Find the sum

Solution:

(i) Here. first term $(a)=1$ and common difference $(d)=(-2)-1=-3$

$\because$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow$ $S_{n}=\frac{n}{2}[2 \times 1+(n-1) \times(-3)]$

$\Rightarrow$ $S_{n}=\frac{n}{2}(2-3 n+3) \Rightarrow S_{n}=\frac{n}{2}(5-3 n)$ ...(i)

We know that, if the last term $(l)$ of an AP is known, then

$l=a+(n-1) d$

$\Rightarrow \quad-236=1+(n-1)(-3) \quad[\because l=-236$, given $]$

$\Rightarrow \quad-237=-(n-1) \times 3$

$\Rightarrow \quad n-1=79 \Rightarrow n=80$

Now, put the value of $n$ in Eq. (i), we get

$S_{n}=\frac{80}{2}[5-3 \times 80]=40(5-240)$

$=40 \times(-235)=-9400$

Hence, the required sum is $-9400$. Alternate Method

Given, $a=1, d=-3$ and $l=-236$

$\therefore$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[a+l]$

$=\frac{80}{2}(1+(-236)$ $[\because n=80]$

$=40 \times(-235)=-9400$

(ii) Here, first term, $a=4-\frac{1}{n}$

Common difference, $d=\left(4-\frac{2}{n}\right)-\left(4-\frac{1}{n}\right)=\frac{-2}{n}+\frac{1}{n}=\frac{-1}{n}$

$\because$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow$  $S_{n}=\frac{n}{2}\left[2\left(4-\frac{1}{n}\right)+(n-1)\left(\frac{-1}{n}\right)\right]$

$=\frac{n}{2}\left\{8-\frac{2}{n}-1+\frac{1}{n}\right\}$

$=\frac{n}{2}\left(7-\frac{1}{n}\right)=\frac{n}{2} \times\left(\frac{7 n-1}{n}\right)=\frac{7 n-1}{2}$

(iii) Here, first term $(A)=\frac{a-b}{a+b}$

and common difference, $D=\frac{3 a-2 b}{a+b}-\frac{a-b}{a+b}=\frac{2 a-b}{a+b}$

$\because$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow \quad S_{n}=\frac{n}{2}\left\{2 \frac{(a-b)}{(a+b)}+(n-1) \frac{(2 a-b)}{(a+b)}\right\}$

$=\frac{n}{2}\left\{\frac{2 a-2 b+2 a n-2 a-b n+b}{a+b}\right\}$

$=\frac{n}{2}\left(\frac{2 a n-b n-b}{a+b}\right)$

$\therefore$$S_{11}=\frac{11}{2}\left\{\frac{2 a(11)-b(11)-b}{a+b}\right\}$

$=\frac{11(11 a-6 b)}{a+b}=\frac{11}{2}\left(\frac{22 a-12 b}{a+b}\right)$