Question 1

Find the first five terms of the sequence in which the n^ th term is given by (i) t n=3 n-4, (ii) t n=2-5 n, and (iii) t n=n^2-2 n+3 for n ≥ 1.

Accepted Answer

Substituting into the rule. A sequence given by a rule for t n is like a machine: feed in a position number n, and out comes the term that sits in that position. "The first five terms" therefore means put n = 1, 2, 3, 4, 5 into the rule, one at a time. (i) t n = 3n - 4 t 1 = 3(1) - 4 = -1, t 2 = 3(2) - 4 = 2, t 3 = 3(3) - 4 = 5, t 4 = 3(4) - 4 = 8, t 5 = 3(5) - 4 = 11. (ii) t n = 2 - 5n t 1 = 2 - 5(1) = -3, t 2 = 2 - 5(2) = -8, t 3 = 2 - 5(3) = -13, t 4 = 2 - 5(4) = -18, t 5 = 2 - 5(5) = -23. (iii) t n = n^2 - 2n + 3 t 1 = 1 - 2 + 3 = 2, t 2 = 4 - 4 + 3 = 3, t 3 = 9 - 6 + 3 = 6, t 4 = 16 - 8 + 3 = 11, t 5 = 25 - 10 + 3 = 18. Putting the three answers side by side: Worth noticing. In (i) each term is 3 more than the one before, and in (ii) each is 5 less than the one before — both are arithmetic progressions, and the common difference is exactly the number multiplying n in the rule. In (iii) the gaps are 1, 3, 5, 7, which keep changing, so that sequence is not an AP. Answer. (i) -1, 2, 5, 8, 11 (ii) -3, -8, -13, -18, -23 (iii) 2, 3, 6, 11, 18

Question 2

Find the 10^ th and 15^ th terms of the sequence t n=5 n-3 for n ≥ 1.

Accepted Answer

Substituting into the rule. The rule t n = 5n - 3 works for any position n, so there is no need to write out the terms one by one until we reach the tenth. We simply put n = 10 and n = 15 directly into the rule. For the 10^ th term, take n = 10: t 10 = 5(10) - 3 = 50 - 3 = 47. For the 15^ th term, take n = 15: t 15 = 5(15) - 3 = 75 - 3 = 72. Check. The sequence starts 2, 7, 12, 17, … — it goes up by 5 each step. From t 1 = 2, reaching t 10 takes 9 steps of 5, so t 10 = 2 + 9 × 5 = 47. Going on from t 10 to t 15 takes 5 more steps: 47 + 5 × 5 = 72. Both values agree. Answer. t 10 = 47 and t 15 = 72.

Question 3

Determine whether 97 and 172 are terms of the sequence t n=5 n-3 for n ≥ 1.

Accepted Answer

Working backwards from the rule. A number is a term of this sequence only if some positive whole number n makes 5n - 3 equal to it. So for each number we set up the equation and see whether the n that comes out is a counting number. A fractional n would mean the number falls between two terms and so is not in the sequence at all. Is 97 a term? Put t n = 97: 5n - 3 = 97 ⇒ 5n = 100 ⇒ n = 20. Here n = 20 is a positive whole number, so yes. Checking: 5(20) - 3 = 100 - 3 = 97. ✓ Is 172 a term? Put t n = 172: 5n - 3 = 172 ⇒ 5n = 175 ⇒ n = 35. Again n = 35 is a positive whole number, so yes. Checking: 5(35) - 3 = 175 - 3 = 172. ✓ Another way to see it. Every term 5n - 3 leaves remainder 2 on division by 5 (since 5n - 3 = 5(n-1) + 2 ). Both 97 = 5 × 19 + 2 and 172 = 5 × 34 + 2 leave remainder 2, and both are bigger than the first term 2, so both must appear. A number such as 100 or 96, which leaves a different remainder, could never be a term. Answer. Both are terms: 97 is the 20^ th term and 172 is the 35^ th term.

Question 4

Which term of the sequence t n=5 n-3 for n ≥ 1 is 607?

Accepted Answer

Working backwards from the rule. We are told the value of a term and asked for its position. So put t n = 607 and solve for n: 5n - 3 = 607. Add 3 to both sides: 5n = 610. Divide both sides by 5: n = 122. Since 122 is a positive whole number, 607 really does occupy a position in this sequence. Check. t 122 = 5(122) - 3 = 610 - 3 = 607. ✓ Answer. 607 is the 122^ nd term.

Question 5

A sequence is given by the recursive rule t 1=-5, t n+1=t n+3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

Accepted Answer

Recursive rule, then a closed form. A recursive rule tells you where to start and how to take one step. Here the start is t 1 = -5, and the step t n+1 = t n + 3 says: to get the next term, add 3 to the one you have. The first five terms. Adding 3 each time: t 1 = -5, t 2 = -5 + 3 = -2, t 3 = -2 + 3 = 1, t 4 = 1 + 3 = 4, t 5 = 4 + 3 = 7. So the sequence begins -5, -2, 1, 4, 7, … — an AP with first term -5 and common difference 3. Is 52 a term? Stepping up by 3 all the way to 52 would take a long time, so first turn the recursive rule into a direct one. Going from t 1 to t n takes (n-1) steps, and each step adds 3: t n = -5 + (n-1)× 3 = 3n - 8. Now set this equal to 52: 3n - 8 = 52 ⇒ 3n = 60 ⇒ n = 20. n = 20 is a positive whole number, so 52 does occur. Check. t 20 = 3(20) - 8 = 60 - 8 = 52. ✓ (Also, t 5 = 3(5) - 8 = 7, which matches the fifth term found above, so the closed form is the right one.) Answer. The first five terms are -5, -2, 1, 4, 7. Yes, 52 is a term — it is the 20^ th term.

Question 6

Let T 1=1, T 2=2, T 3=4, and T n = T n -1 + T n -2 + T n -3 for n ≥ 4. Find T 4, T 5, T 6, T 7, and T 8.

