Question

Question 21 (ii)
Find the sum:
$(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+\cdots uptonterms$

Answer 1

Here, first term, $a=4-\frac{1}{n}$
Common difference , $d=(4-\frac{2}{n})-(4-\frac{1}{n})=\frac{-2}{n}+\frac{1}{n}=\frac{-1}{n}$
$\because$ Sum of n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow S_n=\frac{n}{2}[2(4-\frac{1}{n})+(n-1\left)\left(\frac{-1}{n}\right)\right]$
$=\frac{n}{2}\{8-\frac{2}{n}-1+\frac{1}{n}\}$
$=\frac{n}{2}(7-\frac{1}{n})=\frac{n}{2}\times \left(\frac{7n-1}{n}\right)=\frac{7n-1}{2}$