We use
, for the nth term of an A.P.
>>1.given that the tenth, fourth and first terms of an A.P<<
>>are the three consecutive terms of the G.P.<<
or
or
We have the system:
Solve that system by solving one for a variable and
substituting in the other. (Lot of messy work).
The two solutions are:
(d,r) = (0,1) or (4/3, 1/2)
The first solution, (d,r)=(0,1) gives the trivial sequence
4,4,4,4,4,4,4... for both the A.P. with d=0 and a G.P. with r=1.
4, 4, 4, 4, 4, 4, 4, 4, 4, 4 = A.P. with common difference 0
4, 4, 4 = G.P. which common ratio 1...
The second solution (d,r) = (4/3, 1/2) gives the A.P
4, 16/3, 20/3, 8, 28/3, 32/3, 12, 40/3, 44/3, 16 = A.P.
4, 8, , 16 = G.P.
The sum of the first 6 terms is given by the sum formula:
------------------
In the trivial case, 4,4,4,4,4,4,..., the sum of the first
6 terms is 24.
Edwin