Question 662303
The common ratio, is the ratio between consecutive terms in a geometric sequence, and it is constant, so
{{{2p/(p-1)=(4p+6)/2p}}} --> {{{4p^2=(p-1)(4p+6)}}} equating the cross-products
{{{4p^2=(p-1)(4p+6)}}} --> {{{4p^2=4p^2+2p-6}}} --> {{{0=2p-6}}} --> {{{2p=6}}} --> {{{highlight(p=3)}}}
 
The first three terms are
{{{p-1=3-1=2}}}
{{{2p=2*3=6}}} and
{{{4p+6=4*3+6=12+6=18}}}
The common ratio is {{{6/2=highlight(3)}}}
 
The sum of the first {{{n}}} terms of a geometric sequence with first term {{{A}}}
and common ratio {{{r}}} is
{{{a*((r^n-1)/(r-1))}}}
In this case, it is {{{2*((3^10-1)/(3-1))=2*(59049-1)/2=highlight(59048)}}}