Geometric Progression II

it is an infinite series here a=1,r=1/2 n=15 and it obviously a(r)to the Power n-1=1/16384

Change the term into $1,\frac{2}{1},\frac{4}{1},\frac{8}{1},\frac{16}{1},...$

$a=1$

$r=2$

$n=15$

$15^{th} term = ar^{n-1}$

$15^{th} term = 1\times 2^{15-1}$

$15^{th} term = 2^{14}$

$15^{th} term = 16384$

So the $15^{th}$ term is $\boxed{\frac{1}{16384}}$

$1\times \left ( \frac{1}{2} \right )^{15-1} = \frac{1}{16384}$

In the above sequence the numerator remains the same and the formula is 2^position. But here the first number is 1 and its denominator is also 1. So, we have to subtract 1 from 15 which is 14.

= 2^14

=16384

= 1/16384

Here, a=1 ; common ratio, r = (a {2}/a {1}) = (1/2)/1 =1/2 and n=15. Thus, nth term = a(r^(n-1)) = 1((1/2)^(15-1))= 1/16384

since 1/1 is already equivalent to 1 then it is divided to 2 fifteen times, the answer is 1/16384