1st, let's see what kind of PA we have:for i=1 , the 1st term =a₁ = 9
for I=2 . the 2nd term =a₂ = 11
for I=3 , the 3rds term =a₃ =13, we notice that a₂-a₁ =a₃-a₂ = 2 =d.
So we know the value of a₁ =9 & the common difference d=2 & we know also that the number of term n = 18 (I goes from 1 to 18). Let; calculate the value of the 18th term (the last term): Last term = a₁ + (n-1)d, ,Plug the know values: Last term = 9 + (18-1) (2) ==> L=43
And the sum = (a₁ + L) (n/2) ==> S=(9+43)(18/2) = 52x9 = 468
