Sequences and Series

Chapter 5 · Algebra · 2 questions in the paper · 12 min read

Sequences and Series

What the syllabus asks for 3
  • Arithmetic progression
  • Geometric progression
  • Special sums and mixed problems

Two questions a year come from sequences, and the Dean's Office model had a seven-day "1 paisa, 2 paisa, 4 paisa" savings problem.

The question is almost always one of five things: the nth term of an AP or GP, a sum, an AM or GM, a "which term is", or a sum to infinity. Memorise the formulas below and each takes under a minute.

What the paper asks from this chapter
  • AP: a_n=a+(n-1)d, S_n=\frac{n}{2}[2a+(n-1)d], three numbers in AP, insert means.
  • GP: a_n=ar^{n-1}, S_n=\frac{a(r^n-1)}{r-1}, S_\infty=\frac{a}{1-r}.
  • Special sums: \sum n, \sum n^2, \sum n^3, odd and even numbers.
  • Means: AM=\frac{a+b}{2}, GM=\sqrt{ab}, and AM\ge GM.

5.1 Arithmetic progression

  • a_n=a+(n-1)d; number of terms n=\frac{l-a}{d}+1.
  • S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l).
  • a,b,c in AP \iff 2b=a+c. For three numbers in AP write a-d,a,a+d (sum =3a).
  • To insert m means between a and b: d=\frac{b-a}{m+1}.
  • If S_n is given, a_n=S_n-S_{n-1} and a_1=S_1.

diagram
Figure 5.1. Gauss's trick drawn as a rectangle. Writing the sum 1 + 2 + ... + n twice, once forwards and once backwards, makes a rectangle n wide and n + 1 tall, so S = n(n + 1)/2; for n = 10 that is 10 x 11 / 2 = 55 and for n = 100 it is 5,050. The bars on the right are an AP with a = 3, d = 4: the 10th term is 3 + 9 x 4 = 39 and the sum of ten terms is 10/2 x (3 + 39) = 210.

5.2 Geometric progression

  • a_n=ar^{n-1}, r=\frac{a_2}{a_1}. S_n=\frac{a(r^n-1)}{r-1} for r\ne1; for |r|<1, S_\infty=\frac{a}{1-r}.
  • a,b,c in GP \iff b^2=ac. Three numbers in GP: \frac{a}{r},a,ar (product =a^3).
  • Recurring decimal: 0.\overline{6}=\frac{0.6}{1-0.1}=\frac{2}{3}.
  • Harmonic progression: the reciprocals of its terms form an AP.

diagram
Figure 5.2. A geometric progression grows or shrinks by a constant FACTOR. (a) 2, 6, 18, ... with r = 3: the sixth term is 2 x 3^5 = 486 and the sum of six terms is 2(3^6 - 1)/(3 - 1) = 728. (b) With |r| < 1 the sum settles: 1 + 1/2 + 1/4 + ... = a/(1 - r) = 2, drawn as a strip that is halved again and again. So 0.777... = 7/9 and 0.333... = 1/3.

5.3 Special sums and mixed problems

Sum Value
1+2+\dots+n \frac{n(n+1)}{2}
1+3+5+\dots (n odd numbers) n^2
2+4+\dots+2n n(n+1)
1^2+2^2+\dots+n^2 \frac{n(n+1)(2n+1)}{6}
1^3+2^3+\dots+n^3 \left[\frac{n(n+1)}{2}\right]^2
Doubling savings

Problem. You save 1 paisa on day 1, 2 on day 2, 4 on day 3, and so on. Total for a week?

Solution. GP with a=1, r=2, n=7: S=\frac{2^7-1}{2-1}=127 paisa.

photograph
(a) Fibonacci (Leonardo of Pisa)
photograph
(b) Statue of Fibonacci, Pisa
photograph
(c) Sunflower: spirals in 34 and 55
Plate 5.1. Fibonacci in two places. (a) Leonardo of Pisa, called Fibonacci (about 1170-1250), in an engraving: his book Liber Abaci (1202) carried Hindu-Arabic numerals to Europe and posed the rabbit problem that gives 1, 1, 2, 3, 5, 8, ... (b) His statue in the Camposanto of Pisa. (c) A sunflower head: its seeds spiral in 34 and 55 curves, two neighbouring Fibonacci numbers.Photo credits: (a) Unknown, Public domain; (b) Hans-Peter Postel, CC BY 2.5; (c) George Chernilevsky, CC BY 4.0.

