The sum of the second, third, fourth, and fifth integers is 192: \[ (n+2) + (n+4) + (n+6) + (n+8) = 192 \] \[ 4n + 20 = 192 \] \[ 4n = 172 \] \[ n = \frac{172}{4} = 43 \] The first integer is \(n = 43\). The last integer is \(n+10 = 43+10 = 53\). The sum of the first and last integers is \(43 + 53 = 96\).
Step 4: Final Answer:
The sum of the first and last integers is 96.