Question 1104925
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)
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Then we can say,
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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)
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we are asked to prove
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(a^3 + b^3 + c^3)/(b^3 + c^3 + d^3) = a/d
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let's work with the left sie of the = sign
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(p^3 + p^3r^3 + p^3r^6) / (p^3r^3 + p^3r^6 + p^3r^9) =
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p^3(1 + r^3 + r^6) / p^3(r^3 + r^6 + r^9) =
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(1 + r^3 + r^6) / r^3(1 + r^3 + r^6) =
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1 / r^3
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now we look at the right side of the =
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a / d = p / pr^3 = 1 / r^3
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The proof is complete
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