Find the sum
(i) `1+(-2)+(-5)+(-8)+ … +(-236)`
(ii) `(4-(1)/(n))+(4-(2)/(n))+(4-(3)/(n))+ …` upto `n` terms.
(iii) `(a-b)/(a+b)+(3a-2b)/(a+b)+(5a-3b)/(a+b)+ … ` to 11 terms.

Explanation

(i) Here, first term (a) = 1 and common difference (d)`= (- 2)-1= -3`
` :' ` Sum of `n` terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]`
`implies S_(n)=(n)/(2)[2xx1+(n-1)(-3)]`
`implies S_(n)=(n)/(2)(2-3n+3)impliesS_(n)=(n)/(2)(5-3n) " " ` …(i)
We know that, if the last term `(l)` of an AP is known, then
` " " l=a+(n-1)d`
`implies -236=1+(n-1)(-3) " " [ :' l= -236 "given"]`
`implies -237= -(n-1)xx3`
`implies n-1=79impliesn=80`
Now, put the value of `n` in Eq. (i), we get
` S_(n)=(80)/(2)[5-3xx80]=40(5-240)`
` " " =40xx(-235) = -9400`
Hence, the required sum is - 9400.
Alternate Method
Given, `a=1,d= -3` and `l= -236`
` :. ` Sum of `n` terms of an AP, `S_(n)=(n)/(2)[a+l]`
`S_(n)=(80)/(2)(1+(-236) " " [ :' n=80]`
`" " =40xx(-235)= -9400`
(ii) Here, first term, `a=4-(1)/(n)`
Common difference, `d=(4-(2)/(n))-(4-(1)/(n))=(-2)/(n)+(1)/(n)=(-1)/(n)`
` :' ` Sum of `n` terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]`
`implies S_(n)=(n)/(2)[2xx(4-(1)/(n))+(n-1)((-1)/(n))]`
` " " =(n)/(2){8-(2)/(n)-1+(1)/(n)}`
` " " =(n)/(2)(7-(1)/(n))=(n)/(2)xx((7n-1)/(n))=(7n-1)/(2)`
(iii) Here, first term `(A)=(a-b)/(a+b)`
and common difference, `D=(3a-2b)/(a+b)-(a-b)/(a+b)=(2a-b)/(a+b)`
` :' ` Sum of `n` terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]`
`implies S_(n)=(n)/(2){2((a-b))/((a+b))+(n-1)((2a-b))/((a+b))}`
`" "=(n)/(2){(2a-2b+2an-2a-bn+b)/(a+b)}`
`" "=(n)/(2)((2an-bn-b)/(a+b))`
` :. S_(n)=(11)/(2){(2a(11)-b(11)-b)/(a+b)}`
`" " =(11(11a-6b))/(a+b)=(11)/(2)((22a-12b)/(a+b))`