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

A. $$\frac{(2n + 5)}{2}$$

B. (2n + 2)

C. (n + 3)

D. $$\frac{(2n + 9)}{2}$$

Answer: Option A

Solution(By Examveda 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)