Accepted Answer

Building a recursive sequence step by step. This rule needs three earlier terms to make a new one: each term from the fourth onwards is the sum of the three terms immediately before it. That is why the book has to hand us three starting values, T 1 = 1, T 2 = 2, T 3 = 4. We just apply the rule T n = T n-1 + T n-2 + T n-3 one term at a time, always using the three most recent values. T 4 = T 3 + T 2 + T 1 = 4 + 2 + 1 = 7, T 5 = T 4 + T 3 + T 2 = 7 + 4 + 2 = 13, T 6 = T 5 + T 4 + T 3 = 13 + 7 + 4 = 24, T 7 = T 6 + T 5 + T 4 = 24 + 13 + 7 = 44, T 8 = T 7 + T 6 + T 5 = 44 + 24 + 13 = 81. Laid out in order: Notice that the sequence is neither an AP (the differences 1, 2, 3, 6, 11, 20, 37 are not constant) nor a GP (the ratios are not constant either). It is a close cousin of the Fibonacci sequence, which adds the previous two terms instead of three. Answer. T 4 = 7, T 5 = 13, T 6 = 24, T 7 = 44, T 8 = 81.

Question 7

Find the 10^ th and 26^ th terms of the AP: 3, 8, 13, 18,....

Accepted Answer

n^ th term of an AP. First check that this really is an AP by looking at consecutive differences: 8 - 3 = 5, 13 - 8 = 5, 18 - 13 = 5. The difference is the same every time, so it is an AP with first term a = 3 and common difference d = 5. To reach the n^ th term from the first, you take (n-1) steps of size d — not n steps, because the first term is already there before any step is taken. That gives the standard rule t n = a + (n-1)d = 3 + (n-1)× 5 = 5n - 2. The 10^ th term: t 10 = 3 + (10 - 1)× 5 = 3 + 45 = 48. The 26^ th term: t 26 = 3 + (26 - 1)× 5 = 3 + 125 = 128. Check with the short rule. 5(10) - 2 = 48 and 5(26) - 2 = 128. ✓ Also 5(1) - 2 = 3 and 5(4) - 2 = 18, matching the first and fourth listed terms. Answer. t 10 = 48 and t 26 = 128.

Question 8

Which term of the AP: 21, 18, 15,... is - 81? Also, is 0 a term of this AP? Give reasons for your answer.

Accepted Answer

n^ th term of an AP, then solving for the position. The differences are 18 - 21 = -3 and 15 - 18 = -3, so this is an AP with a = 21 and d = -3. A negative common difference simply means the terms are going down. Its n^ th term is t n = a + (n-1)d = 21 + (n-1)(-3) = 21 - 3n + 3 = 24 - 3n. Which term is -81? Set t n = -81: 24 - 3n = -81 ⇒ -3n = -105 ⇒ n = 35. Since 35 is a positive whole number, -81 is genuinely a term. Check: t 35 = 24 - 3(35) = 24 - 105 = -81. ✓ Is 0 a term? Set t n = 0: 24 - 3n = 0 ⇒ 3n = 24 ⇒ n = 8. Again n = 8 is a positive whole number, so yes, 0 is a term. The reason it works is that 24 happens to be exactly divisible by 3; had the equation given a fraction such as n = 8.5, the value 0 would have fallen between two terms and would not be in the AP. We can see it directly by writing the AP out: 21, 18, 15, 12, 9, 6, 3, 0, -3, … and 0 is indeed in the eighth place. ✓ Answer. -81 is the 35^ th term. Yes, 0 is a term of this AP — it is the 8^ th term, because 24 - 3n = 0 gives the whole number n = 8.

Question 9

Find the n^ th term of the AP: 11, 8, 5, 2... Write the recursive rule for this AP.

Accepted Answer

n^ th term of an AP, plus its recursive rule. Check the differences first: 8 - 11 = -3, 5 - 8 = -3, 2 - 5 = -3. Constant, so this is an AP with first term a = 11 and common difference d = -3. Explicit ( n^ th -term) rule. From the first term you take (n-1) steps of -3 to arrive at the n^ th term: t n = a + (n-1)d = 11 + (n-1)(-3) = 11 - 3n + 3 = 14 - 3n. Check it on the given terms: t 1 = 14 - 3 = 11, t 2 = 14 - 6 = 8, t 3 = 14 - 9 = 5, t 4 = 14 - 12 = 2. ✓ All four match. Recursive rule. A recursive rule says where to start and how to take one step. Here you start at 11 and each new term is 3 less than the previous one: t 1 = 11, t n+1 = t n - 3 for n ≥ 1. The two rules describe exactly the same sequence. The recursive one is handy for writing the next few terms; the explicit one is what you want when someone asks for the 100^ th term without listing the other 99. Answer. t n = 14 - 3n; recursive rule: t 1 = 11, t n+1 = t n - 3.

Question 10

An AP consists of 50 terms in which the 3^ rd term is 12 and the last term is 106. Find the 29^ th term. (Hint: If ' a ' is the first term and ' d ' the common difference, then we arrive at the equations a+2 d=12 and a+49 d=106. Solve this pair of linear equations for ' a ' and ' d '.)

Accepted Answer

Two unknowns, two equations. Every term of an AP is built from just two numbers, the first term a and the common difference d, through t n = a + (n-1)d. So two pieces of information are enough to pin the whole AP down. The 3^ rd term is 12: a + (3-1)d = 12 ⇒ a + 2d = 12. … (1) The AP has 50 terms, so the last term is the 50^ th, and it is 106: a + (50-1)d = 106 ⇒ a + 49d = 106. … (2) Solve the pair. Subtracting (1) from (2) removes a altogether: (a + 49d) - (a + 2d) = 106 - 12 ⇒ 47d = 94 ⇒ d = 2. Put d = 2 back into (1): a + 2(2) = 12 ⇒ a + 4 = 12 ⇒ a = 8. Now the 29^ th term: t 29 = a + (29-1)d = 8 + 28 × 2 = 8 + 56 = 64. Check. With a = 8, d = 2: the 3^ rd term is 8 + 4 = 12 ✓ and the 50^ th term is 8 + 49 × 2 = 8 + 98 = 106 ✓. Both given facts are reproduced. Answer. a = 8, d = 2, and the 29^ th term is 64.

