Question

If the mean of the n, n+2, n+4 , n+6 and
n+8 is 11.find the value of n.​

Answer 1

$\Huge{\underline{\underline{\mathfrak{ Solution \colon }}}}$

$mean \: = \frac{(n + 2 +)(n + 4) + (n + 6)(n + 8) }{5}$

$11 = \frac{5n + 20}{5}$

5n + 20 = 55

5n = 55 - 20

5n = 35

$n = \frac{35}{5}$

n = 7

Answer 2

Answer:

$\large\boxed{\sf{n=7}}$

Step-by-step explanation:

Given numbers are n , n+2, n+4, n+6 and n+8.

Also, their mean = 11

Total number of given numbers = 5

We know that, mean of numbers is the ratio of sum of all numbers to that of number of all numbers.

Therefore, we will get

$= > \frac{n + n + 2 + n + 4 + n + 6 + n + 8}{5} = 11 \\ \\ = > \frac{5n + 20}{5} = 11 \\ \\ = > 5n + 20 = 11 \times 5 \\ \\ = > 5n + 20 = 55 \\ \\ = > 5n = 55 - 20 \\ \\ = > 5n = 35 \\ \\ = > n = \frac{35}{5} \\ \\ = > n = 7$