Answer: Option D
Answer: (d)
Given,
$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8)"/9 = n$
$=> \text"(9a + 36)"/9 = n$
=> a + 4 = n -------------------- ( 1 )
If the next 2 numbers are included , let new average = k
$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8 + a + 9 + a + 10)"/11 = k$
$\text"(11a + 55)"/11 = k$
a + 5 = k -------------------- ( 2 )
Subtracting ( 1 ) from ( 2 ) , we get :
a + 5 - a - 4 = k - n
=> k - n = 1
=> k = n + 1
Therefore the new average is n + 1