Question

The average of first 51 multiples of 6 is 52 K. Then the value of K is​

Answer 1

$\fontsize{18}{10}{\textup{\textbf{The value of K is 3.}}}$

Step-by-step explanation:

We know that if S is the sum of given n numbers, then the average of these n number is

$A=\dfrac{S}{n}.$

Also, the sum of first m natural numbers is given by

$S_m=\dfrac{m(m+1)}{2}.$

Given that the average of first 51 multiples of 6 is 52 K.

So, we have

$\dfrac{6\times1+6\times2+~.~.~.~+6\times51}{51}=52K\\\\\\\Rightarrow \dfrac{6(1+2+~.~.~.~+51)}{51}=52K\\\\\Rightarrow 6\times\dfrac{51(51+1)}{2}=51\times52K\\\\\Rightarrow 3\times51\times52=51\times52K\\\\\Rightarrow K=3.$

Thus, the required value of K is 3.

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Answer 2

The value of K is equal to 3.

Step-by-step explanation:

The first 51 multiples of 6 are:

6, 12, 18, 24, ...

The given sequence are in AP.

Here, first term (a) = 6, common difference (d) = 12 - 6 = 5 and

the number of terms (n) = 51

To find, the value of K = ?

We know that,

The sum of nth term of an AP.

$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$

The sum of 51th term of an AP.

$S_{51}=\dfrac{51}{2}[2(6)+(51-1)6]$

⇒$S_{51}=\dfrac{51}{2}[12+300]$

⇒ $S_{51}=\dfrac{51}{2}[312]$

∴ The average of first 51 multiples of 6

⇒ $\dfrac{51}{2\times 51}[312]$ = 52 K

⇒ 52 K = 156

⇒ K = 3

Thus, the value of K is equal to 3.