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If the sum of the 100 terms of an A.P. is `-1` and the sum of the even numbered terms lying in first 100 terms is 1, thenA. common difference of the sequence is `(-3)/(50)`B. first term of the sequence is `(-149)/(50)`C. `100^(th)` term of the sequence is`(74)/(25)`D. arithmetic mean of the odd numbered terms is `(1)/(25)` |
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Answer» Correct Answer - B::C `(100)/(2)(2a+99d)=-1,(50)/(2)(2(a+d)+49xx2d)=1` solving we get `d=((3)/(50)),a=(-149)/(50)` `t_(100)=(74)/(25),AM=-(1)/(25)` |
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