Given that Sn=nA+n2B
Putting n = 1, 2, 3, ..... ......, we get
S1=A+B,S2=2A+4B,S3=3A+9B
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Therefore T1=S1=A+B,T2=S2−S1=A+3B,T3=S3−S2=A+5B,
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Hence the sequence is (A+B)(A+3B),(A+5B),.....
Here a = A + B and common difference d = 2B.