Answer to Question #217655 in Algebra for Ali Haider

Question

(b) For the given sequence 1/n, 1/2n, 1/4n,…  Find the 16th term. Find sum of first sixteen terms where n is your arid number i.e. 19-arid-234 take n=234

Answer 1

Given sequence is

$\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}...$

To find the 16th term and sum of first sixteen terms where n = 234.

We have the given sequence is

$\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}...$

Now let us consider the terms of the sequence to be named as a_k

Then

$a_1= \frac{1}{n}, a_2 = \frac{1}{2n}, a_3=\frac{1}{4n}$

Next

$\frac{a_2}{a_1}= \frac{ \frac{1}{2n} }{\frac{1}{n}} \\and \\\frac{a_3}{a_2}=\frac{\frac{1}{4n}}{\frac{1}{2n}}=\frac{1}{2} \\$

hence the common ratio of the terms of the given sequence are equal so the given sequence is a geometric sequence whose first term is $a_1 = \frac{1}{n}$ and common ratio r is $r=\frac{1}{2}$

As the given sequence is a geometric sequence so its kth term is given by $a_k=a_1r^{k-1}$ i.e $a_k=\frac{1}{n}\Big(\frac{1}{2}\Big)^{k-1}$

For the 16th term we consider k=16 in $a_k=\frac{1}{n}\Big(\frac{1}{2}\Big)^{k-1}$ then we have

$a_{16}=\frac{1}{n}\Big(\frac{1}{2}\Big)^{16-1} \\ a_{16}=\frac{1}{n}\Big(\frac{1}{2}\Big)^{15} \\ a_{16}= \frac{1}{32768n}$

Next the sum of first sixteen terms of the given sequence will be given by $S_{16}=\frac{a_1(1-r^{16})}{1-r}$ as the formula for the sum of the first k terms with first term a₁ and common ratio r<1 is given by $S_{k}=\frac{a_1(1-r^{k})}{1-r}$

For $a_1 = \frac{1}{n}$ and $r=\frac{1}{2}$ in $S_{16}=\frac{a_1(1-r^{16})}{1-r}$ we get

$S_{16}=\frac{\frac{1}{n}(1-(\frac{1}{2})^{16})}{1-\frac{1}{2}} \\ S_{16}=\frac{\frac{1}{n}(\frac{65535}{65536})}{\frac{1}{2}} \\ S_{16}=\frac{65535}{32768n}$

So the 16th term and sum of first sixteen terms of the sequence $\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}…$ are $\frac{1}{32768n}$ and $\frac{65535}{32768n}$ respectively.

As n = 234 so substituting n = 234 in the answers we get the 16th term to be $\frac{1}{7667712}$ and sum of first sixteen terms to be $\frac{21845}{2555904}$

Hence the required answers for n = 234 are 16th term is $\frac{1}{7667712}$ and sum of first sixteen terms is $\frac{21845}{2555904}$