Some people above have mentioned this method but no one has illustrated properly so here goes:
Taking $a_{n-1}$ to LHS,
$a_{n} - a_{n-1} = 6n^{2} + 2n$
Taking summation,
$\sum a_{n} - \sum a_{n-1} = 6 \sum n^{2} + 2 \sum n = (n)(n+1) (2n+1) + n(n+1)$
We can write like this
$\sum a_{n} - \sum a_{n-1}= a_{n}+\sum a_{n-1} - \sum a_{n-1}=a_{n}$
Example for the above statement:
For the $AP:2,4,6,8 $ to get the $4^{th}$ term, $(2+4+6+8)- (2+4+6) = 8$
For the $GP: 3,9,27,81$ for the $3^{rd}$ term, $(3+9+27) - (3+9) = 27$
so our equation reduces to: $a_{n} = (n)(n+1) (2n+1) + n(n+1)$
Putting n = 99 as asked in the question,
$a_{99} = 99*100*199 + 99*100 = 1980000 = K * 10^{4} $
So K =198. Now wasn't this easy.😀
---------------------------------------------------------------------------
p.s.:
$1+2+3+.....+n=\sum{n}=\frac{n(n+1)}{2}$
$1^{2}+2^{2}+3^{2}+.....+n^{2}=\sum{n^{2}}=\frac{n(n+1)(2n+1)}{6}$
$1^{3}+2^{3}+3^{3}+.....+n^{3}=\sum{n^{3}}=\left[\frac{n(n+1)}{2}\right]^{2}=\frac{n^{2}(n+1)^{2}}{4}$