SOLUTION: if a, b, c and d are in continuous proportion then prove that: (a^3 + b^3 + c^3)/(b^3 + c^3 + d^3) = a/d

Question 1104925: if a, b, c and d are in continuous proportion then prove that:
(a^3 + b^3 + c^3)/(b^3 + c^3 + d^3) = a/d
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
If a,b, c,d are in continuous proportion, a/b = b/c = c/d, i.e a,b,c and d are in a geometric progression(GP)
:
Then we can say,
:
a = p, b=pr, c=pr^2 and d=pr^3 (p is the first term and r is the common ratio of a GP series)
:
we are asked to prove
:
(a^3 + b^3 + c^3)/(b^3 + c^3 + d^3) = a/d
:
let's work with the left sie of the = sign
:
(p^3 + p^3r^3 + p^3r^6) / (p^3r^3 + p^3r^6 + p^3r^9) =
:
p^3(1 + r^3 + r^6) / p^3(r^3 + r^6 + r^9) =
:
(1 + r^3 + r^6) / r^3(1 + r^3 + r^6) =
:
1 / r^3
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now we look at the right side of the =
:
a / d = p / pr^3 = 1 / r^3
:
The proof is complete
:
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