Answer to Question #217556 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

This is a geometric progression.

Let the 1st term be $a$ and the common ratio $r$.

Then $a=\frac{1}{n}$ and $r=\frac{1}{2n} ÷ \frac{1}{n} = \frac{1}{2n} × \frac{n}{1} = \frac{1}{2}.$

Let the nth term be $U_n.$ By the formula for the nth term of a geometric progression,

$U_n=ar^{n-1}$

So the 16th term of the geometric progression is

$U_{16}=ar^{15}=\frac{1}{n} × \left(\frac{1}{2}\right)^{15} \\=\frac{1}{n}×\frac{1}{32768} = \frac{1}{32768n}$

Let the sum of the first $n$ terms of the sequence be $S_{16}.$ Then

$S_{16}=\dfrac{a\left(1-r^{n}\right)}{1-r}$

So the sum of the first 16 terms is

$S_{16}=\dfrac{a\left(1-r^{n}\right)}{1-r} \\ = \dfrac{1}{n}\dfrac{\left(1-\left(\dfrac{1}{2}\right)^{n}\right)}{1-\dfrac{1}{2}} \\$

$=\dfrac{1}{n}×2×\dfrac{65535}{65536} \\ =\dfrac{65535}{32768n}$

If $n = 156$ then

$S_n=\dfrac{65535}{511808}$