If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence an is given by:

an = a1 + (n-1)d

a3 = 4

a1 + (4-1)d = 4

a1 + 3d = 4

a17 = - 52

a1 + (17-1)d = -52

a1 + 16d = -52

by solving the system of equations

a1 + 3d = 4

a1 + 16d = -52

we find

a1 = 220/13

d = -56/13

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a9 = a1 + (9-1)d

a9 = 220/13 + 8*(-56/13)

a9 = -228/13

a9 = -17 7/13

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the nineth term is -228/13 or -17 7/13.