Answer
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Hint: We start solving the problem by calculating the first six multiples of 4 by multiplying 4 with the integers 1, 2, 3, 4, 5 and 6 respectively. We then substitute the obtained values in the formula of mean of n numbers $\dfrac{{{a}_{1}}+{{a}_{2}}+......+{{a}_{n}}}{n}$ and make necessary calculations to get the required value of the mean of the numbers.
Complete step by step answer:
According to the problem, we need to find the mean of the first six multiples of 4.
Let us first find the six multiples of 4 and then find the mean of those numbers.
We know that the first six multiples of 4 are found by multiplying 4 with 1, 2, 3, 4, 5, 6.
So, we get \[4\times 1=4\].
$\Rightarrow 4\times 2=8$.
$\Rightarrow 4\times 3=12$.
$\Rightarrow 4\times 4=16$.
$\Rightarrow 4\times 5=20$.
$\Rightarrow 4\times 6=24$.
So, we have found the first six multiples of 4 as 4, 8, 12, 16, 20, 24.
Now, let us find the mean of the obtained numbers 4, 8, 12, 16, 20, 24. Let us assume it as m.
We know that the mean of the numbers ${{a}_{1}}$, ${{a}_{2}}$,……, ${{a}_{n}}$ is $\dfrac{{{a}_{1}}+{{a}_{2}}+......+{{a}_{n}}}{n}$.
So, we get mean (m) = $\dfrac{4+8+12+16+20+24}{6}$.
$\Rightarrow $ mean (m) = $\dfrac{84}{6}$.
$\Rightarrow $ mean (m) = $14$.
So, we have found the mean of the first six multiples of 4 as 14.
∴ The mean of the first six multiples of 4 is 14.
So, the correct answer is “Option C”.
Note: We should not consider negative integers or zero while taking the integral multiples of 4, as the multiples are always assumed to start with the integer 1 and move forwards. We should consider multiplying 4 with 0 or negative integers or fractions if it is mentioned in the problem. We can also solve this problem by finding the mean of the numbers 1, 2, 3, 4, 5 and 6 first and then multiplying this mean with 4.
Complete step by step answer:
According to the problem, we need to find the mean of the first six multiples of 4.
Let us first find the six multiples of 4 and then find the mean of those numbers.
We know that the first six multiples of 4 are found by multiplying 4 with 1, 2, 3, 4, 5, 6.
So, we get \[4\times 1=4\].
$\Rightarrow 4\times 2=8$.
$\Rightarrow 4\times 3=12$.
$\Rightarrow 4\times 4=16$.
$\Rightarrow 4\times 5=20$.
$\Rightarrow 4\times 6=24$.
So, we have found the first six multiples of 4 as 4, 8, 12, 16, 20, 24.
Now, let us find the mean of the obtained numbers 4, 8, 12, 16, 20, 24. Let us assume it as m.
We know that the mean of the numbers ${{a}_{1}}$, ${{a}_{2}}$,……, ${{a}_{n}}$ is $\dfrac{{{a}_{1}}+{{a}_{2}}+......+{{a}_{n}}}{n}$.
So, we get mean (m) = $\dfrac{4+8+12+16+20+24}{6}$.
$\Rightarrow $ mean (m) = $\dfrac{84}{6}$.
$\Rightarrow $ mean (m) = $14$.
So, we have found the mean of the first six multiples of 4 as 14.
∴ The mean of the first six multiples of 4 is 14.
So, the correct answer is “Option C”.
Note: We should not consider negative integers or zero while taking the integral multiples of 4, as the multiples are always assumed to start with the integer 1 and move forwards. We should consider multiplying 4 with 0 or negative integers or fractions if it is mentioned in the problem. We can also solve this problem by finding the mean of the numbers 1, 2, 3, 4, 5 and 6 first and then multiplying this mean with 4.
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