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

NIFT PG 2019 Question Paper with solutions

Question : 22 of 150

Marks: +1, -0

The average of 5 consecutive integers starting with

m

n

. What is the average of 6 consecutive integers starting with

(m+2) ?

Solution:

Let 5 consecutive integers starting with

m

m

m+1, m+2, m+3, m+4

According to the question,

m+m+1+m+2+m+3+m+4=5 × n

⇒ \;\; 5 m+10=5 n

⇒ \;\; m+2=n

.....(i)
Now, average of 6 consecutive numbers starting with

[m+2+m+3+m+4+m+5+m+6+m+7]/{6}=\;{6 m+27}/{6}

=\;{2 m+9}/{2}=\;{2(n-2)+9}/{2}[ ∵

from Eq. (i),

m+2=n]

=\;{2 n+5}/{2}

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