A) \[A-B\]
B) \[A+B\]
C) \[2A\]
D) \[2B\]
Correct Answer: D
Solution :
Given that \[=\frac{n(n+1)}{2}\left[ \frac{n(n+1)}{2}+\frac{2n+1}{3} \right]\] Putting \[n=1,\ 2,\ 3,\ .............,\]we get \[{{S}_{1}}=A+B,\,{{S}_{2}}=2A+4B,\,\,{{S}_{3}}=3A+9B\] .............................................................. .............................................................. Therefore \[{{T}_{1}}={{S}_{1}}=A+B,\ {{T}_{2}}={{S}_{2}}-{{S}_{1}}=A+3B,\] \[{{T}_{3}}={{S}_{3}}-{{S}_{2}}=A+5B\], .............................................................. .............................................................. Hence the sequence is \[(A+B),(A+3B),\ (A+5B),........\] Here \[a=A+B\] and common difference \[d=2B\].
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