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Find the sum:
$ 1+(-2)+(-5)+(-8)+\ldots+(-236) $
To do:
We have to find the given sums.
Solution:
(i) The given A.P. is \( 1+(-2)+(-5)+(-8)+\ldots+(-236) \).
Here,
$a_1=1, d=-2-1=-3$
We know that,
$a_n=a+ (n-1)d$
$S_n=\frac{n}{2}(a+l)$
This implies,
$l=a_n= 1 + (n-1)(-3)$
$-236= 1-3n+3$
$-236= 4-3n$
$3n= 4 + 236$
$3n =240$
$n = 80$
Therefore,
$S_n=\frac{80}{2}[1+(-236)]$
$=40(-235)$
$=-9400$
Therefore, \( 1+(-2)+(-5)+(-8)+\ldots+(-236)=-9400 \).
(ii) Let the number of terms of the given A.P. be $n$, first term be $a$ and the common differnce be $d$.
First term $a_1=a=4-\frac{1}{n}$
Second term $a_2= 4-\frac{2}{n}$
Common difference $d=a_2-a_1=4-\frac{2}{n}-(4-\frac{1}{n})=\frac{-2+1}{n}=\frac{-1}{n}$
We know that,
Sum of $n$ terms $S_{n} =\frac{n}{2}(2a+(n-1)d)$
$=\frac{n}{2}[2(4-\frac{1}{n})+(n-1)(\frac{-1}{n})]$
$=\frac{n}{2}[\frac{8n-2-n+1}{n}]$
$=\frac{n}{2}(\frac{7n-1}{n})$
$=\frac{7n-1}{2}$
Hence, the sum of the $n$ terms of the given series is $\frac{7n-1}{2}$.
(iii) In the given sequence,
First term $a_1=\frac{a-b}{a+b}$
Common difference $d=\frac{3 a-2 b}{a+b}-\frac{a-b}{a+b}$
$=\frac{2 a-b}{a+b}$
Sum of $n$ terms of an AP $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$S_{n} =\frac{n}{2}[2 \frac{(a-b)}{(a+b)}+(n-1) \frac{(2 a-b)}{(a+b)}]$
$=\frac{n}{2}[\frac{2 a-2 b+2 a n-2 a-b n+b}{a+b}]$
$=\frac{n}{2}(\frac{2 a n-b n-b}{a+b})$
$S_{11}=\frac{11}{2}[\frac{2 a(11)-b(11)-b}{a+b}]$
$=\frac{11}{2}(\frac{22 a-12 b}{a+b})$
$=\frac{11(11 a-6 b)}{a+b}$
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