# describe the sequence 0,1,0,1/3,0,1/5 and give the next 3 terms.

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Alternate terms are zero starting from the 1st term.

Alternate terms are 1/(2n-1) in the 2n postion.

So

T2n+1 = 0. Or the terms at odd places are zero. n=1,2,3 etc

T2n = 1/(2n-1)

So T6 = 1/5 is 6th given term

**T7 = 0**

**T8 = 1/(2*4-1) = 1/7** is the 8 th term

**T9 = 0.**

T10 = 1/(2*5-1) =1/9 ie the 10 th term.

And so on.....

Therefore T7,T8 and T9 are 0,1/7 and 0 respectively.