diagram
Figure 5.3. Square sums, computed. For n = 10: the sum of n is 55, the sum of n^2 is 10 x 11 x 21 / 6 = 385, and the sum of n^3 is 3,025 = 55^2, so the sum of cubes is the SQUARE of the sum of integers. Partial sums of cubes for n = 1 to 6 are 1, 9, 36, 100, 225, 441, the squares of 1, 3, 6, 10, 15, 21. Also 1 + 3 + 5 + ... (n odd numbers) = n^2.

diagram
Figure 5.4. Fibonacci squares and the golden angle. (a) Squares of side 1, 1, 2, 3, 5, 8, 13 fit together into a rectangle 21 x 13. The ratios of neighbouring terms 2, 1.5, 1.667, 1.6, 1.625, 1.615 close in on the golden ratio 1.618. (b) Seeds placed at a turn of 137.5 degrees each (the golden angle) form the sunflower spirals; 300 seeds are plotted by the cell.

5.4 Eight traps in sequences and series

Progression questions are formula questions with a hidden counting step. The usual errors are off-by-one slips in n, or applying an infinite-sum formula to a ratio that is too big. These eight traps are checked against the figures above.

  1. The n-th term has n-1, not n. For an AP, a_n=a+(n-1)d, so the 10th term of 3,7,11,\ldots is 3+9\times4=39 and not 3+10\times4 (Fig 5.1). The same applies to a GP: a_n=ar^{n-1}, so the sixth term of 2,6,18,\ldots is 2\times3^5=486. Always count how many steps you take from the first term.
  2. Count the terms before you sum. The number of terms in a,a+d,\ldots,l is \frac{l-a}{d}+1. The list 7,10,13,\ldots,100 has \frac{93}{3}+1=32 terms, not 31. Then S=\frac n2(a+l) is the quickest way to add them.
  3. Choose the symmetric form for three or four terms in AP. Three terms are a-d,a,a+d, so their sum is 3a at once. Four terms are a-3d,a-d,a+d,a+3d with common difference 2d. The arithmetic mean of two numbers is \frac{x+y}{2}, and if x,y,z are in AP then 2y=x+z.
  4. The sum to infinity needs |r|<1. S_\infty=\frac{a}{1-r} is valid only when the ratio is between -1 and 1 (Fig 5.2). For r=3 the terms keep growing and there is no sum. Recurring decimals are good examples: 0.777\ldots=\frac{0.7}{1-0.1}=\frac79.
  5. Geometric mean uses a square root and three terms in GP use \frac ar,a,ar. The geometric mean of x and y is \sqrt{xy}, so the GM of 4 and 9 is 6, while the AM is 6.5. For positive numbers AM is never less than GM. If three numbers in GP have product 216, the middle number is \sqrt[3]{216}=6.
  6. Know the three standard sums and when to subtract. \sum n=\frac{n(n+1)}{2}, \sum n^2=\frac{n(n+1)(2n+1)}{6} and \sum n^3=\left[\frac{n(n+1)}{2}\right]^2 (Fig 5.3). For a sum that starts at 6 and ends at 20, use S(20)-S(5), not S(20)-S(6). The sum of the first n odd numbers is n^2.
  7. Identify the type before using a formula. A constant difference means AP and a constant ratio means GP, so check two gaps first. Some series are neither: \frac{1}{1\cdot2}+\frac1{2\cdot3}+\cdots telescopes, because \frac1{n(n+1)}=\frac1n-\frac1{n+1}, so the sum of n terms is \frac{n}{n+1}. Differences of differences point to a polynomial formula.
  8. Fibonacci questions use "add the last two". The sequence 1,1,2,3,5,8,13,\ldots has F_n=F_{n-1}+F_{n-2}, and the ratio \frac{F_{n+1}}{F_n} tends to 1.618 (Fig 5.4). Check whether the question starts counting at F_1=1 or F_0=0. In money problems, equal yearly raises are an AP while percentage raises are a GP, so a salary of Rs 30,000 with a Rs 2,000 raise is an AP, and with a 5 percent raise it is a GP.

5.4.1 Worked checks

Q. Find the 20th term and the sum of the first 20 terms of 3,7,11,\ldots

Answer. a=3,d=4, so a_{20}=3+19\times4=79. S_{20}=\frac{20}{2}(3+79)=10\times82=820. Check with the other formula: \frac{20}{2}[6+19\times4]=10\times82=820.

