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Chapter 8 · Predicting What Comes Next: Exploring Sequences and Progressions

Common ratio, and why the nth term is ar^(n−1)

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Geometric progressions10 min

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Also recorded in Hindi.Englishहिन्दी

3, 6, 12, 24 — and the test you have used all chapter fails on it. Stop subtracting consecutive terms and start dividing them.

The idea

A geometric progression is built by the same step-counting argument as an arithmetic one, with multiplication where the addition used to be: reaching position n takes n − 1 multiplications by r, so the exponent is n − 1 and the first term arrives with r raised to the power zero, untouched. The deeper change is in the test. An AP is diagnosed by subtracting consecutive terms, a GP by dividing them, and once you are looking at ratios instead of gaps the family turns out to include things that do not look like growth at all — sequences that shrink toward nothing, and sequences that flip sign at every step. Constant ratio, not increase, is the whole definition.

What you should be able to do

  • Test a sequence for a constant ratio and decide whether it is a geometric progression
  • Identify the first term and the common ratio of a given GP
  • Derive tₙ = arⁿ⁻¹ by counting the multiplications between the first term and the nth
  • Explain why the exponent is n − 1 rather than n
  • Write both the explicit and the recursive rule for a given GP
  • Handle common ratios that are whole, fractional, negative and irrational
  • Find which term of a GP equals a stated value, by matching powers
  • Contrast a GP with an AP on the same data and say which test settles it

Words to know

TermDefinition in one lineFirst introduced
geometric progressiona sequence in which consecutive terms have a fixed ratioprinted in bold in §8.6, p. 186
GPthe chapter's own abbreviation for a geometric progressionprinted in bold in §8.6, p. 186
common ratiothe fixed multiplier applied at each step, written rprinted in bold in §8.6, p. 187
constant multiplierthe chapter's plainer name for the same quantityprinted in bold in §8.6, p. 187
first termthe term in position 1, written a in the general formprinted in bold in §8.6, p. 187
nth termthe term in position n, given here by arⁿ⁻¹printed in bold in §8.6, p. 187
ratio testdeciding whether a sequence is a GP by dividing consecutive termsan added term; the chapter carries out the division without naming the test

Where people slip up

  • **"The nth term is arⁿ."** Test it on the first term: that gives ar, one multiplication too many. Standing on the first term you have multiplied by nothing.
  • "A geometric progression grows." Example 9's ratio is 3/4 and its terms fall; Example 8's ratio is −1 and its terms alternate between two values forever. Both are GPs. Growth is a consequence of r being bigger than 1, not part of the definition.
  • "Look at the differences." For 3, 6, 12, 24 the differences are 3, 6, 12 — they are not constant, and a student who only knows the AP test concludes there is no pattern. The chapter's Think and Reflect on p. 186 is engineered to trigger exactly that moment; use it.
  • "The ratio only has to work once." Every consecutive pair must give the same value. The page divides five pairs for a reason; 1, 2, 4, 7 passes the first test and fails the second.
  • "A negative ratio means the terms get smaller." With r = −2 the terms grow in size while flipping sign. Size and sign are separate consequences of r.
  • "An irrational ratio is not allowed." Exercise Set 8.3 item 6 uses √2. Nothing in the definition restricts r to fractions.
  • "An AP with a big common difference beats a GP." Only for a while. Put 1, 5, 9, 13 beside 3, 6, 12, 24 and let the second overtake — the exponent wins eventually, whatever the difference is.
  • **"r can be anything at all."** A ratio of 0 would collapse every later term to zero and there would be nothing to divide by; the chapter's examples never use it.
Transcript1,424 words

Here is a tile pattern, four stages of it. Every stage is three squares wide. What changes is the number of rows. One row, then two, then four, then eight. The rows double each time. So the counts are three, six, twelve, twenty-four. Carry it on and you get forty-eight, then ninety-six. Now the usual question. What is the rule? The last few patterns were built by adding the same amount over and over. This one is not.

Start the way you would with any pattern. Look at what changes from one term to the next. Three to six is a jump of three. Six to twelve is a jump of six. Twelve to twenty-four is a jump of twelve. Every one of those is an addition. That much is true. But the amount added is different every time. Three, then six, then twelve, then twenty-four. The differences are not constant, so there is no fixed amount to add and no rule.

A student who only knows the difference test stops here and says there is no pattern. There plainly is one. So the test is wrong for this job, not the pattern. Try the other operation. Instead of subtracting each term from the next, divide. Six divided by three is two. Twelve divided by six is two. Twenty-four divided by twelve is two. Forty-eight divided by twenty-four is two. Ninety-six divided by forty-eight is two.

Five divisions, and every one of them gives the same answer. That is the pattern the subtraction could not see. Not a fixed amount added, a fixed number multiplied by. A sequence like this, where consecutive terms have a fixed ratio, is called a geometric progression. The fixed multiplier has a name too. It is the common ratio, and here it is two. One warning before going further. The division has to work every time, not once.

