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Find two consecutive even integers such that the sum of their reciprocals is 9/40. a. 8 and 9 b. 6 and 8 c. 8 and 10 d. 10 and 12
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Let your first even integer be \(n\). Then \(n+2\) is the second number. You want the sum of their reciprocals to equal \(\frac{9}{40}\) \[\frac{1}{n}+\frac{1}{n+2} = \frac{9}{40}\]Solve for \(n\) and you've got the first number. You'll get two solutions, but only one of them is an even integer...
As this is that bane of good instruction, the multiple-choice question, you could also just try plugging each answer choice into the formula and see which one works. Better if you aren't under some incredible time pressure to actually go through the steps of solving it, in my opinion.
Thank you so much!
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