If the sum of first 11 terms of an AP a1,a2,a3,..........is 0(a1is not equal to 0)then the sum of a1,a3,a5,.........a23 is ka1, where k is equal to:

Answers (1)

Solution: 

\sum ak=0\ \Rightarrow 11a+55d=0\ \Rightarrow a+5d=0

Now,

\ \a1+a3+..........+a23=ka1\ \Rightarrow 12a+d(a1+a3+.........+a23)=ka1\ \Rightarrow 12a+d(2+4+6+.....+22)=ka\ \Rightarrow 12a+2d	imes 66=ka\12(a+11d)=ka\ \Rightarrow 12(a+11(-a/5))=ka\ \Rightarrow 12(1-11/5)\ \ k=-72/5

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