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, then

Options

A. 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)`

Answer

B, C

Explanation

`(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)`