Prove that sequence contains an infinite number of three-term geometric sequences with a common ratio.

An arithmetic sequence S has terms $$t_{1}$$ , $$t_{2}$$, $$t_{3}$$ , . . ., where $$t_1 = a$$ and the common difference is $$d$$. The terms $$t_{5}$$,$$t_{9}$$, and $$t_{16}$$ form a three-term geometric sequence with common ratio $$r$$. Prove that S contains an infinite number of three-term geometric sequences, all having the same common ratio $$r$$.

This is a question from Waterloo Euclid EWokshop. EWorkShop

I have some problems understanding the solution:

• Why does the solution add the extra condition that a and b must
• both be congruent to 1 modulo 3 or
• both be congruent to 2 modulo 3
• The solution forgets to mention that in case $d=0$, the claim (with $r=1$) trivially holds for arbitrary triples. -- Also, strictly we cannot choose $a$ when $a$ is already defined as starting term Feb 8, 2019 at 3:11
• Another gap in the solution: The triples produced do not necessarily have ratio $r$ Feb 8, 2019 at 3:19
• So, is this question wrong in a way, or it's just the solution provided is not good enough? Feb 8, 2019 at 3:20
• I believe using $a$ to mean $2$ things in the same answer is a bit confusing also. However, note that the ratio is $r = \frac{7}{4}$, so it seems it will work, as long as the values of $a$ (as in the second part) and $b$ are chosen appropriately to also ensure this value of $r$ is met, with it being $r = \frac{7}{4} = \frac{a}{b}$ as the answer now shows. Feb 8, 2019 at 3:31

As in the given solution, we either find $$d=0,$$ in which case $$r=1$$ and any triple from the (constant) sequence is geometric with ratio $$r$$.
Or we find $$t_k =\frac a4(3k+1).$$ Note that this makes $$r=\frac{t_9}{t_5}=\frac{\frac a4(3\cdot 9+1)}{\frac a4(3\cdot 5+1)}=\frac 74.$$ But how can we find many suitable index triple $$i,j,k$$ such that $$\frac{t_k}{t_j}=\frac{t_j}{t_i}=\frac 74$$? The condition boils down to $$\frac{3k+1}{3j+1}=\frac{3j+1}{3i+1}=\frac 74$$, or that for suitable $$n$$, we have $$3k+1=49n,\quad 3j+1=28n, \quad 3i+1=16n.$$ As $$16,28,49\equiv 1\pmod 3$$ and so are the left hand sides, making $$n\equiv 1\pmod 3$$ suffices. In other words, we let $$n=3m+1$$ for arbitrary natural $$m$$ and then \begin{align} k&=\frac{49(3m+1)-1}3=49m+16\\ j&=\frac{28(3m+1)-1}{3}=28m+9\\ i&=\frac{16(3m+1)-1}{3}=16m+5.\end{align} (The original triple corresponds to $$m=0$$, obviously).