Sequences with a Surd Common Ratio

Every sheet of paper in your printer belongs to a sequence. A5 is one hundred and forty-eight millimetres across, A4 is two hundred and ten, A3 is two hundred and ninety-seven, and A2 is four hundred and twenty. Those sizes are designed so that each step up multiplies both sides by the same fixed number, then the millimetres are rounded off. That multiplier is not one point four, and it is not one point four one four either. It is the square root of two, and no decimal will ever write it down exactly. Sequences built on a multiplier like that turn up on Higher papers with the root left in, and the common slip is to meet one and immediately flatten it into a decimal. Every term after that point is an approximation, and the exact chain the question wanted is gone. One note before the maths starts. This material is Higher tier only, so if your papers are Foundation, the video called Finding the nth Term of a Linear Sequence is a better use of the next few minutes.

Start with a sequence that behaves itself, because it is the cleanest place to watch the method work. It begins two, then two root three, then six, then six root three. Geometric means one fixed multiplier sits between every pair of neighbouring terms, and that multiplier is called the common ratio. You find it by dividing a term by the one immediately before it. When that ratio is a plain whole number, the video called Recognising Special Sequences handles it. Here it is a root, which changes the arithmetic, not the idea. So look at just the first two terms, two and two root three. Which multiplier takes you from two to two root three? A, root three. B, three. C, two root three. Pick one. I'll wait. The answer is A, root three, because two times root three is two root three. Written as a division, that is two root three divided by two, and the two at the front cancels, leaving root three on its own. Now confirm it on the next pair by multiplying forwards rather than dividing again. Two root three times root three gives two times three, which is six, and six is exactly the term sitting there. Forwards is the friendlier check, because dividing six by two root three means rationalising a denominator, and the video called Rationalising the Denominator owns that job. Both neighbouring pairs agree, so the common ratio is root three, and it stays written as a root.

Finding the ratio is half the job. Using it is the other half, and this is where the roots start doing something worth watching. The sequence had reached six root three, so multiply that by the ratio, root three. Root three times root three is three, so six root three times root three is six times three, which is eighteen. The root has vanished completely. That is the mechanism underneath this whole topic. A square root multiplied by itself gives back the number underneath it, so root two times root two is two. Which is exactly why these sequences alternate. Whole number, surd, whole number, surd, all the way along, because each multiplication either brings a root in or cancels one out. Carry on from eighteen and you multiply by root three once more, giving eighteen root three. So the sequence continues eighteen, then eighteen root three. Your turn now. Take that sequence one step further and find the term that comes after eighteen root three. Take your time. I'll wait right here. Eighteen root three times root three is eighteen times three, which is fifty-four. Be precise about the form the exam wants. In surd form means you write root three, not one point seven three two, and a term that comes out whole, like eighteen or fifty-four, is written as a plain number with no root attached. Tidy any surd you produce, as well. If a multiplication leaves you with something like root eight, the expected form is two root two, and the video called Simplifying Surds carries that rule. If a question genuinely asks for a decimal, round once at the very end, never in the middle of the chain, because a rounded term multiplied on four more times drifts a long way off. Walking a sequence forward one term at a time is all this video does. A formula that jumps straight to any position belongs to the video called Finding the nth Term of an Exponential Sequence.

The second example dangles an escape route in front of you, so take it deliberately and see where it goes. It begins six, then eighteen root two, then one hundred and eight, then three hundred and twenty-four root two. You are asked for the common ratio in surd form, and for the term after that last one. Here is the tempting move. Ignore the surd terms and compare the whole numbers instead: one hundred and eight divided by six is eighteen, with no root anywhere near it. Do the same with the two surd terms and eighteen turns up again. Multiply one hundred and eight by eighteen and you land on one thousand nine hundred and forty-four, which really is the next term. It worked, and it is still the wrong answer to the question that was asked. Eighteen is not the common ratio. Eighteen is the common ratio used twice. Ask for the ratio itself and you have to square root eighteen to get back to it, landing on three root two. That is the surd you were avoiding, arriving late. The skip needs the term two places back, as well. Hand it a single pair of neighbours and it has nothing at all to divide. So do it the reliable way. Using neighbours, find the common ratio of six, eighteen root two, one hundred and eight. Have a go at this one. I'll wait. Eighteen root two divided by six is three root two. Check it forwards: eighteen root two times three root two is fifty-four times two, which is one hundred and eight. The neighbours agree, so three root two is the ratio. Now the term they asked for. Three hundred and twenty-four root two times three root two: multiply the whole numbers, three hundred and twenty-four times three is nine hundred and seventy-two, then root two times root two is two, and nine hundred and seventy-two times two is one thousand nine hundred and forty-four. The same one thousand nine hundred and forty-four as the shortcut gave, reached by the method the question was actually testing. Here is the line to carry into the exam hall. Next door, not next but one, and the root stays in.

A Higher paper sat in November two thousand and twenty-two caught every version of this happening at once. Here are two lines from the examiner report for that paper, about a question near the end of it where a sequence had common ratio two root five. The words are theirs. For a question towards the end of the paper part (a) was answered reasonably well. Some students started by using consecutive terms to find the common ratio. Having identified the common ratio as two root five many worked out the next term as four thousand but four hundred root five times two root five equals eight thousand was a common error. Instead of finding the common ratio some students used alternating terms of the sequence and the calculation two hundred times, two hundred divided by ten, gave the next term without them having to deal with surds. Take the middle of that again. Those students started correctly, from consecutive terms, and then the multiplication itself went wrong. Four hundred root five times two root five is four hundred times two, which is eight hundred, and root five times root five is five, so eight hundred times five is four thousand. The error reported there, eight thousand, is double that. The surd multiplication is not decoration around the question. It is the thing being examined, sitting right in the middle of it. And that second line is the shortcut you have just watched. Two hundred divided by ten is twenty, and twenty is what you get when you skip a term. Twenty is not the common ratio there. Twenty is two root five squared, because two squared is four and root five squared is five. Skip a term and you get the ratio squared, every single time. So take the ratio from neighbours, write it in surd form, and let the working show the roots multiplying. A decimal on the page hides the one step the question set out to look at.

Say that line once more: next door, not next but one, and the root stays in. Here is the run-through behind it. The common ratio comes from dividing a term by its neighbour, and the answer stays a root: root three, or three root two, never one point seven three two. Confirm it by multiplying forwards onto the next term, because two agreeing pairs are what make the sequence geometric at all. Then extend by multiplying by that surd itself, remembering that a root times itself gives back the number underneath, which is why the terms alternate between whole numbers and surds. And if you skip a term to dodge the roots, the number you get is the ratio squared, not the ratio, so the surd is waiting for you either way. You have got this when four terms with a surd ratio hand you the ratio in surd form and the next term, with no calculator anywhere in sight.

Next in the chapter: Finding the nth Term of a Linear Sequence, where the rule stops being a multiplier and becomes a formula for any position you like.

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