A) 201
B) - 201
C) - 198
D) + 198
Correct Answer: C
Solution :
(c): We have, \[Sn=3n-{{n}^{2}}\] \[\] \[{{S}_{1}}=2,\text{ }{{S}_{2}}=2;\text{ }{{S}_{3}}=0;\text{ }{{S}_{4}}=-4;\text{ }{{S}_{5}}=-10\] Thus, common difference is d = ? 2 Hence the 101st term of the sequence is \[=a+\left( 101-1 \right)d=2+100\times \left( -2 \right)\text{ }=-198\]You need to login to perform this action.
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