Geometric Progression question (1 Viewer)

[laser6628]

Hi i'm new here...
Anyways this was from a past paper in my school and there are no posted solutions. So...
In a Geometric series, the sum of the 1st and 3rd terms is 39 whilst the sum of the 4th and 6th term is 131.625. Determine first term and common ratio.
Thanks

 

[michaeljennings]

I have common ratio of 1.5 and the first term is 12 Ill post working in a sec

Let the first term be 'a' and let the common ratio be 'd'

Equation 1

Equation 2

Equation 1 (factorised)
Equation 2 (factorised)

Then to eliminate (1+d^2) we get



From here 'a' can be eliminated and make 'd' the subject. Therefore d is 1.5

Sub d=1.5 back into equation 1 and a = 12

 

[MATHSCAPE]

T1 = a, T3 = ar^3, T4 = ar^3 and T6 = ar^5
T1 + T3 = a(1+r^2) = 39
T4 + T6 = ar^3(1+r^2) = 131.625
T4 + T6 = r^3(T1+T2) = 131.625
solve for r
r = 1.5
sub r=1.5 into T1 + T3 to solve for a
a = 12