1.) Three numbers form a geometric sequence whose common ratio is 0.5. If the first is reduced to 10 more than one quarter its value, the second decreased by 10, and the third increased by 10 more than twice its value, the resulting three numbers form an arithmetic sequence. Determine the original three numbers.

Let x be the first number in the sequence, so the first three numbers are

{x, 0.5x, 0.5²x}

Then

{x/4 + 10, 0.5x - 10, 2(0.5²x) + 10}

is arithmetic, so there is some constant c such that

0.5x - 10 = x/4 + 10 + c   ==>   x/2 - 10 = x/4 + 10 + c

2(0.5²x) + 10 = 0.5x - 10 + c   ==>   x/2 + 10 = x/2 - 10 + c

Solve the second equation for c :

x/2 + 10 = x/2 - 10 + c

c = 20

Substitute this into the first equation and solve for x :

x/2 - 10 = x/4 + 10 + 20

x/4 = 40

x = 160

Then the terms are

{160, 80, 40}

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

w'x is 0.

hope it