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Find the sum of n terms of the series$ \left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots . $
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 $\frac{7n-1}{2}$.
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