Take one, two, four, seven. Two divided by one is two. Four divided by two is two. So far it looks like a doubling. But seven divided by four is one and three quarters, and the run is broken. Two agreeing pairs and one that does not is not a geometric progression. And checking the two ends is not a shortcut either. One, two, ten, twenty. The first division is two and the last division is two, and the one in the middle is five.

Every consecutive pair, or you have not run the test. Back to the tiles. Leave the three alone and show what was done to it. Three. Then three times two. Then three times four. Then three times eight. Now rewrite the multipliers as powers of two. Two is two to the one. Four is two squared. Eight is two cubed. So the terms are three, three times two to the one, three times two squared, three times two cubed.

Look at the exponents on their own. Nothing, one, two, three. The first term has no power of two attached, and every step after it adds one to the exponent. Which means the nth term is three times two to the power n minus one. Where does that n minus one come from? Count the multiplications. To get from the first term to the fourth you multiply by two, then again, then again. Three multiplications.

To reach the tenth term is nine multiplications, landing on one thousand five hundred and thirty-six. And standing on the first term, you have multiplied by nothing at all. Zero multiplications. So the number of multiplications behind you is always one less than the position you are standing in. That is the whole reason for the minus one. Test the rival version. If the nth term were a times r to the n, the first term would be a times r, one multiplication too many.

For the tiles it would give six where the answer is three. Wrong at the very first term, and wrong at every term after it. So the general shape is this. Call the first term a and the ratio r. Then the terms run a, a r, a r squared, a r cubed, and so on. The nth term is a times r to the power n minus one. There is a second way to write the same progression, and it is often easier to use.

Say what the first term is, and say what each term does to the one before it. First term three; after that, each term is two times the one before. That gives you the sequence one step at a time. The other gives you any term you like directly. One more fencepost. Bacteria doubling from thirty stand at a hundred and twenty at the end of hour two, and four hundred and eighty at hour four.

That is thirty times two to the power h, not h minus one, because the original thirty is sitting at hour nothing. Three sequences. Say whether each one is a geometric progression, and what its ratio is. One, two, four, eight, sixteen. Yes. Every division gives two. One, three, nine, twenty-seven, eighty-one. Yes. Every division gives three. One, minus one, one, minus one, one. This one is the interesting case, and the answer is still yes. Every division gives minus one.

It never grows and never shrinks. It only ever holds two values, and it still passes the test. That last one is a hint that growing is not part of the definition. Here is the proof. Five, fifteen quarters, forty-five sixteenths, one hundred and thirty-five sixty-fourths. Those terms are getting smaller. Run the test anyway. Fifteen quarters divided by five is three quarters. Forty-five sixteenths divided by fifteen quarters is three quarters.

One hundred and thirty-five sixty-fourths divided by forty-five sixteenths is three quarters. Three divisions, one answer. First term five, common ratio three quarters, and the nth term is five times three quarters to the power n minus one. The terms fall forever and never reach nothing. A ratio below one makes a progression that shrinks. Now let the ratio be negative and watch two things at once. Take one with a ratio of minus two. The terms are one, minus two, four, minus eight.

Ignore the signs and look at the sizes. One, two, four, eight. They are getting bigger. Now look at the signs on their own. Plus, minus, plus, minus. So the terms grow while flipping. A negative ratio does not mean the terms get smaller. Size and sign are two separate consequences of the ratio. Here is one that does both. Three, minus three halves, three quarters, minus three eighths, with a ratio of minus one half.

Shrinking and flipping together, and there is nothing wrong with it. Two things the formula buys you. First, a distant term without going back to the start. Suppose you know the eighth term of a doubling progression is one hundred and ninety-two, and you want the twelfth. Twelve is four positions past eight, so four multiplications by two. One hundred and ninety-two times sixteen is three thousand and seventy-two, and you never needed the first term.

Second, finding which term a given value is. In two, six, eighteen, where does four thousand three hundred and seventy-four sit? Divide it by the first term and you get two thousand one hundred and eighty-seven, which is three to the seventh. Seven multiplications, so position eight. And the ratio does not have to be a fraction. Take two, two root two, four, stepping by root two each time. A hundred and twenty-eight is two to the seventh, and twelve steps of half an exponent add up to six, so a hundred and twenty-eight sits at position thirteen.

One last comparison, and it is the reason all of this matters. Put a progression that adds beside one that multiplies. A taxi fare starting at two hundred and forty and going up by forty each time. Against the tiles, three, six, twelve, twenty-four. The fare is far ahead at the start. At position eight the fare is five hundred and twenty and the tiles are three hundred and eighty-four. Still winning.

At position nine the fare is five hundred and sixty and the tiles are seven hundred and sixty-eight. The lead has changed hands. By position ten it is six hundred against one thousand five hundred and thirty-six, and the fare never gets back in front. That is what the exponent does. However big the amount added, the multiplying one goes past it and stays past it.

Where this fits

Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.

Builds on

Comes up again in

Either side of this one

The book

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