Question

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

A group of bowling has won the award and $7600 has been divided between three participants, in proportional parts and is in geometric progression. 1,600.00 is the first participant.
How much did the second participant get?

Answer 1

The second recipient got `$2400`

This was built using a computer screen. Small screen mobile's mess up the formatting.

Explanation:

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`