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Question:

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12

Solution:

By using nth termtn=a+(n−1;)dwe haveAs givena+2d=16...2ndtermAnda+6d−(a+4d)=12By using abovee equations,⇒d=6∴a+2(6)=16, thena=4Then A.P isa,a+d,a+2d,a+3d...So,4,4+6,4+2(6),4+3(6),4+4(6)..4,10,16,22,28,34