Precalculus, Chapter 9, 9.2, Section 9.2, Problem 38

a_3=19
a_15=-1.7
Let a_1 be the first term and d be the common difference of the sequence.
a_15=a_1+14d
a_1+14d=-1.7 ---------- (1)
a_3=a_1+2d
a_1+2d=19 ----------- (2)
Now let's solve the equations 1 and 2 to get the a_1 and d,
Subtract equation 2 from equation 1,
14d-2d=-1.7-19
12d=-20.7
d=-20.7/12
d=-1.725
Plug the value of d in equation 2,
a_1+2(-1.725)=19
a_1-3.45=19
a_1=19+3.45
a_1=22.45
a_2=a_1+d
a_2=22.45+(-1.725)
a_2=20.725
a_3=a_2+d
a_3=20.725+(-1.725)
a_3=19
a_4=a_3+d
a_4=19+(-1.725)
a_4=17.275
a_5=a_4+d
a_5=17.275+(-1.725)
a_5=15.55
So the first five terms of the sequence are 22.45,20.725,19,17.275 and 15.55