#### i)Find the sum:1 + (-2) + (-5) + (-8) + ... + (-236)ii)Find the sum:$4-\frac{1}{n}+4-\frac{2}{n}+4-\frac{3}{n}+\ldots$ up to n termsiii) Find the sum:$\frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+.$ to 11 terms

i)

The given series is

$1 + (-2) + (-5) + (-8) + \ldots + (-236)\\$

Here

$a\textsubscript{1}=a = 1 , a\textsubscript{2}= -2\\$

$\\d = a\textsubscript{2} - a\textsubscript{1} = - 2 - 1 = - 3\\ a\textsubscript{n} = - 236\\ a + (n - 1) \times d = - 236 ( a\textsubscript{n} = a + (n - 1)d)\\ 1 + (n - 1) \times (-3) = - 236\\ (n - 1) \times (-3) = - 236 - 1\\ n - 1 = 79\\ n = 80\\$

The formula of the sum is :

${{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\$

$\mathrm{S}_{\mathrm{n}}=\frac{80}{2}[2 \times(1)+(80-1) \times(-3)]$
$= 40 [2 + 79 \times (-3)] \\ = 40 [2 - 237] \\ = 40 \times (-235) \\ S_{n} = -9400$

ii)

$\\The given series is \left[ 4-\frac{1}{n} \right]+\left[ 4-\frac{2}{n} \right]+\left[ 4-\frac{3}{n} \right]+.... `\\ First\,term(a)=\left[ 4-\frac{1}{n} \right] \\ d=\left[ 4-\frac{2}{n} \right]-\left[ 4-\frac{1}{n} \right] \\ =\frac{-2}{n}+\frac{1}{n} \\ =\frac{-1}{n} \ \\ \because {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\$

$\\S_n=\frac{n}{2}[2(4-\frac{1}{n})+\frac{1}{n}-1]\\\\=\frac{7n-1}{2}$

iii)

$\text{The given series is } \frac{a-b}{a+b} + \frac{3a-2b}{a+b} + \frac{5a-3b}{a+b} + \ldots \\ \text{Here } a_{1} = \frac{a-b}{a+b}, a_{2} = \frac{3a-2b}{a+b} \\ d = a_{2} - a_{1} \text{ here } d = \text{common difference} \\ \text{Given that } n = 11 \\$

\begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}\left[2 \mathrm{a}_{1}+(11-1) \mathrm{d}\right] &\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right)\right.\\ &=\frac{11}{2}\left[\frac{2(\mathrm{a}-\mathrm{b})}{\mathrm{a}+\mathrm{b}}+\frac{10(2 \mathrm{a}-\mathrm{b})}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{2 \mathrm{a}-2 \mathrm{~b}+20 \mathrm{a}-10 \mathrm{~b}}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{22 \mathrm{a}-12 \mathrm{~b}}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{2(11 \mathrm{a}-6 \mathrm{~b})}{\mathrm{a}+\mathrm{b}}\right] \\ & \mathrm{S}_{11}=\frac{11(11 \mathrm{a}-6 \mathrm{~b})}{(\mathrm{a}+\mathrm{b})} \end{aligned}