Question

The mean of 1,2,3....K is 6K/11 find K

Answer 1

Answer:

k=11

Step-by-step explanation:

sum of n. natural number=

n(n+1)/2

1+2+....+k/k=6k/11

k(k+1)/2k=6k/11

12k=11k+11

k=11

Answer 2

Answer:

k = 11

Step-by-step explanation:

Given:

series of number = 1,2,3,4...k

mean = $\frac{6k}{11}$

To find: The value of k

Formula Used:

Mean of number 1,2,3, ...n = $\frac{1+2+3+..+n}{n}$

Sum of n natural number 1,2,3,...n = $\frac{n(n+1)}{2}$

Solution:

We know, that the mean of the given number, 1,2,3, ...k = $\frac{1+2+3+..+k}{k}$

∴ $\frac{1+2+3+...+k}{k}$ = $\frac{6k}{11}$  ------(i)

Now, the sum of n natural numbers  = $\frac{n(n+1)}{2}$

∴ putting the values in equation (i), we get

=> $\frac{k(k+1)}{2k}$ = $\frac{6k}{11}$  [ ∵ 1+ 2+ 3+ ...+ k = $\frac{k(k+1)}{2}$ ]

=> $\frac{k+1}{2}$ = $\frac{6k}{11}$

=> 11$k^{}$ + 11$^{}$ = 12$k^{}$

=>  $k^{}$ - 11= 0

∴ k = 11

Hence, The value of k = 11. Ans

#SPJ3