niks18 wrote:

Consider a sequence of seven consecutive integers. The average of the first five integers is \(n\). The average of all the seven integers is:

A. \(n\)

B. \(n+1\)

C. \(k*n\), where \(k\) is a function of \(n\)

D. \(n+\frac{2}{7}\)

E. \(n+2\)

In case of a set of consecutive numbers, Middle number = Average of setSet of 5 numbers - a, a+1,

a+2, a+3, a+4

Middle number = Average = a+2

but we are given that average for first 5 numbers is = n

hence a+2 = n

Set of 7 numbers - a, a+1, a+2,

a+3, a+4, a+5, a+6

Middle number = Average = a+3 = (a+2) + 1 =

n+1**********************************************************************

Consecutive numbers are in A.P. and sum for such a set = number of terms * (first term + last term) / 2

If we divide the sum by number of terms then we get average. hence average = (first term + last term) / 2Set of 5 numbers - a, a+1,

a+2, a+3, a+4

Average = (first term + last term) / 2 = (a + a+4) / 2 = a+2

Average = a+2 = n

Set of 7 numbers - a, a+1, a+2,

a+3, a+4, a+5, a+6

Average = (first term + last term) / 2 = (a + a+6) / 2 = a+3 = (a+2)+1

Average =

n + 1
_________________

---------------------------------------------------------------

Target - 720-740

http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html

http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html