Question

d) First 51 multiples of 6 1) 154 2) 158 3) 160 4) 156 ​

Answer 1

Answer:

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}.A=

n

S

.

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

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

m

=

2

m(m+1)

.

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

So, we have

\begin{gathered}\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.\end{gathered}

51

6×1+6×2+ . . . +6×51

=52K

⇒

51

6(1+2+ . . . +51)

=52K

⇒6×

2

51(51+1)

=51×52K

⇒3×51×52=51×52K

⇒K=3.

Thus, the required value of K is 3.

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