Question

The AM of the first ten odd natural numbers is:
[a] 10
[b] 100
[c] 1000
[d] 1

Answer 1

Hint: Use the formula $AM=\dfrac{Sum\text{ }of\ observations}{Number\ of\ observations}$ and ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. Find the sum of the first ten natural numbers and divide it by 10. This will give you the value of AM. Alternatively, you can use that if $\left\{ {{a}_{n}} \right\}$ is an A.P then the mean of first n terms of the A.P is $a+\dfrac{n-1}{2}d$. Substitute a = 1, n = 10 and d =2 in the above result to get the answer.

Complete step-by-step answer:

We know that \[1+3+5+\ldots +2n-1={{n}^{2}}\].
Put n = 10, we get
$1+3+5+\ldots +19={{10}^{2}}=100$
Hence 1+ 3 +5 +…+19 = 100.
Now, we know that $AM=\dfrac{Sum\text{ }of\ observations}{Number\ of\ observations}$.
Here the sum of observations =100.
Number of observations = 10
Hence AM $=\dfrac{100}{10}=10$.
Hence the arithmetic mean of the first ten odd natural numbers is 10.
Hence option [a] is correct.

Note: Alternatively of $\left\{ {{a}_{n}} \right\}$ is an A.P then the mean of first n terms of the A.P is $a+\dfrac{n-1}{2}d$.
Proof: We know that in an A.P ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
Hence A.M $=\dfrac{{{S}_{n}}}{n}=\dfrac{\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)}{n}=\dfrac{1}{2}\left( 2a+\left( n-1 \right)d \right)$
We know that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$.
Using the above identity, we get
AM$=\dfrac{2a}{2}+\dfrac{\left( n-1 \right)d}{2}=a+\dfrac{n-1}{2}d$
Hence proved.
[2] We can use the above-mentioned property to solve the question.
The set of first n natural numbers forms an A.P with a = 1, n = 10 and d= 2.
Hence we have AM$=1+\dfrac{10-1}{2}\times 2=1+9=10$, which is the same result as obtained above.
[3] The sequence of A.Ms of an A.P also is an A.P with first term same as the first term of the original A.P and common difference half the common difference of the original A.P