Algebra (Olympiad)hardFree

Ratio Recurrence Sequence: Find (a1 a2 a10)/(32 a7) | IOQM

JEE Maths question with a full step-by-step solution.

Question

Let

\{a_n\}

be a sequence of positive integers such that

a_1=a_2=2

. For

n\ge1

a_{n+2}a_n-a_{n+1}^2=a_{n+1}a_n.

Determine the value of

\dfrac{a_1a_2a_{10}}{32\,a_7}

Solution

Answer: 63

Step 1: Divide the recurrence by

a_{n+1}a_n

(each term is positive, so this is allowed):

\frac{a_{n+2}a_n}{a_{n+1}a_n}-\frac{a_{n+1}^2}{a_{n+1}a_n}=\frac{a_{n+1}a_n}{a_{n+1}a_n}\ \Rightarrow\ \frac{a_{n+2}}{a_{n+1}}-\frac{a_{n+1}}{a_n}=1.

Step 2: Let

b_n=\dfrac{a_{n+1}}{a_n}

. The relation above says

b_{n+1}-b_n=1

, so

\{b_n\}

is an arithmetic progression with common difference

1

. Its first term is

b_1=\frac{a_2}{a_1}=\frac{2}{2}=1.

Step 3: Therefore

b_n=1+(n-1)\cdot1=n

, that is

\frac{a_{n+1}}{a_n}=n\quad\text{for every }n.

In particular,

\dfrac{a_8}{a_7}=7

\dfrac{a_9}{a_8}=8

and

\dfrac{a_{10}}{a_9}=9

. Step 4: Write the required expression as a telescoping product of consecutive ratios:

\frac{a_1a_2a_{10}}{32\,a_7}=\left(\frac{a_{10}}{a_9}\right)\left(\frac{a_9}{a_8}\right)\left(\frac{a_8}{a_7}\right)\cdot\frac{a_1a_2}{32}=9\times8\times7\times\frac{2\times2}{32}.

Step 5: Evaluate:

9\times8\times7\times\frac{4}{32}=504\times\frac{1}{8}=63.

Answer: 63

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