If a:b=b:c=c:d=d:e=3:1,then what is the value of (c+e)/(b+d)?

Method I - brute force, which I think is easier. It's quick, too. Find the pattern.

\(\frac{a}{b}\) = \(\frac{3}{1}\) --> \(\frac{b}{c}\) = \(\frac{3}{1}\)-->\(\frac{c}{d}\) = \(\frac{3}{1}\) --> \(\frac{d}{e}\) = \(\frac{3}{1}\)
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\(\frac{a}{b}\) = \(\frac{3}{1}\)
3b = a ---> b = \(\frac{a}{3}\)

\(\frac{b}{c}\) = \(\frac{3}{1}\)
3c = b ---> c = \(\frac{b}{3}\) ...
d = \(\frac{c}{3}\)
e = \(\frac{d}{3}\)

Each term gets divided by 3 to yield the next term. So the greatest term is a and the smallest term is e. Because there is division by 3 a few times, pick a semi-large power of 3 for a. Or be precise:

We have \(\frac{1}{3}\)*\(\frac{1}{3}\)*\(\frac{1}{3}\)*\(\frac{1}{3}\)= \(\frac{1}{81}\)

Let a = 81. Then b = 27, c = 9, d = 3, and e = 1

(c + e)/(b + d) =

(9 + 1)/(27 + 3) =

10/30 = 1/3

Method II - sequence

Find the ratio:
\(\frac{a}{b}\) = \(\frac{3}{1}\)
a = 3b

\(\frac{b}{c}\) = \(\frac{3}{1}\)
b = 3c . . .

This is a geometric sequence, with common ratio of 3 (if \(A_1\) is e). Let e = \(A_1\), and call it x

\(A_{n}\) = \(A_1\) * \(r^{n-1}\), where n is the nth term and r is the common ratio.

\(A_{n}\) = \(A_1 * 3^{n-1}\)

\(A_1\) = e = x
\(A_2\) = d = x * 3 = 3x
\(A_3\) = c = x * \(3^2\) = 9x
\(A_4\) = b = x * \(3^3\) = 27x
\(A_5\) = a = x * \(3^4\) = 81x

Value of (c+e)/(b+d)?

(9x + x)/(27x + 3x) =

\(\frac{10x}{30x}\) = \(\frac{1}{3}\)

ANSWER C