Answer:
T_12 = 5(T_3)
Step-by-step explanation
T_17 = 13(T_2)
T_n = a + (n-1)d; where a=first term and d=common difference
T_17 = a + 16d
T_2 = a + d
Therefore,
a + 16d = 13(a + d)
a + 16d = 13a + 13d
16d - 13d = 13a - a
3d = 12a
d = 4a
To prove that T_12 = 5(T_3)
a + 11d = 5(a + 2d)
a + 11d = 5a + 10d
since d = 4a
a + 11(4a) = 5a + 10(4a)
a + 44a = 5a + 40a
45a = 45a
Since LHS = RHS
Therefore; T_12 = 5(T_3)