Question 11

How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Accepted Answer

Counting the terms of an AP, then adding them. The 2 -digit numbers run from 10 to 99. The smallest of these divisible by 3 is 12, and the largest is 99 (since 9 + 9 = 18 is divisible by 3 ). Between them the multiples of 3 go up in steps of 3, so they form the AP 12, 15, 18, …, 99, a = 12, d = 3. How many are there? Let 99 be the n^ th term: a + (n-1)d = 99 ⇒ 12 + (n-1)3 = 99 ⇒ (n-1)3 = 87 ⇒ n - 1 = 29 ⇒ n = 30. So there are 30 such numbers. Their sum. Pair the first with the last, the second with the second-last, and so on: 12 + 99 = 111, 15 + 96 = 111, 18 + 93 = 111, … every pair adds to the same total. With 30 numbers there are 15 such pairs, so Sum = 15 × 111 = 1665. This pairing is exactly the standard formula S n = n/2(a + l), where l is the last term: S 30 = 30/2(12 + 99) = 15 × 111 = 1665. Check a different way. The multiples of 3 from 12 to 99 are 3(4 + 5 + + 33). Now 4 + 5 + + 33 = (30(4 + 33))/2 = 15 × 37 = 555, and 3 × 555 = 1665. ✓ The two routes agree. Answer. There are 30 two-digit numbers divisible by 3, and their sum is 1665.

Question 12

Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Accepted Answer

Salary as an AP. The salary rises by the same amount, ₹20,000, every year, so the yearly salaries form an AP with a = 5, 00, 000 (salary in the 1st year), d = 20, 000. The salary in the n^ th year is t n = a + (n-1)d = 5, 00, 000 + (n-1)× 20, 000. Set this equal to ₹7,00,000: 5, 00, 000 + (n-1)× 20, 000 = 7, 00, 000. Subtract 5, 00, 000 from both sides: (n-1)× 20, 000 = 2, 00, 000. Divide both sides by 20, 000: n - 1 = 10 ⇒ n = 11. Reading the answer carefully. n = 11 means the salary is ₹7,00,000 in his 11^ th year of work. He began the 1^ st year on ₹5,00,000, so he needed 10 increments — that is, 10 years of service — for the salary to climb to ₹7,00,000. Check. 10 increments add 10 × 20, 000 = 2, 00, 000, and 5, 00, 000 + 2, 00, 000 = 7, 00, 000. ✓ Answer. After 10 years — his salary is ₹7,00,000 in the 11^ th year.

Question 13

A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

Accepted Answer

Sum of an AP. Row 1 has 1 marble, row 2 has 2, row 3 has 3, and so on up to row 25. The total number of marbles is therefore 1 + 2 + 3 + + 25, which is the sum of an AP with a = 1, d = 1, n = 25 and last term l = 25. The pairing trick. Write the sum forwards and backwards, one under the other: Adding the two lines column by column, every column gives 26, and there are 25 columns: 2S = 25 × 26 = 650 ⇒ S = 650/2 = 325. This is the formula S n = n/2(a + l) = 25/2(1 + 25) = (25 × 26)/2 = 325. Check on a small case. The same method for just 4 rows gives (4 × 5)/2 = 10, and indeed 1 + 2 + 3 + 4 = 10. ✓ Adding row by row up to 25 also gives 325. Answer. The child uses 325 marbles in all.

Question 14

Find the 12^ th term of a GP with common ratio 2, whose 8^ th term is 192.

Accepted Answer

Stepping along a GP. In a GP each term is the previous one multiplied by the common ratio r. Going from the 8^ th term to the 12^ th term means taking 12 - 8 = 4 steps, and each step multiplies by r = 2. So t 12 = t 8 × r^4 = 192 × 2^4 = 192 × 16 = 3072. Why r^4 and not r^5? Because the 8^ th term is already in hand; the multiplications carry us 8 → 9 → 10 → 11 → 12, which is four multiplications. Check by finding the whole GP. Using t n = a r^ n-1, t 8 = a × 2^7 = 128a = 192 ⇒ a = 192/128 = 1.5. Then t 12 = 1.5 × 2^11 = 1.5 × 2048 = 3072. (The GP is 1.5, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072 — the eighth term is 192 and the twelfth is 3072, as required.) Answer. t 12 = 3072.

Question 15

Find the 10^ th and n^ th terms of the GP: 5, 25, 125,....

Accepted Answer

n^ th term of a GP. Check that the ratio of consecutive terms is constant: 25/5 = 5, 125/25 = 5. It is, so this is a GP with first term a = 5 and common ratio r = 5. The n^ th term. To reach the n^ th term from the first you multiply by r exactly (n-1) times: t n = a r^ n-1 = 5 × 5^ n-1 = 5^ n. The powers collapse neatly here because the first term and the ratio happen to be the same number. So the sequence is simply 5^1, 5^2, 5^3, … — which matches 5, 25, 125, … ✓ The 10^ th term. Put n = 10: t 10 = 5^10 = 9 765 625. Check on that power. 5^5 = 3125, and 5^10 = (5^5)^2 = 3125 × 3125 = 9 765 625. ✓ Answer. t n = 5^ n and t 10 = 5^10 = 9, 765, 625.

Question 16

A sequence is given by the recursive rule t 1=2, t n+1=3 t n-2 for n ≥ 1. Which term of the sequence is 730?

