(i) `7+10(1)/(2)+14+....+84`
Here, a=7,
`d=10(1)/(2)-7=14-10(1)/(2)=(7)/(2)`
Let `l=a_(n)=84`
`rArr 7+(n-1)(7)/(2)=84 rArr (n-1)(7)/(2)=77`
`rArr n-1=22 rArr n=23`
Now, `S_(n)=(n)/(2)(a+l)`
`rArr S_(23)=(23)/(2)(7+84)=(2093)/(2)`
(ii) `34+32+30+...+ 10`
Here, a=34,
`d=32-34=30-32=-2`
Let `l=a_(n)=10`
`rArr34+(n-1)(-2)=10) rArr (n-1)(-2)=-24`
`rArr n-1=12 rArr n=13`
Now, `S_(n)=(n)/(2)(a+l)`
`rArr S_(13)=(13)/(2)(34+10)=13xx22=286`