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:

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  • 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
  • $\begingroup$ 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 $\endgroup$ Feb 8, 2019 at 3:11
  • 1
    $\begingroup$ Another gap in the solution: The triples produced do not necessarily have ratio $r$ $\endgroup$ Feb 8, 2019 at 3:19
  • $\begingroup$ So, is this question wrong in a way, or it's just the solution provided is not good enough? $\endgroup$
    – Kevin
    Feb 8, 2019 at 3:20
  • $\begingroup$ 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. $\endgroup$ Feb 8, 2019 at 3:31

1 Answer 1


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).


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