Originally Posted by
nikhil let get the general solution by which particular solution can be obtained.
let there b n terms with t(n) as the nth term. let
s=(-11)+(-8)+(-5)+....................................+t(n) (1)
s= ____(-11)+(-8)+(-5)+..............................+t(n) (2)(shift right by 1 term)
subtracting 2 from 1 we have
0=-11+3+3+3+..................................-t(n) or
t(n)=-11+3+3+....n-1 times therefor
t(n)=-11+3(n-1) or
t(n)=3n-14
now we are suppose to add t(1),t(2) till t(n) therefor
S(t(n))=S(3n)-S(14) S stands for summation
S(t(n))=3[(n(n+1))/2]-14n.................(after operating summation)
now for particular solution put n=19
S(t(19))=3(19)(10)-126=444
therefor solution is 444