(1) This is an arithmetic progression (AP) with a common difference of -8.
The first term is 156 (when n=1)=156-8(n-1)=156-8n+8=164-8n. Formula is 164-8n.
(2) First term to become negative is when 164-8n<0, 164<8n, 41<2n, n>41/2. So n>20.5. The first integer is n=21. 164-8×21=164-168=-4.
The first negative term is the 21st term (=-4).
(3) T(n)=an²+bn+c is the quadratic formula such that:
Assuming n starts at 1, T(2)-T(1)=156, the first term of the AP.
T(3)-T(2)=148, T(4)-T(3)=140, T(5)-T(4)=132.
T(5)=-24, so -24-T(4)=132, T(4)=-156.
So T(3)=T(4)-140=-296.
T(2)=T(3)-148=-444.
T(1)=T(2)-156=-600.
We need 3 equations to find 3 unknowns a, b, c.
(1) n=1: a+b+c=T(1)=-600;
(2) n=2: 4a+2b+c=T(2)=-444;
(3) n=3: 9a+3b+c=T(3)=-296.
(4)=(2)-(1): 3a+b=156,
(5)=(3)-(1): 8a+2b=304, 4a+b=152,
(6)=(5)-(4): a=-4, so b=152+16=168.
c=-600-a-b=-600+4-168=-764.
T(n)=-4n²+168n-764.
CHECK according to the formula for T(n):
T(1), T(2), T(3), T(4), T(5)=-600,-444,-296,-156,-24.
First differences: 156, 148, 140, 132 which is the given AP.