[wu::riddles] Geometric In Arithmetic

[wu::riddles] Geometric In Arithmetic
7
Jan

Today we share our analysis of the following problem, which kindly comes to us from Colombian engineer Ariel Nunez, a former Iberoamericano Math Olympiad Bronze medalist!

Problem:

Given a real arithmetic progression 1, a_2, a_3, \ldots, some terms are deleted to produce a geometric sequence 1, b_2, b_3, \ldots with ratio q. Find all possible values of q.

Our Solution:

Define a_n := 1 + (n-1) \cdot d for n \in \mathbb{N}_+ and d \in \mathbb{R}. The geometric subsequence is b_k := a_{n_k} = q^{k-1} for k \in \mathbb{N}_+, where n_k is the indexing sequence. Evaluating q = \frac{b_{k+1}}{b_k},

    \begin{align*} q &= \frac{1+ ({n_{k+1}} - 1) d}{ 1 + (n_k - 1) d} \\ q ( 1 + (n_k - 1) d ) & = 1+ ({n_{k+1}} - 1)d \\ q ( 1/d + n_k - 1 ) & = 1/d + {n_{k+1}} - 1 \\ q \left( n_k - \frac{d-1}{d} \right) & = \left( {n_{k+1}} - \frac{d-1}{d} \right) \end{align*}

Let c_k := n_k - \frac{d-1}{d}. From the above recurrence, c_k = \alpha q^k for some constant \alpha. From the initial condition n_1 = 1,
c_1 = \alpha q = 1 - \frac{d-1}{d}, which implies \alpha = \frac{1}{d q}. Equating these expressions for c_k,

    \begin{align*} n_k - \frac{d-1}{d} &= \left(\frac{1}{dq} \right) q^k \\ n_k &= \frac{1}{d} q^{k-1} + \frac{d-1}{d} \\ &= 1 + \frac{ q^{k-1} - 1 }{d} \\ n_k - 1 &= \frac{ (q-1)(q^{k-2} + q^{k-1} + ... + q + 1 ) }{d} \qquad \forall k \in \mathbb{N}_+. \end{align*}

Since n_k \in \mathbb{N}_+, each k imposes a divisibility relationship between d and q.
From the case of k=2, d | (q-1), so q-1 = m d for some m \in \mathbb{Z}. Consequently,

    \[ P(k) := m (q^{k-2} + q^{k-1} + ... + q + 1 ) \in \mathbb{N}_0 \qquad \forall k \in \mathbb{N}_+, k \geq 2. \]

m and q must both be positive for P(k) to be non-negative for all k \geq 2. By induction since P(k-1) \in \Nb_0, it follows that

    \[ P(k) - P(k-1) = mq^{k-2} \in \mathbb{N}_0 \qquad \forall k \geq 2. \]

This constraint cannot be satisfied if q is irrational. Suppose q \in \mathbb{Q}_+ \setminus \mathbb{N}_+, such that q = r/s for non-commensurate r,s \in \mathbb{N}_+. From the case of k=3, mq \in \mathbb{N}_0 implies that m = p s for some p \in \mathbb{N} not divisible by s. However, from the case of k=4, mq^2 \in \mathbb{N}_0 can be rewritten as p r^2 / s \in \mathbb{N}, which implies that p is divisible by s — which is a contradiction. Therefore, q can only be a natural number. q=1 is achievable when d=0, and all other natural numbers for q are achievable when d=1.

Do you have other approaches for solving the problem? Let us know!

QED_arith_geo_subseq

 

For more puzzles:

Leave A Reply

Your email address will not be published. Required fields are marked *