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:

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\times 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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