[Chrissy1994]

So I have a math group project that my group and I have completed all questions except this one. I hope I'll be able to get some help here. I'm assuming that I'll have to get a system of simultaneous equations in order to solve for r, d and at least an a. My problem is finding these equations in the first place to start with. I'll post below what I have so far.

The question:

Question 10 [5 mks]
Given the following concerning an arithmetic series and a geometric series:
- The fourth and seventh terms of the arithmetic series are the same as the second and third terms of the geometric series respectively.
- The first term of the geometric series exceeds the first term of the arithmetic
series by 64/3.
- The sum of the first three terms of the arithmetic series, S_AP-3 and the sum of
the first two terms of the geometric series, S_GP-2 are related by the formula
S_AP-3 + 3S_GP-2 + 21 = 0.
What is the total of the sum of the first four terms of the arithmetic series and the sum
of the first three terms of the geometric series?

What I have:

Let a_n be the arithmetic series
Let g_n be the geometric series

We are given that:
a₄ = g₂
a₇ = g₃
g = a + 64/3

from the general formulas for specific elements we have:

g₂ = g * r = a₄ = a + 3d
g3 = g* r² = a₇ = a + 6d

Therefore, we get:
(a + 64/3)r = a + 3d = ar + (64/3)r - a - 3d = 0
(a + 64/3)r² = a + 6d = ar² + (64/3)r² - a - 6d = 0
r = (a + 6d) - (a+3d)

[ebaines]

First, please note that you misused equals signs in two of the last three equations - those should be:

(a + 64/3)r = a + 3d --> ar + (64/3)r - a - 3d = 0
(a + 64/3)r^2 = a + 6d --> ar^2 + (64/3)r^2 - a - 6d = 0

How did you arrive at the last equation. If using a7-a4,that would yield g(r^2-r).

You have not yet tried to apply the final condition that you were given, which is:

A+(A+d)+(A+2d) + 3(g+gr) +21 = 0 --> 3A + 3d +3(A+64/3)(1+r) + 21 = 0

That gives you an additional equation, so now you should be able to solve for A, d, and r, and then back substitute for g