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a, b, x, y, . . . Each term of the sequence above is 9 more than
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04 Dec 2023, 21:15
There's 2 ways to approach this question: 1) Try plugging in 54 as a, and work down to check whether y = 5/6x or 2) set up the equations and solve for a. I think Method 1 would be faster, but I'll explain both:
Method 1: Set a = 54
The sequence described can be written as \( Term = \frac{1}{3}(Previous Term) + 9\). So if a = 54, then:
\(b = \frac{1}{3}(54) + 9 = 27\)
\(x = \frac{1}{3}(27) + 9 = 18\)
\(y = \frac{1}{3}(18) + 9 = 15\)
Now, we have to check whether y = 5/6x.
\(\frac{5}{6}x = \frac{5}{6}(18) = 15 = y\)
So, using this method, we know that a must be 54.
Method 2: setting up equations to solve for a
\(b = \frac{1}{3}a + 9\)
\(x = \frac{1}{3}(b) + 9 =\frac{ 1}{3}(\frac{1}{3}a+9) + 9 = \frac{1}{9}a + 12\)
\(y = \frac{1}{3}(x) + 9 = \frac{1}{3(}\frac{1}{9}a+12) + 9 = \frac{1}{27}a + 13\)
We now have x and y in terms of a. Now we plug those into y = 5/6x to solve for a:
\(\frac{1}{27}a + 13 = \frac{5}{6}(\frac{1}{9}a+12)\)
\(5a + 540 = 2a + 702\)
\(3a = 162\)
\(a = 54\)