What is the solution to this problem (Geometric and Arithmetic progression)?

Question

Numbers $a , b,c , 64$ are consecutive members of a geometric progression.
Numbers $a,b,c$ are respectively the first, fourth, eighth members of an arithmetic progression.

Calculate $a + b - c$

Answer 1

Set $a,b=aq,c=aq^2,64=aq^3$ consecutive members of the geometric progression. Clearly, $a,q\neq0$ because the sequence contains a non-zero term $64.$

If $r$ is the ratio of the arithmetic progression, then $$\begin{aligned}b&=a+3r=aq&\Rightarrow \quad3r&=a(q-1)\\c&=a+7r=aq^2&\Rightarrow \quad 7r&=a(q-1)(q+1)\end{aligned}$$ It is sure that $q\neq 1$ (from the given constraints).
From the two last equations we elliminate $r$ to get $$q+1={7\over 3}$$

I am sure you can finish the solution.

Answer 2

We can also work out the details from the requirements that $$ a_1 \ = \ a \ \ , \ \ a_4 \ = \ a + (4 - 1)·d \ = \ b \ \ , \ \ a_8 \ = \ a + (8 - 1)·d \ = \ c \ \ , $$ where $ \ d \ $ is the difference between terms of the arithmetic progression, so $ \ a + b - c \ = \ a + (a + 3d) - (a + 7d) \ = \ a - 4d \ \ . $

That these numbers also form a geometric progression requires that the ratio between terms is given by $$ r \ \ = \ \ \frac{b}{a} \ = \ \frac{c}{b} \ \ \Rightarrow \ \ \frac{a \ + \ 3d}{a} \ = \ \frac{a \ + \ 7d}{a \ + \ 3d} \ \ \Rightarrow \ \ a^2 \ + \ 7ad \ \ = \ \ a^2 \ + \ 6ad \ + \ 9d^2 $$ $$ 9d^2 \ - \ ad \ \ = \ \ d·(9d - a) \ \ = \ \ 0 \ \ ; $$ since $ \ d \neq 0 \ \ , $ we have $ \ a = 9d \ \ , $ indicating that $ \ a \neq 0 \ \ $ also.

Thus far, our geometric progression is then $ \ a \ \ , \ \ a \ + \ 3·\frac{a}{9} \ = \ \frac43 a \ \ , \ \ a \ + \ 7·\frac{a}{9} \ = \ \frac{16}{9} \ = \ \left(\frac43 \right)^2 a $ $ , \ \ 64 \ \ ; \ $ furthermore, $ \ a + b - c \ = \ a \ - \ 4·\frac{a}{9} \ = \ \frac59·a \ \ . $

The terms in the geometric progression thus tell us that $$ \left(\frac43 \right)^3 · a \ \ = \ \ \frac{64}{27}·a \ \ = \ \ 64 \ \ \Rightarrow \ \ a \ \ = \ \ 27 \ \ \Rightarrow \ \ a + b - c \ = \ \frac{5}{9}·27 \ \ = \ \ 15 \ \ . $$ [We can check this by calculating $$ \ d \ = \ \frac{27}{9} \ = \ 3 \ \ \Rightarrow \ \ b \ \ = \ \ 27 \ + \ 3·3 \ = \ \ 36 \ \ = \ \ \frac43·27 \ \ , $$ $$ c \ \ = \ \ 27 \ + \ 7·3 \ \ = \ \ 48 \ \ = \ \ \frac{16}{9}·27 \ \ \Rightarrow \ \ a + b - c \ \ = \ \ 27 + 36 - 48 \ \ = \ \ 15 \ \ . \ ]$$