We are given the details of an Arithmetic Progression (A.P. series):
We need to find the common difference ($d$) of this A.P. series.
The formula for the n-th term of an A.P. is:
$a_n = a_1 + (n-1)d$
Where $a_1$ is the first term and $d$ is the common difference.
Using this formula, we can set up two equations based on the given information:
$a_5 = a_1 + (5-1)d$
$52 = a_1 + 4d \quad \text{(Equation 1)}$
$a_{17} = a_1 + (17-1)d$
$184 = a_1 + 16d \quad \text{(Equation 2)}$
To find the common difference ($d$), we can subtract Equation 1 from Equation 2:
$ (a_1 + 16d) - (a_1 + 4d) = 184 - 52 $
Simplify the equation:
$ a_1 + 16d - a_1 - 4d = 132 $
$ 12d = 132 $
Now, isolate $d$ by dividing both sides by 12:
$ d = \frac{132}{12} $
$ d = 11 $
Therefore, the common difference of the A.P. series is 11.
The value of $1 + (\frac{1}{2^1} + \frac{1}{3}) + (\frac{1}{2^2} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + ... + (\frac{1}{2^9} + ... + \frac{1}{1023})$ lies between