The average of 5 consecutive integers starting with ‘m’ is n…

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The average of 5 consecutive integers starting with ‘m’ is n.
What is the average of 6 consecutive integers starting with (m + 2)?

  1. A. $$\frac{(2n + 5)}{2}$$
  2. B. (2n + 2)
  3. C. (n + 3)
  4. D. $$\frac{(2n + 9)}{2}$$

Answer: Option A

Here is complete explanation of The average of 5 consecutive integers starting with ‘m’ is n….

Solution(By ExamCraze Team)

According to the question,
Let M = 1
∴ 5 consecutive integers are = 1, 2, 3, 4, 5
∴ $$frac{1 + 2 + 3 + 4 + 5}{5}$$   = n
n = $$frac{15}{5}$$ = 3
∴ 6 consecutive integers starting with (m + 2) are = 3, 4, 5, 6, 7, 8
∴ $$frac{3 + 4 + 5 + 6 + 7 + 8 }{6}$$    = $$frac{33}{6}$$  = $$frac{11}{2}$$
Now check from option to put n = 3
Option : (A) $$frac{(2n + 5)}{2}$$
= $$frac{2 × 3 + 5}{2}$$   = $$frac{11}{2}$$ (satisfied)

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