Question

The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?​

Answer 1

Answer:

n+4

Step-by-step explanation:

Given that first number is m and the average of 5 numbers is n. So numbers are m,(m+1),(m+2),(m+3),(m+4).

As the middle number is average which is m+3.

Therefore n=m+3.

We can write m=n-2.

Now the first 9 consecutive numbers starting from m+2 are:

(m+2),(m+3),(m+4),(m+5),(m+6),(m+7),(m+8),(m+9),(m+10).

Average is m+6 (as it is 5th term is middle and hence it is average)

From the above we have the value of m so (n-2)+6=n+4.

Answer 2

The average of 9 consecutive integers starting with (m + 2 )  is  n + 4

Step-by-step explanation:

Given as :

The average of 5 consecutive integers starting with m as the first integer is n

Let The average of 9 consecutive integers starting with ( m + 2 ) = x

According to question

first integer = m

Second integer = m + 1

Third integer = m + 2

fourth integer = m + 3

fifth integer = m + 4

Average of 5 consecutive integers =$\dfrac{m+(m+1)+(m+2)+(m+3)+(m+4)}{5}$

Or, n =  $\dfrac{m+(m+1)+(m+2)+(m+3)+(m+4)}{5}$

Or, m + ( m +1 ) + ( m +2 ) + ( m +3 ) + ( m + 4 ) = 5 n

Or, 5 m + 10 = 5 n

Or,  5 ( m + 2 ) = 5 n

∴          m + 2 = $\dfrac{5n}{5}$

i.e        m + 2 = n

or,        m = n - 2

So, The value of m = n - 2

Again

The average of 9 consecutive integers starting with ( m + 2 )

i.e  first  integer = m + 2 = n - 2 + 2 = n

Second integer = ( m + 2) + 1 = n - 2 +2 + 1 = n + 1

Third integer = ( m + 2) + 2 = n - 2 + 2 + 2= n + 2

fourth integer = ( m + 2) + 3 = n - 2 + 2 + 3 = n + 3

fifth integer = ( m + 2) + 4 = n - 2 + 2 + 4 = n + 4

sixth integer= ( m + 2) + 5 = n - 2 + 2 + 5 = n + 5

seventh integer = ( m + 2) + 6 = n - 2 + 2 + 6 = n + 6

eight integer = ( m + 2) + 7 = n - 2  + 2 + 7 = n + 7

ninth integer  = ( m + 2) + 8 = n - 2 + 2 + 8 = n + 8

So, The nine consecutive integer are n , n + 1 , n + 2 , n + 3 , n + 4 , n + 5 , n + 6 , n + 7 , n + 8

Average of 9 consecutive integers starting with ( m + 2 ) = $\dfrac{n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n +6 + n + 7 + n+ 8}{9}$

Or,   x = $\dfrac{9n+36}{9}$

∴      x = n + 4

So, The average of 9 consecutive integers starting with ( m + 2 ) = n + 4

Hence, The average of 9 consecutive integers starting with is (n + 4)  Answer