Help! Maths! Geometric and arithmetic sequences.?

Question

The 4th, 8th, and 14th terms of an arithmetical sequence with common difference 0.5 are in a geometrical sequence. Find the first term of the arithmetical sequence and the common ratio of the geometrical sequence.

Answer 1

The terms can be expressed as:

(4th) a +3(0.5)
(8th) a +7(0.5)
and
(14th) a+13(0.5)

They are also in a geometric sequence.

(* NOTE : you don't specify if they are consecutive terms in that geometric sequence so there can be multiple answers if we don't make the following assumption.)

If the three are consecutive geometric terms, then we know that the pairwise ratios between them are equal:

[a +3(0.5)] / [a +7(0.5)] = [ a +7(0.5)]/[a+13(0.5)]

(a + 1.5)/ (a + 3.5) = (a + 3.5)/(a + 6.5)

cross multiply....
(a+1.5) x (a+6.5) = (a+3.5)^2

a^2 + 8a + 9.75 = a^ + 7a + 12.25

(8a -7a) = (12.25 - 9.75)

a = 2.5 is the first term of the arithmetic sequence
and
r= (a + 3.5)/ (a + 1.5) = (6)/(4) = 3/2 = 1.5 is the common ratio.

Answer 2

Let x = the first term
Then x+1.5 = 4th term
x+ 3.5 = 8th term
x+ 6.5 = 14th term
So (x+1.5)/(x+3.5)= (x+3.5)/(x+6.5)
Solving gives x = 2.5 = 1st term of arithmetic sequenc
The 4th, 8th, and 14th terms are 4,6, and 9 which is a geometric series with common ration 1.5.