Find the sum:
\( 1+(-2)+(-5)+(-8)+\ldots+(-236) \)

To do:

We have to find the given sums.


(i) The given A.P. is \( 1+(-2)+(-5)+(-8)+\ldots+(-236) \).


$a_1=1, d=-2-1=-3$ 

We know that,

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


This implies,

$l=a_n= 1 + (n-1)(-3)$

$-236= 1-3n+3$

$-236= 4-3n$

$3n= 4 + 236$

$3n =240$

$n = 80$





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





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 $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{n} =\frac{n}{2}[2 \frac{(a-b)}{(a+b)}+(n-1) \frac{(2 a-b)}{(a+b)}]$

$=\frac{n}{2}[\frac{2 a-2 b+2 a n-2 a-b n+b}{a+b}]$

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

$S_{11}=\frac{11}{2}[\frac{2 a(11)-b(11)-b}{a+b}]$

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

$=\frac{11(11 a-6 b)}{a+b}$

Updated on: 10-Oct-2022


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