Question
Find the sum:
$4-\frac{1}{\text{n}}+4-\frac{2}{\text{n}}+4-\frac{3}{\text{n}}+......\text{upto n terms}.$
$4-\frac{1}{\text{n}}+4-\frac{2}{\text{n}}+4-\frac{3}{\text{n}}+......\text{upto n terms}.$
✓
Answer
Here, first term, $\text{a}=4-\frac{1}{\text{n}}$
Common difference, $\text{d}=\bigg(4-\frac{2}{\text{n}}\bigg)-\bigg(4-\frac{1}{\text{n}}\bigg)$
$\text{d}=-\frac{2}{\text{n}}+\frac{1}{\text{n}}=-\frac{-1}{\text{n}}$
$\because$ Sum of n terms of an AP,
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg[2\bigg(4-\frac{1}{\text{n}}\bigg)+\big(\text{n}-1\big)\bigg(\frac{-1}{\text{n}}\bigg)\bigg]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\left\{8-\frac{2}{\text{n}}-1+\frac{1}{\text{n}}\right\}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg(7-\frac{1}{\text{n}}\bigg)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg(\frac{7\text{n}-1}{\text{n}}\bigg)$
$\Rightarrow\text{S}_\text{n}=\frac{7\text{n}-1}{\text{2}}$
Common difference, $\text{d}=\bigg(4-\frac{2}{\text{n}}\bigg)-\bigg(4-\frac{1}{\text{n}}\bigg)$
$\text{d}=-\frac{2}{\text{n}}+\frac{1}{\text{n}}=-\frac{-1}{\text{n}}$
$\because$ Sum of n terms of an AP,
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg[2\bigg(4-\frac{1}{\text{n}}\bigg)+\big(\text{n}-1\big)\bigg(\frac{-1}{\text{n}}\bigg)\bigg]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\left\{8-\frac{2}{\text{n}}-1+\frac{1}{\text{n}}\right\}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg(7-\frac{1}{\text{n}}\bigg)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\bigg(\frac{7\text{n}-1}{\text{n}}\bigg)$
$\Rightarrow\text{S}_\text{n}=\frac{7\text{n}-1}{\text{2}}$
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