Given that 6,9,131/2,201/4 form a sequence of numbers. Find the sum of its 13 terms

Answer 1

I don't know the answer so help me

Answer 2

6,9,131/2,201/4
the way you typed it, I see no pattern but ..

6,9,13 1/2,20 1/4, ...
= 6, 9, 27/2, 81/2, ..
a GP with a = 6, r = 3/2
sum(13) = a(r^13 - 1)/(r-1)
= 6( (3/2)^13 - 1)/(3/2 - 1)
= appr 2323.434