Here, a1=5, a2=143, a3=133 and a4=4
a2−a1=143−5=14−153=−13,
a3−a2=133−143=−13,
a4−a3=4−133=12−133=−13
∴ a2−a1=a3−a2=a4−a3
Since the difference between successive terms are same, it forms an AP.
The next three terms are,
a5=a1+4d=5+4(−13)=5−43=113
[∵ an=a1+(n−1)d]
a6=a1+5d=5+5(−13)=5−53=103
a7=a1+6d=5+6(−13)=5−2=3