The average of a group of 8 consecutive even numbers is 41. What is the sum of the smallest and largest numbers in the group?

Correct — C.
Eight consecutive even numbers form an AP with difference 2
In an AP, average = (smallest + largest) ÷ 2
So smallest + largest = 2 × 41 = 82

Check by listing: smallest = a, largest = a + 14, average = a + 7
a + 7 = 41, so a = 34 and the largest is 48
34 + 48 = 82 → option (c)

Why the others are wrong

• (a)
  78 — 78 halves to 39, so it belongs to eight consecutive evens averaging 39 (32 to 46). This group averages 41, so its ends must sum to 82.
• (b)
  80 — 80 halves to 40, an even number. Eight consecutive evens average a + 7 with a even, which is always odd, so 80 cannot be first + last for any such group.
• (d)
  84 — 84 halves to 42, which is one of the eight numbers (34 to 48), not their mean. With an even count the mean falls between the two middle numbers, 40 and 42.

Consecutive even numbers are an arithmetic progression with common difference 2. In any AP the terms pair off from the ends: first + last = second + second-last, and each pair equals 2 × average.

So you never need the list itself: first + last = 2 × 41 = 82.

With an even count of terms, the mean sits between the two middle terms — here 40 and 42 — which is why it can be odd.

Key facts

• In an AP, average = (first term + last term) ÷ 2
• n consecutive even numbers starting at a end at a + 2(n − 1)
• With an even number of terms, the average is the mean of the two middle terms

Common traps

• Writing the largest as a + 16 instead of a + 14 — eight terms have only seven gaps of 2
• Setting the average equal to one of the numbers, when with an even count it falls between the two middle ones

No directly related past PYQ was found.