Geometric Series

The sum of the first 2011 terms is 200 and of the next 2011 terms is 380 - 200 = 180(given).

From 4023 to 6033 the next 2011 terms being in the geometric progression have sum= 180 (180/200), i.e. 162.

Hence the required sum of 6033 terms is 380 + 162 = 542.

The sum up to the $n$ -th term is given by $f\cdot\frac{1-c^n}{1-c}$ for some $f,c$ .

So as the first 2011 terms equal 200 we get: $f=\frac{200\cdot(1-c)}{1-c^{2011}}$ .

Now the first 4022 terms sum up to

$380=f\cdot\frac{1-c^{4022}}{1-c}=\frac{200\cdot(1-c)}{1-c^{2011}}\cdot \frac{1-c^{4022}}{1-c}=200(1+c^{2011})$

yielding $1+c^{2011}=\frac{380}{200}=\frac{19}{10}$ .

Hence we get $c^{2011}=\frac9{10}$

Now the interesting sum of the first 6033 terms is $f\cdot\frac{1-c^{6033}}{1-c}=\frac{200\cdot(1-c)}{1-c^{2011}}\cdot \frac{1-c^{6033}}{1-c} = \frac{200\cdot(1-c^{6033})}{1-c^{2011}}=200\cdot\frac{1-\left(\frac9{10}\right)^3}{1-\frac9{10}}=542$ .