Find the sum:
$ 4-\frac{1}{n}+4-\frac{2}{n}+4-\frac{3}{n}+\ldots $ upto $ n $ terms

Given:

Given series is \( \left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots . \)

To do:

We have to find the sum of $n$ terms of the series.

Solution:

Let the number of terms of the given A.P. be $n$, first term be $a$ and the common differnce be $d$.

First term $a_1=a=4-\frac{1}{n}$

Second term $a_2= 4-\frac{2}{n}$

Common difference $d=a_2-a_1=4-\frac{2}{n}-(4-\frac{1}{n})=\frac{-2+1}{n}=\frac{-1}{n}$

We know that,

Sum of $n$ terms $S_{n} =\frac{n}{2}(2a+(n-1)d)$

$=\frac{n}{2}[2(4-\frac{1}{n})+(n-1)(\frac{-1}{n})]$

$=\frac{n}{2}[\frac{8n-2-n+1}{n}]$

$=\frac{n}{2}(\frac{7n-1}{n})$

$=\frac{7n-1}{2}$

Hence, the sum of the $n$ terms of the given series is $\frac{7n-1}{2}$.