Accepted Answer

Turning a recursive rule into a pattern. The rule is t 1 = 2 and t n+1 = 3t n - 2. The straightforward route is simply to generate terms until 730 appears: t 1 = 2, t 2 = 3(2) - 2 = 4, t 3 = 3(4) - 2 = 10, t 4 = 3(10) - 2 = 28, t 5 = 3(28) - 2 = 82, t 6 = 3(82) - 2 = 244, t 7 = 3(244) - 2 = 730. So 730 is the 7^ th term. A neater way to see why. This is not a GP — the ratios 4/2 = 2 and 10/4 = 2.5 are different — but it is nearly one. Subtract 1 from both sides of the rule: t n+1 - 1 = 3t n - 3 = 3 (t n - 1). So the shifted numbers t n - 1 do form a GP, with first value t 1 - 1 = 1 and common ratio 3: t n - 1 = 1 × 3^ n-1 ⇒ t n = 3^ n-1 + 1. Now set t n = 730: 3^ n-1 + 1 = 730 ⇒ 3^ n-1 = 729. Since 729 = 3^6, we get n - 1 = 6, so n = 7. Check. t 7 = 3^6 + 1 = 729 + 1 = 730 ✓, and the closed form also reproduces the earlier terms: 3^0+1 = 2, 3^1+1 = 4, 3^2+1 = 10. ✓ Both methods agree. Answer. 730 is the 7^ th term.

Question 17

Which term of the GP: 2, 6, 18,... is 4374? Write the explicit formula as well as the recursive formula for the n^ th term.

Accepted Answer

n^ th term of a GP. The ratios of consecutive terms are 6/2 = 3 and 18/6 = 3, so this is a GP with a = 2 and r = 3. Explicit formula. Reaching the n^ th term takes (n-1) multiplications by 3: t n = a r^ n-1 = 2 × 3^ n-1. Recursive formula. Start at 2, and each new term is three times the one before: t 1 = 2, t n+1 = 3 t n for n ≥ 1. Which term is 4374? Set t n = 4374: 2 × 3^ n-1 = 4374 ⇒ 3^ n-1 = 4374/2 = 2187. Now express 2187 as a power of 3: 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729, 3^7 = 2187. So 3^ n-1 = 3^7, giving n - 1 = 7 and n = 8. Check. t 8 = 2 × 3^7 = 2 × 2187 = 4374 ✓. Listing the GP: 2, 6, 18, 54, 162, 486, 1458, 4374 — the eighth term is indeed 4374. Answer. 4374 is the 8^ th term. Explicit formula: t n = 2 × 3^ n-1; recursive formula: t 1 = 2, t n+1 = 3t n.

Question 18

A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60\% of the height from which it fell. It continues bouncing in this way-each time rising to 60\% of the previous height. (i) What height does the ball reach after the 5^ th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6^ th time?

Accepted Answer

Bounce heights as a GP. Each bounce leaves the ball rising to 60 % — that is, 0.6 times — of the height it fell from. So the peak heights after successive bounces form a GP with first term 80 × 0.6 and common ratio 0.6: h 1 = 80 × 0.6 = 48, h 2 = 48 × 0.6 = 28.8, h 3 = 17.28, h 4 = 10.368, h 5 = 6.2208 (all in metres). In general, the height reached after the k^ th bounce is h k = 80 × (0.6)^k. (i) Height after the 5^ th bounce. h 5 = 80 × (0.6)^5. Now (0.6)^2 = 0.36, (0.6)^4 = 0.36 × 0.36 = 0.1296, and (0.6)^5 = 0.1296 × 0.6 = 0.07776. Hence h 5 = 80 × 0.07776 = 6.2208 m. This matches the step-by-step list above. ✓ (ii) Total vertical distance up to the 6^ th hit on the ground. Trace the journey carefully, because the ball travels each bounce height twice — once going up and once coming back down — while the very first 80 m drop is travelled only once. Notice that only the first five bounce heights are involved: the sixth bounce happens after the sixth hit, so it does not count. Total = 80 + 2 (48 + 28.8 + 17.28 + 10.368 + 6.2208). The bracket is the sum of 5 terms of a GP with a = 48 and r = 0.6: S 5 = (a (1 - r^5))/(1 - r) = (48 (1 - 0.07776))/0.4 = (48 × 0.92224)/0.4 = 120 × 0.92224 = 110.6688. (Adding the five numbers directly gives 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688 as well. ✓) Therefore Total = 80 + 2 × 110.6688 = 80 + 221.3376 = 301.3376 m. Answer. (i) 6.2208 m (about 6.22 m). (ii) 301.3376 m (about 301.34 m).

Question 19

Which term of the sequence 2,2 √ 2, 4, … is 128?

Accepted Answer

Spot the common ratio. Divide each term by the one before it: (2√ 2 )/2=√ 2, 4/(2√ 2 )=2/(√ 2 )=√ 2. The same number √ 2 comes out both times, so this is a GP with first term a=2 and common ratio r=√ 2, and its n^ th term is a n=a r^ n-1=2(√ 2 )^n-1. Turn everything into a power of 2. Comparing an expression containing √ 2 with 128 is awkward, but both are powers of 2, and that makes the comparison easy. Since √ 2 =2^1/2, a n=2^1·(2^1/2)^n-1=2^1+n-1/2=2^(n+1)/2. Set it equal to 128. Now 128=2^7, so we need 2^(n+1)/2=2^7. Two powers of the same base 2 are equal only when the exponents are equal, so (n+1)/2=7 ⇒ n+1=14 ⇒ n=13. Check. The 13^ th term is a 13=2(√ 2 )^12=2[(√ 2 )^2]^6=2· 2^6=2· 64=128. 128 is the 13^ th term of the sequence.

Question 20

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Look at Fig. 8.12 and try to answer the following questions. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the n^ th stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the n^ th stage. What happens to this area as n, the number of stages, goes on increasing?

