Consider the sequence 1,3,3,3,5,5,5,5,5,7,7,7,7,7,7,7.............. and evaluate it's 2016th term
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If you observe, you'll find that the NUMBER of terms are in an AP, starting with 1 and having a common difference 2. The sum of the number of terms will be equal to 2016.
So, [tex] \frac{n}{2}(2.1+(n-1).2)=2016 n(2+2n-2)=4032 2n^2=4032 n^2=2016 n= \sqrt{2016} n=44.8=45. So, the 2016th term is 45. [/tex]
So, [tex] \frac{n}{2}(2.1+(n-1).2)=2016 n(2+2n-2)=4032 2n^2=4032 n^2=2016 n= \sqrt{2016} n=44.8=45. So, the 2016th term is 45. [/tex]
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