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 {an}\{a_n\} be a sequence of positive integers such that a1=a2=2a_1=a_2=2. For n≥1n\ge1,
an+2an−an+12=an+1an.a_{n+2}a_n-a_{n+1}^2=a_{n+1}a_n.
 Determine the value of a1a2a1032 a7\dfrac{a_1a_2a_{10}}{32\,a_7}.
Solution
Answer: 63
Step 1: Divide the recurrence by an+1ana_{n+1}a_n (each term is positive, so this is allowed):
an+2anan+1an−an+12an+1an=an+1anan+1an ⇒ an+2an+1−an+1an=1.\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 bn=an+1anb_n=\dfrac{a_{n+1}}{a_n}. The relation above says bn+1−bn=1b_{n+1}-b_n=1, so {bn}\{b_n\} is an arithmetic progression with common difference 11. Its first term is
b1=a2a1=22=1.b_1=\frac{a_2}{a_1}=\frac{2}{2}=1.
 Step 3: Therefore bn=1+(n−1)⋅1=nb_n=1+(n-1)\cdot1=n, that is
an+1an=nfor every n.\frac{a_{n+1}}{a_n}=n\quad\text{for every }n.
 In particular, a8a7=7\dfrac{a_8}{a_7}=7, a9a8=8\dfrac{a_9}{a_8}=8 and a10a9=9\dfrac{a_{10}}{a_9}=9. Step 4: Write the required expression as a telescoping product of consecutive ratios:
a1a2a1032 a7=(a10a9)(a9a8)(a8a7)⋅a1a232=9×8×7×2×232.\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:
9×8×7×432=504×18=63.9\times8\times7\times\frac{4}{32}=504\times\frac{1}{8}=63.
 Answer: 63
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