How solve it ? How much did the second participant get?

Let the geometric term be #ar^n#

so we have #ar^0; ar; ar^2;ar^3...#

Note that #r^0# has the same value as #1#

Given that #ar^0=1600color(white)("d")->axx1=1600#

Thus #a=1600#

It is also given that #ar^0+ar+ar^2=7600" "..Eqn(1)#

Factor out #a# giving #a(1+r+r^2)=7600#

#a(1+r+r^2)=7600color(white)("ddd")->color(white)("ddd")1600(1+r+r^2)=7600#

Divide both sides by 1600

#color(white)("dddddddddddddddddddd")->color(white)("ddd")1+r+r^2color(white)("d.")=(76cancel(00))/(16cancel(00))#

Subtract #76/16# from both sides.

#color(white)("dddddddddddddddddddd")->color(white)("ddd")r^2+r-60/16=0" "Eqn(2)#

Note that #(60-:4)/(16-:4)=15/4#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Now we have a quadratic and only 1 of the solutions to "Eqn(2)" should work in "Eqn(1))#

Compare #y=0=ax^2+bx+c color(white)("d")->color(white)("d")y=0=r^2+r-15/4#

#x=(-b+-sqrt(b^2-4ac))/(2a)color(white)("d")->color(white)("d")r=(-1+-sqrt(1^2-4(1)(-15/4)))/(2(1)) #

#color(white)("dddddddddddddddddd")->color(white)("d")r=-1/2+-sqrt(16)/2#

#color(white)("dddddddddddddddddd")->color(white)("d")r=-1/2+-2#

#color(white)("DDDDDDDdddddddd")->color(white)("d")r=-5/2 and + 3/2#

The #-5/2# will not work as #1600xx(-5/2)# gives a negative value so it is not logical.

Testing #(+3/2)#

#1600+[1600xx3/2]+[1600xx(3/2)^2] #

#1600+[2400]+[3600] = 7600" as required"#

So the second recipient got #$2400#