PRMO 2017 question 5

5. Let u,v,wu, v, w be real numbers in geometric progression such that u>v>wu > v > w. Suppose
u40=vn=w60u^{40} = v^n = w^{60}. Find the value of nn.

If u,v,wu,v,w form a geometric progression with common ratio r,r, then we can write

v=ur,w=ur2.v = u r, \quad w = u r^2.

The conditions u40=vn=w60u^{40} = v^n = w^{60} become

u40=(ur)nandu40=(ur2)60.u^{40} = (ur)^n \quad\text{and}\quad u^{40} = (ur^2)^{60}.

From the second equation,

(ur2)60  =  u60 r120  =  u40,(ur^2)^{60} \;=\; u^{60}\,r^{120} \;=\; u^{40},

which simplifies (dividing both sides by u40u^{40}) to

u20 r120  =  1    ⟹    u20=r−120    ⟹    u=r−6.u^{20}\,r^{120} \;=\; 1 \;\;\Longrightarrow\;\; u^{20} = r^{-120} \;\;\Longrightarrow\;\; u = r^{-6}.

Plug this into the first equation,

(ur)n  =  (r−6⋅r)n  =  r−6n+n  =  r−5n(ur)^n \;=\; (r^{-6}\cdot r)^n \;=\; r^{-6n + n} \;=\; r^{-5n}

must equal u40=(r−6)40=r−240.u^{40} = (r^{-6})^{40} = r^{-240}. Hence

r−5n  =  r−240    ⟹    −5n=−240    ⟹    n=48.

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