Accepted Answer

Follow the rule, one stage at a time. The paragraph tells us exactly what happens at each step: every red (shaded) square is cut by trisection into 3× 3=9 equal small squares, the middle one is thrown away, and the remaining 8 stay red. So going from one stage to the next, every red square is replaced by 8 red squares, each with side one-third as long. (i) Red squares in Stages 0 to 3. - Stage 0: the whole sheet is one red square, so 1 red square. - Stage 1: the one square becomes 8, so 8 red squares. - Stage 2: each of those 8 becomes 8, so 8× 8=64. - Stage 3: each of those 64 becomes 8, so 8× 64=512. (ii) Predicting Stages 4 and 5. The rule "multiply by 8" does not change, so Stage 4: 8× 512=4096, Stage 5: 8× 4096=32768. (iii) A rule for the n^ th stage. The counts 1,8,64,512,… form a GP with first term 1 (at Stage 0) and common ratio 8. Writing S n for the number of red squares at Stage n, Explicit: S n=8^ n, Recursive: S 0=1, S n=8 S n-1 (n≥ 1). The explicit formula lets you jump straight to any stage; the recursive one says in symbols what the construction actually does. (iv) Area of the red region. Take the Stage 0 square to have area 1 square unit, so its side is 1 unit. Trisecting the side gives small squares of side 13, and area ( 13)^2= 19 of the square they came from. Eight of the nine pieces are kept, so at every stage red area → 8×1/9×( red area )=8/9×( red area ). Hence, with A n the red area at Stage n: A 1=8/9, A 2=64/81, A 3=512/729. A 4=4096/6561, A 5=32768/59049. The same numbers come out if you count directly: at Stage n there are 8^n red squares, each of side ( 13)^n, so the red area is 8^n·(1/(3^n))^2=(8^n)/(9^n). That is a useful second check. Explicit: A n=(8/9)^n, Recursive: A 0=1, A n=8/9A n-1 (n≥ 1). What happens as n increases? Each stage multiplies the area by 89, and 89<1, so the red area gets smaller every single time and never stops shrinking. Multiplying by 89 again and again drives ( 89)^n as close to 0 as we like — for instance ( 89)^50≈ 0.003. So the red area tends to 0: in the "limit" the carpet has more and more squares but less and less area. (Note it never actually equals 0 at any finite stage.) Answers: (i) 1, 8, 64, 512. (ii) 4096 and 32768. (iii) S n=8^n; S 0=1, S n=8S n-1. (iv) 8/9, 64/81, 512/729, then 4096/6561 and 32768/59049; A n=(8/9)^n with A 0=1, A n=8/9A n-1; and the red area shrinks towards 0 as n grows.

Question 21

Find the 31^ st term of an AP whose 11^ th term is 38 and 16^ th term is 73.

Accepted Answer

Two terms are enough to find d. In an AP the n^ th term is a n=a+(n-1)d. To get from the 11^ th term to the 16^ th term you add d five times, so a 16-a 11=5d ⇒ 73-38=5d ⇒ 5d=35 ⇒ d=7. Now find the first term. a 11=a+10d=38 ⇒ a+70=38 ⇒ a=-32. The 31^ st term. a 31=a+30d=-32+30(7)=-32+210=178. Check a second way. Go from the 16^ th term instead, adding d fifteen times: a 31=a 16+15d=73+15(7)=73+105=178. The 31^ st term is 178.

Question 22

Determine the AP whose third term is 16 and whose 7^ th term exceeds the 5^ th term by 12.

Accepted Answer

Use the gap between two terms to get d. Going from the 5^ th term to the 7^ th term adds d twice, so a 7-a 5=2d. We are told this gap is 12, so 2d=12 ⇒ d=6. Then use the third term to get a. a 3=a+2d=16 ⇒ a+12=16 ⇒ a=4. Write out the AP. With a=4 and d=6: 4, 10, 16, 22, 28, 34, … and in general a n=4+(n-1)6=6n-2. Check. a 3=6(3)-2=16, and a 7-a 5=(6· 7-2)-(6· 5-2)=40-28=12. The AP is 4, 10, 16, 22, 28, … (first term 4, common difference 6), with a n=6n-2.

Question 23

How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)

Accepted Answer

Find the first and last three-digit multiples of 7. Three-digit numbers run from 100 to 999. - Smallest: 100÷ 7=14 remainder 2, so 100 is not a multiple; the next multiple is 7× 15=105. - Largest: 999÷ 7=142 remainder 5, so the largest one is 7× 142=994. These form an AP. The multiples of 7 in between go up in steps of 7: 105, 112, 119, …, 994, a=105, d=7. Count the terms. Let 994 be the n^ th term: a+(n-1)d=994 105+(n-1)7=994 (n-1)7=889 n-1=127 ⇒ n=128. Check another way. The multiples are 7× 15, 7× 16, …, 7× 142, so we are counting the whole numbers from 15 to 142, and there are 142-15+1=128 of them. There are 128 three-digit numbers divisible by 7.

Question 24

How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)

Accepted Answer

Find the first and last multiples of 4 in the range. - The first multiple of 4 after 10 is 12 (since 10÷ 4=2 remainder 2, and 4× 3=12 ). - The last multiple of 4 before 250 is 248 (since 250÷ 4=62 remainder 2, and 4× 62=248 ). (Neither 10 nor 250 is itself a multiple of 4, so it makes no difference here whether "between" includes the end numbers.) These form an AP. 12, 16, 20, …, 248, a=12, d=4. Count the terms. Let 248 be the n^ th term: 12+(n-1)4=248 (n-1)4=236 n-1=59 ⇒ n=60. Check another way. The numbers are 4× 3, 4× 4, …, 4× 62, that is the whole numbers from 3 to 62, and there are 62-3+1=60 of them. 60 multiples of 4 lie between 10 and 250.

