1.

If the mean of first n natural numbers is \(\frac{5n}{9},\) then n =A. 5 B. 4 C. 9 D. 10

Answer»

First n natural numbers are 

1, 2, 3, 4, …, n 

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations. 

and, we have given series 

1, 2, 3, …, n 

Clearly the above series is an AP(Arithmetic progression) with 

first term, a = 1 and 

common difference, d = 1 

And no of terms is clearly n. 

And last term is also n. 

We know, sum of terms of an AP if first and last terms are known is:

\(S_n=\frac{n}{2}(a+a_n)\)

Putting the values in above equation we have sum of series i.e.

\(1+2+3+...+n=\frac{n}{2}(1+n)\)

\(=\frac{n(n+1)}{2}...[1]\)

As,

Mean \(=\frac{Sum\,of\,all\,terms}{no\,of\,terms}\)

\(=\frac{1+2+3+...+n}{n}\)

⇒ Mean \(\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\)

Given, 

mean \(=\frac{5n}{9}\)

\(\Rightarrow \frac{n+1}{2}=\frac{5n}{9}\)

⇒ 9n + 9 = 10n 

⇒ n = 9


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