Given that:
Sq= 63q – 3q2
Put q = 1
S1 = T1 = 63 – 3 = 60
Put q = 2
S2 = 63 x 2 – 3 x (2)2 = 126 – 12 = 114
T2 = S2 – S1 = 114 – 60 = 54
Put q = 3
S3 = 63 x 3 – 3(3)2
= 189 – 27 = 162
T3 = S3 – S2
= 162 – 114 = 48
Therefore, first term of this A.P. is 60 and Common difference is 54 - 60 = -6
Tp = a + (p – 1) d
-60 = 60 + (p – 1) (-6)
-120 = (p – 1) (-6)
(p – 1) = 20
p = 21
Now 11th term of this A.P
T11 = 60 + (11 – 1) (-6)
= 60 – 60 = 0