Question 25

Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Accepted Answer

Let the GP be a, ar, ar^2, ar^3, ar^4, … In a GP neither a nor r can be 0, which is what lets us divide by them below. Use the second condition first — it gives r on its own. "The fifth term is 4 times the third term" means ar^4=4 ar^2. Divide both sides by ar^2 (allowed, since a≠ 0 and r≠ 0 ): r^2=4 ⇒ r=2 or r=-2. Both signs must be kept — squaring loses the sign, so we test each. Now use the first condition. "The sum of the first two terms is -4 " means a+ar=a(1+r)=-4. Case r=2: a(1+2)=-4⇒ 3a=-4⇒ a=-4/3. The GP is -4/3, -8/3, -16/3, -32/3, -64/3, … Check: - 43- 83=-12/3=-4; fifth term -64/3=4×(-16/3). Case r=-2: a(1-2)=-4⇒ -a=-4⇒ a=4. The GP is 4, -8, 16, -32, 64, … Check: 4+(-8)=-4; fifth term 64=4× 16. Two GPs satisfy both conditions: 4, -8, 16, -32, 64, … (with a=4, r=-2 ) and -4/3, -8/3, -16/3, -32/3, … (with a=-4/3, r=2 ). Either one is a correct answer to "find a GP".

Question 26

Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Accepted Answer

Set it up as an AP with common difference 1. Suppose 100 is the sum of k consecutive natural numbers starting at a: a+(a+1)+(a+2)+ +(a+k-1)=100, a≥ 1, k≥ 1. This is an AP with first term a, common difference 1 and k terms, so its sum is k/2 [2a+(k-1) ]=100 ⇒ k(2a+k-1)=200. Two facts cut the search down to almost nothing. Fact 1 — one factor must be odd. The numbers k and 2a+k-1 differ by 2a-1, an odd number, so one of them is even and the other is odd. Since 200=2^3× 5^2, the only odd divisors of 200 are 1, 5, 25. So one of k and 2a+k-1 is 1, 5 or 25. Fact 2 — k is the smaller factor. Because a≥ 1, 2a+k-1 ≥ k+1 k, so k· k<k(2a+k-1)=200, giving k^2<200 and therefore k≤ 14. Check each possibility. If k is the odd one: k∈ 1,5,25, and k≤ 14 rules out 25. - k=1: 2a+0=200⇒ a=100. This is just "100" by itself — a single number, not really a sum. - k=5: 5(2a+4)=200⇒ 2a+4=40⇒ a=18. Gives 18+19+20+21+22. If 2a+k-1 is the odd one: 2a+k-1∈ 1,5,25, so k=200, 40 or 8; only k=8 satisfies k≤ 14. - k=8: 2a+7=25⇒ 2a=18⇒ a=9. Gives 9+10+11+12+13+14+15+16. Verify both sums. 18+19+20+21+22=100 (five terms averaging 20) 9+10+11+12+13+14+15+16=(8(9+16))/2=4× 25=100 A quick computer search over all starting values also finds exactly these two (plus the one-term case), which confirms nothing was missed. There are exactly two ways to write 100 as a sum of two or more consecutive natural numbers: 100=18+19+20+21+22 100=9+10+11+12+13+14+15+16 (If a "sum" of a single number is allowed, 100=100 is a third, trivial, case.)

Question 27

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2^ nd hour, 4^ th hour and n^ th hour?

Accepted Answer

Doubling means a GP with r=2. Start with 30 bacteria. Each hour the count is multiplied by 2, so the counts form a GP: 30 start, 60 1st hour, 120 2nd hour, 240 3rd hour, 480 4th hour, … Notice the count at the end of the n^ th hour has been doubled n times, so the exponent is n itself (not n-1 ) — the starting count 30 belongs to hour 0, before any doubling. End of the 2^ nd hour. 30× 2^2=30× 4=120. End of the 4^ th hour. 30× 2^4=30× 16=480. End of the n^ th hour. 30× 2^ n. Check. Put n=2: 30× 4=120. Put n=0: 30× 1=30, the original count. 120 bacteria after 2 hours, 480 after 4 hours, and 30× 2^ n after n hours.

Question 28

The sum of the 4^ th and 8^ th terms of an AP is 24 and the sum of the 6^ th and 10^ th terms is 44. Find the first three terms of the AP.

Accepted Answer

Turn each statement into an equation in a and d. With a n=a+(n-1)d: a 4+a 8=(a+3d)+(a+7d)=2a+10d=24 ⇒ a+5d=12 1 a 6+a 10=(a+5d)+(a+9d)=2a+14d=44 ⇒ a+7d=22 2 (Each equation was halved, which keeps the numbers small.) Subtract to eliminate a. Doing (2)-(1): (a+7d)-(a+5d)=22-12 2d=10 ⇒ d=5. Back-substitute into (1). a+5(5)=12 ⇒ a+25=12 ⇒ a=-13. First three terms. a=-13, a+d=-8, a+2d=-3. Check. a 4=-13+3(5)=2 and a 8=-13+7(5)=22, so a 4+a 8=24. Also a 6=12 and a 10=32, so a 6+a 10=44. The first three terms are -13, -8, -3 (with a=-13 and d=5 ).

Question 29

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Accepted Answer

Use the formula for the sum of the first n natural numbers. 1+2+3+ +n=(n(n+1))/2. We need the smallest n with (n(n+1))/2 1000 ⇒ n(n+1) 2000. Estimate first, then test. Since n(n+1) is a little more than n^2, and √ 2000 ≈ 44.7, the answer must be near 44 or 45. So test both rather than guessing: n=44: (44× 45)/2=1980/2=990, 990 1000. n=45: (45× 46)/2=2070/2=1035, 1035 1000. Why no smaller n can work. The sums (n(n+1))/2 increase as n increases (each step adds the positive number n+1 ). So if n=44 already falls short, every n≤ 44 falls short too. Therefore 45 is the smallest. The smallest value is n=45, and then 1+2+ +45=1035 1000.

Question 30

Which term of the GP: 2, 8, 32,... is 131072? Write the explicit formula as well as the recursive formula for the n^ th term.

