Let S_n denote the sum of first n-terms of an arithmetic progression.

$\mathrm{S}_{10}=530 \Rightarrow \frac{10}{2}\{2 \mathrm{a}+9 \mathrm{~d}\}=530$

$\Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=106$

and $S_{5}=140 \Rightarrow \frac{5}{2}\{2 a+4 d\}=140$

$\Rightarrow 2 \mathrm{a}+4 \mathrm{~d}=56$

$\Rightarrow 5 \mathrm{~d}=50 \Rightarrow \mathrm{d}=10 \Rightarrow \mathrm{a}=8$

Now, $\mathrm{S}_{20}-\mathrm{S}_{6}=\frac{20}{2}\{2 \mathrm{a}+19 \mathrm{~d}\}-\frac{6}{2}\{2 \mathrm{a}+5 \mathrm{~d}\}$

$=14 a+175 d$

$=(14 \times 8)+(175 \times 10)$

$=1862$