ElectricPavlov

a1 = x     r = 1/2 x^2

Sn =  a1 (1-r^n)/(1-r) = a1/(1-r)         =    x /  ( 1 - 1/2x^2)  = -12

                                                                x = -12 + 6x^2

                                                                0 = 6x^2 -x - 12               x = 18/12     and     -16/12

                                                                                                          added together = 2/12 = 1/6           

Since you do not know HOW you are supposed to solve this:

   As a follow on to cphill 's answer.....

IF   7x+8   is a factor....then   -8/7 is a root   you can do the synthetic division...or just substitute  -8/7 in to the equation to see if it will make it = zero

7 (-8/7)^3 - 55(-8/7)^2 - 135(-8/7) - 72       DOES = 0   

          so   7x + 8   IS a factor     (this is real easy if you have a calculator with a memory to store (-8/7) and then substitute it in.

You can do the same with the second one...... cphill already found   r-8 is NOT a factor.

x/y            (x+6)/y  = x/y + 1      and     x/(y+36) = x/y -1      solve these two equations for x and y

                 x+6   = x + y             

                  6 = y                        and    x/(6+36)=x/6  - 1     results in  x=7

original fraction thus equals   x/y = 7/6         

      add 6 to numerator , it increases by 1     add 36 to denominator   7/42 = 1/6    ...it has decreased by 1 

(vf - vi)/time = Average acceleration       Vf = velocity final   Vi = velocity initial

(30-0)/40sec  = 3/4 m/s^2

b estimate would be the point on the graph which is closest to parallel to the line drawn from 0,0   to 40, 30

       my estimate is 10 or 11 seconds

I think since you know the fake weighs a bit less, it will take 3 tries...

place 6 on each side    the lighter side has the fake

take the lighter side    and place   3   on each side

the lighter side again will have the fake

place one of each of these 3 on opposite sides of the scale......if they balance the one NOT on the scale is the fake....

   if they do NOT balance, the lighter side is the fake.

You COULD get lucky and pick two coins at random and one will weigh less ...if you luckily picked the fake...so in this case it would be ONE try would be the MINIMUM>

https://web2.0calc.com/questions/help_3472

If you look at the generic Sum of a Geometric Series

Sn= a1(1-r^n)/(1-r)      and the equation you are given  (with some manipulation)

  a1 = -6           and    (1-r^n) / (1-r)  =   (1+2^n)/(1+2)         so r = -2

Because   -6 (1+2^n) / (1+2)  =

                  6 (-1 -2^n)/(1+2)

                  2 (-1-2^n)   

-6 * -2 *-2*-2 = 48

log4(x-6) + log4(x-3) =1

log4((x-6)(x-3) = 1

log4(x^2-9x+18) = 1  

x^2-9x+18 = 4

x^2-9x+14 = 0

(x-7)(x-2) = 0      x = 7 and 2   but since there are no logs of negatives     x = 7

The total salt in the final solution has to equal the salt from the component solutions

x = liters of 25%        salt content = .25x

10-x = liters of 5 %    salt content = .05(10-x)

final solution salt content =  .1 (10) = 1

.25x+ .05(10-x) = 1

.25x + .5 - .05x = 1

.20x = .5

x = 2.5 liters of 25%       and  10 -2.5 = 7.5 liters of 5 %