Accepted Answer

Find the common ratio. 8/2=4, 32/8=4, so this is a GP with a=2 and r=4, and a n=2· 4^ n-1. Compare powers of 2. Both 2 and 4 are powers of 2, and so is 131072, so rewrite everything with base 2. Since 4=2^2, a n=2^1·(2^2)^n-1=2^1+2(n-1)=2^ 2n-1. Now factor 131072 as a power of 2 by halving repeatedly: 131072=2^17 ( because 2^10=1024 and 2^7=128, and 1024× 128=131072). Equate the exponents. 2^ 2n-1=2^17 ⇒ 2n-1=17 ⇒ 2n=18 ⇒ n=9. Check. a 9=2· 4^8=2× 65536=131072. Formulas. Explicit: a n=2· 4^ n-1 (=2^ 2n-1), Recursive: a 1=2, a n=4 a n-1 (n≥ 2). 131072 is the 9^ th term; a n=2· 4^ n-1, and a 1=2, a n=4a n-1.

Question 31

The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.

Accepted Answer

Choose the terms symmetrically. For three terms of a GP it is far easier to write them as a/r, a, ar than as a, ar, ar^2 — because then the product collapses at once. Use the product. a/r· a· ar=a^3=-1 ⇒ a=-1, since -1 is the only real number whose cube is -1. Use the sum. With a=-1, -1/r+(-1)+(-1)r=13/12 -(1/r+1+r)=13/12 1/r+r=-13/12-1=-25/12. Multiply through by 12r (valid because r≠ 0 in a GP): 12+12r^2=-25r ⇒ 12r^2+25r+12=0. Factorise. Split the middle term using 9+16=25 and 9× 16=144=12× 12: 12r^2+9r+16r+12=3r(4r+3)+4(4r+3)=(3r+4)(4r+3)=0, so r=-4/3 or r=-3/4. Write out the terms. With a=-1 and r=-3/4: a/r=-1/(-3/4)=4/3, a=-1, ar=(-1)(-3/4)=3/4. So the terms are 4/3, -1, 3/4. The other root r=-4/3 gives the very same three numbers in the reverse order, 3/4, -1, 4/3. Check. 4/3+(-1)+3/4=(16-12+9)/12=13/12 4/3×(-1)×3/4=-1 Common ratio r=-3/4 (or r=-4/3, which just reverses the order), and the three terms are 4/3, -1, 3/4.

Question 32

If the 4^ th, 10^ th and 16^ th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Accepted Answer

Write all three terms from the same first term and ratio. Let the GP have first term A and common ratio R, so its n^ th term is A R^ n-1. (In a GP A≠ 0 and R≠ 0.) Then x=a 4=AR^3, y=a 10=AR^9, z=a 16=AR^15. Method 1 — show the two ratios are equal. Three numbers are in GP exactly when the second divided by the first equals the third divided by the second. Here y/x=(AR^9)/(AR^3)=R^ 9-3=R^6, z/y=(AR^15)/(AR^9)=R^ 15-9=R^6. The two ratios are both R^6, so they are equal. Hence x, y, z form a GP with common ratio R^6. Method 2 — show y^2=xz. This is the same statement rearranged, and needs no division: y^2=(AR^9)^2=A^2R^18, xz=(AR^3)(AR^15)=A^2R^ 3+15=A^2R^18. So y^2=xz, which is precisely the condition for x, y, z to be in GP. Why it works. The term numbers 4, 10, 16 are themselves equally spaced (each 6 apart). Stepping 6 places along a GP always multiplies by R^6, so picking terms at equal gaps out of a GP always leaves you with another GP. Hence x, y, z are in GP, with common ratio R^6. Proved.

Question 33

The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Accepted Answer

Set up both conditions. Let the three terms be a, ar, ar^2. Then a+ar+ar^2=a(1+r+r^2)=26 1 a^2+a^2r^2+a^2r^4=a^2(1+r^2+r^4)=364 2 The key identity. The awkward factor in (2) factorises: 1+r^2+r^4=(1+r+r^2)(1-r+r^2). Expanding the right side confirms it: 1-r+r^2+r-r^2+r^3+r^2-r^3+r^4=1+r^2+r^4 Divide (2) by the square of (1). Squaring (1) gives a^2(1+r+r^2)^2=676. Dividing (2) by that: (a^2(1+r+r^2)(1-r+r^2))/(a^2(1+r+r^2)^2)=364/676 (1-r+r^2)/(1+r+r^2)=364/676=7/13. The whole point of dividing is that a cancels, leaving an equation in r alone. (Cancelling 1+r+r^2 is safe: for every real r it equals (r+ 12)^2+ 34 0, so it is never zero.) Solve for r. Cross-multiplying: 13(1-r+r^2)=7(1+r+r^2) 13-13r+13r^2=7+7r+7r^2 6r^2-20r+6=0 ⇒ 3r^2-10r+3=0. Split the middle term: 3r^2-9r-r+3=3r(r-3)-1(r-3)=(3r-1)(r-3)=0, so r=3 or r=1/3. Find a for each root, using (1). - r=3: a(1+3+9)=26⇒ 13a=26⇒ a=2. Terms 2, 6, 18. - r= 13: a(1+ 13+ 19)=a·13/9=26⇒ a=18. Terms 18, 6, 2 — the same three numbers written backwards. Check. 2+6+18=26 2^2+6^2+18^2=4+36+324=364 The terms are 2, 6, 18 (or 18, 6, 2, the same GP taken in the reverse order).

Question 34

Suppose P 1=1, P 2=2 and for n 2, P n = P 1+ P 2+ + P n -1 +1. Find the values of P 1, P 2, …, P 8. Can you find a simpler recursive formula for P n? Can you give an explicit formula?

