Question 891695
The sum of the first n terms of an AP is given by 

{{{A[n] = (n/2)*(2*a[1] + (n-1)d)}}}
==> {{{A[20] = 10(2*a[1] + 19d) = 420 }}}
and {{{A[50] = 25(2*a[1] + 49d) = 2550}}}

or {{{2*a[1] + 19d = 42 }}} and {{{2*a[1] + 49d = 102 }}}
Solving these equations give

{{{a[1] = d = 2}}}

==> {{{g[11] = g[1]*r^(11-1) = 2*2^10 = 2^11 = 2048}}}