the sequence 3;-2;x;-24...is quadratic sequence. find the value of x

Answer 1

If I understand you correctly, you have the points
(1,3), (2,-2), (3,x), (4,-24)
and those points lie on a quadratic function graph.
let the function by y = ax^2 + bx + c
from (1,3) : a + b + c = 3 --- #1
from (2,-2) : 4a + 2b + c = -2 ---#2
from (4,-24) : 16a + 4b + c = -24 ---#3

#2 - #1: 3a + b = -5
#3 - #1: 15a + 3b = -27 --> 5a + b = -9
subtract those last two:
2a = -4
a = -2
in 3a+b= -5 ---> b = 1
in #1: -2+1+c=3
c = 4

y = -2x^2 + x + 4

so for the missing point (3,x)
x = -18+3+4 = -11