Accepted Answer

Compute the terms one at a time, straight from the rule. The rule for n 2 is: add up all the earlier terms, then add 1. P 1=1, P 2=2 P 3=P 1+P 2+1=1+2+1=4 P 4=P 1+P 2+P 3+1=1+2+4+1=8 P 5=1+2+4+8+1=16 P 6=1+2+4+8+16+1=32 P 7=1+2+4+8+16+32+1=64 P 8=1+2+4+8+16+32+64+1=128 A simpler recursive formula. Each term is clearly double the one before it. Here is why, without checking case by case. For n 3, write the rule twice: P n=(P 1+P 2+ +P n-2+P n-1)+1, P n-1=(P 1+P 2+ +P n-2)+1. Subtracting the second from the first, everything inside the brackets cancels except P n-1: P n-P n-1=P n-1 ⇒ P n=2P n-1. And P 3=4=2× 2=2P 2 directly, while P 2=2=2× 1=2P 1 as well. So Simpler recursive formula: P 1=1, P n=2P n-1 for all n≥ 2. This says the sequence is just a GP with first term 1 and common ratio 2 — the "sum of everything before, plus 1" rule was a disguised doubling. Explicit formula. Starting at 1 and doubling n-1 times, Explicit: P n=2^ n-1. Check it against the original rule. For n 2, the earlier terms are 1+2+4+ +2^ n-2, a GP sum equal to 2^ n-1-1. Adding 1 gives 2^ n-1=P n. (For example P 8=2^7=128, matching the table.) P 1,…,P 8 are 1, 2, 4, 8, 16, 32, 64, 128; the simpler recursion is P n=2P n-1 with P 1=1; and the explicit formula is P n=2^ n-1.

Question 35

Suppose W 1=1, W 2=2 and for n 2, W n = W 1+ W 2+ + W n -2 +2. Find the values of W 1, W 2, …, W 8. Do you recognise this sequence?

Accepted Answer

Read the rule carefully — it stops two terms short. For n 2, W n adds up the terms only as far as W n-2 (not W n-1 ), and then adds 2. W 1=1, W 2=2 W 3=W 1+2=1+2=3 W 4=W 1+W 2+2=1+2+2=5 W 5=W 1+W 2+W 3+2=1+2+3+2=8 W 6=1+2+3+5+2=13 W 7=1+2+3+5+8+2=21 W 8=1+2+3+5+8+13+2=34 Do you recognise it? Yes — 1, 2, 3, 5, 8, 13, 21, 34 are the Fibonacci numbers: from the third term on, each one is the sum of the two before it (1+2=3, 2+3=5, 3+5=8, 5+8=13, 8+13=21, 13+21=34). Why the rule produces Fibonacci. For n 3, write the rule twice: W n=(W 1+ +W n-3+W n-2)+2, W n-1=(W 1+ +W n-3)+2. Subtracting, all the common terms cancel and only W n-2 survives: W n-W n-1=W n-2 ⇒ W n=W n-1+W n-2. That is exactly the Fibonacci rule. It also holds at n=3, since W 3=3=2+1=W 2+W 1. If the Fibonacci numbers are written F 1=F 2=1, F 3=2, F 4=3, F 5=5, …, then our sequence is that list shifted by one place: W n=F n+1. W 1,…,W 8 are 1, 2, 3, 5, 8, 13, 21, 34 — the Fibonacci sequence, obeying W n=W n-1+W n-2 (equivalently W n=F n+1 ).

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$
\(\displaystyle 3n - 4\)	\(\displaystyle -1\)	\(\displaystyle 2\)	\(\displaystyle 5\)	\(\displaystyle 8\)	\(\displaystyle 11\)
\(\displaystyle 2 - 5n\)	\(\displaystyle -3\)	\(\displaystyle -8\)	\(\displaystyle -13\)	\(\displaystyle -18\)	\(\displaystyle -23\)
\(\displaystyle n^{2} - 2n + 3\)	\(\displaystyle 2\)	\(\displaystyle 3\)	\(\displaystyle 6\)	\(\displaystyle 11\)	\(\displaystyle 18\)

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle \mathrm{T}_{n}\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 4$	$\displaystyle 7$	$\displaystyle 13$	$\displaystyle 24$	$\displaystyle 44$	$\displaystyle 81$

Stretch of the journey	Distance (m)
Initial drop, up to hit $\displaystyle 1$	\(\displaystyle 80\)
Up to \(\displaystyle h_{1}\) and back down, up to hit $\displaystyle 2$	\(\displaystyle 2 \times 48 = 96\)
Up to \(\displaystyle h_{2}\) and back down, up to hit $\displaystyle 3$	\(\displaystyle 2 \times 28.8 = 57.6\)
Up to \(\displaystyle h_{3}\) and back down, up to hit $\displaystyle 4$	\(\displaystyle 2 \times 17.28 = 34.56\)
Up to \(\displaystyle h_{4}\) and back down, up to hit $\displaystyle 5$	\(\displaystyle 2 \times 10.368 = 20.736\)
Up to \(\displaystyle h_{5}\) and back down, up to hit $\displaystyle 6$	\(\displaystyle 2 \times 6.2208 = 12.4416\)

Stage \(\displaystyle n\)	Red squares \(\displaystyle S_n\)	Side of each square	Red area \(\displaystyle A_n\)
$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 1$	$\displaystyle 1$
$\displaystyle 1$	$\displaystyle 8$	\(\displaystyle 1/3\)	\(\displaystyle 8/9\approx 0.889\)
$\displaystyle 2$	$\displaystyle 64$	\(\displaystyle 1/9\)	\(\displaystyle 64/81\approx 0.790\)
$\displaystyle 3$	$\displaystyle 512$	\(\displaystyle 1/27\)	\(\displaystyle 512/729\approx 0.702\)
$\displaystyle 4$	$\displaystyle 4096$	\(\displaystyle 1/81\)	\(\displaystyle 4096/6561\approx 0.624\)
$\displaystyle 5$	$\displaystyle 32768$	\(\displaystyle 1/243\)	\(\displaystyle 32768/59049\approx 0.555\)

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle P_n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 4$	$\displaystyle 8$	$\displaystyle 16$	$\displaystyle 32$	$\displaystyle 64$	$\displaystyle 128$

Exercise Set 8.1 1–6 (part 1 of 5)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6