The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?
Options:
A . ${2n+9}/2$
B . (n+3)
C . (n+2)
D . ${2n+5}/2$
Answer: Option D Answer: (d) m + m + 1 + m + 2 + m + 3 + m + 4 = 5n ⇒ 5m + 10 = 5n ⇒ m + 2 = n ....(i) Required average = ${m+2+m+3+m+4+m+5+m+6+m+7}/6$ = ${6m+27}/6$ = ${2m+9}/2$ = ${2(n-2)+9}/2$= ${2n+5}/2$ By (i) [m = n – 2]
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