Answer to Question #217550 in Algebra for Bhatti

Question

For the given sequence 1/n, 1/2n, 1/4n,…  Find the 16th term. Find sum of first sixteen terms where n is 156 

Answer 1

${a_1} = \frac{1}{n} = \frac{1}{{{2^0} \cdot n}}$

${a_2} = \frac{1}{{2n}} = \frac{1}{{{2^1} \cdot n}}$

${a_3} = \frac{1}{{4n}} = \frac{1}{{{2^2} \cdot n}}$

Then

${a_k} = \frac{1}{{{2^{k - 1}} \cdot n}}$

So

${a_{16}} = \frac{1}{{{2^{15}} \cdot n}} = \frac{1}{{{\rm{32768}}n}}$

We find the required sum as the sum of the first 16 terms of the geometric progression with the first term ${a_1} = \frac{1}{n}$ and the denominator $q = \frac{1}{2}$ :

${S_{16}} = \frac{{{a_1}\left( {1 - {q^{16}}} \right)}}{{1 - q}} = \frac{{\frac{1}{n}\left( {1 - {{\left( {\frac{1}{2}} \right)}^{16}}} \right)}}{{1 - \frac{1}{2}}} = \frac{{65535}}{{32768n}}$

If $n=156$ then

${S_{16}} = \frac{{65535}}{{32768 \cdot 156}} = \frac{{21845}}{{1703936}}$

Answer: ${a_{16}} = \frac{1}{{{\rm{32768}}n}}$ , ${S_{16}} = \frac{{21845}}{{1703936}}$