Question

1. The sum of the three consecutive members of the geometric sequence is 13. The proportion of the third and the first member is 9. Determine the members of this sequence.
2. Between the numbers 4 and 60 we insert the two numbers so that the first three consecutive numbers are making geometric sequence and the last three consecutive numbers are an arithmetic sequence. What numbers did we use?

Answer 1

If a = first term and r = common ratio we have

a + ar + ar^2 = 13   and ar^2 / a = r^2 = 9

so r = 3

and a + 3a + 9a = 13 
so a = 1

so they are  1,3 and 9 

2.
in geometric series we have

4 , 4r ,4r^2 , 60

Arithmetic;
 4,  4r , 4r + d ,  4r + 2d

so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30 
so d = 30 - 2r 
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0

2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r  = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24

and the numbers are 4*3 =  12 , 4*3^2 = 36

first 3  are  4 , 12 and 36  ( in geometric)
and last 3 are 12, 36 and 60  ( in arithmetic)

The 2 numbers we ause are 12 and 36.