Question

Find the sum of n terms of the series (4-1/n)+(4-2/n)...

Answer 1

$a = 4 - \frac{1}{n} \\ a2 = 4 - \frac{n}{2} \\ d = a2 - a \\ d = 4 - \frac{2}{n} - 4 + \frac{1}{n} \\ d = \frac{ - 1}{n}$

$sum = \frac{n}{2} (2a + (n - 1)d) \\ = \frac{n}{2} (2(4 - \frac{1}{n}) + (n - 1)( \frac{ - 1}{n})$

$on \: taking \: \frac{1}{n} \: common$

we get,

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

n and 1/n gets cancelled

$\frac{1}{2} (6 + 1 - n) \\ sum= \frac{1}{2} (7 - n)$

Therefore sum of the given progression is

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