Q. Find the sum of the first six terms of 2,6,18,\ldots and the sum to infinity of 8,4,2,\ldots

Answer. The first is a GP with r=3: S_6=\frac{2(3^6-1)}{3-1}=728, which matches Fig 5.2. The second has r=\frac12, so S_\infty=\frac{8}{1-\frac12}=16. The infinite sum exists because |r|<1.

Q. Evaluate 1^2+2^2+\cdots+10^2 and 1^3+2^3+\cdots+10^3.

Answer. \sum n^2=\frac{10\times11\times21}{6}=385. \sum n^3=\left(\frac{10\times11}{2}\right)^2=55^2=3025. These match the bars in Fig 5.3.

Formula and fact sheet

Item AP GP
nth term a+(n-1)d ar^{n-1}
Sum \frac{n}{2}[2a+(n-1)d] \frac{a(r^n-1)}{r-1}
Condition 2b=a+c b^2=ac
Mean \frac{a+b}{2} \sqrt{ab}

Chapter summary

  • Count terms first: n=\frac{l-a}{d}+1.
  • S_n for AP is n times the average of first and last.
  • Use \frac{a}{r},a,ar for three GP numbers and a-d,a,a+d for AP.
  • S_\infty exists only when |r|<1.
  • a_n=S_n-S_{n-1} recovers terms from sums.
  • AM\ge GM, with equality only when the numbers are equal.

Practice questions

Set A — Recall (20 questions)

  1. The 10th term of the AP 2, 5, 8, … isa27b30c32d29
  2. The sum of the first 20 natural numbers isa210b200c420d190
  3. The 5th term of the GP 3, 6, 12, … isa24b96c48d45
  4. The arithmetic mean of 4 and 16 isa8b10c12d20
  5. The geometric mean of 4 and 16 isa10b12c6d8
  6. The sum 1+2+4+8+16+32+64 isa64b128c127d126
  7. The sum of the first n odd numbers isan^2bn(n+1)c\frac{n(n+1)}{2}d2n
  8. The sum to infinity of 1+\frac{1}{2}+\frac{1}{4}+\cdots isa1b2c\frac{3}{2}dinfinite
  9. If a,b,c are in AP, thenab^2=acbb=a+cc2b=a+cdb=ac
  10. The common ratio of 2, -6, 18, … isa3b-2c4d-3
  11. The sum of the first 10 even natural numbers isa100b110c55d120
  12. 1^3+2^3+3^3 equalsa36b14c6d27
  13. The common difference of 7, 4, 1, … isa3b11c-4d-3
  14. 1^2+2^2+3^2+4^2+5^2 equalsa15b225c55d50
  15. In a harmonic progression the reciprocals of the terms formaan APba GPcan HPda constant sequence
  16. The number of terms in the AP 5, 9, 13, …, 41 isa9b10c11d12
  17. The 3rd term of the GP with first term 2 and ratio 3 isa12b9c18d54
  18. The sum of the first n natural numbers isan^2bn(n+1)c\frac{n(n-1)}{2}d\frac{n(n+1)}{2}
  19. The sum to infinity of the GP 3, 1, \frac{1}{3}, … isa\frac{9}{2}b4c6d9
  20. Which of the following is a GP?a1, 3, 5, 7b3, 9, 27, 81c2, 4, 6, 8d1, 4, 9, 16
Answers and explanations — Set A

1. (d) — 2 + 9 × 3 = 29.

2. (a) — 20 × 21 / 2 = 210.

3. (c) — 3 × 2⁴ = 48.

4. (b) — (4 + 16)/2 = 10.

5. (d) — √(4 × 16) = 8; GM is below AM.

6. (c) — \dfrac{2⁷ - 1}{2 - 1} = 127.

7. (a) — 1 + 3 + 5 + … = n².

8. (b) — a/(1 − r) = 1/(1/2) = 2.

9. (c) — the middle term is the average of its neighbours; (a) is for a GP.

10. (d) — 18 / (−6) = −3.

11. (b) — n(n + 1) = 10 × 11 = 110.

12. (a) — 1 + 8 + 27 = 36 = (1 + 2 + 3)².

13. (d) — 4 − 7 = −3.

14. (c) — 5 × 6 × 11 / 6 = 55.

15. (a) — that is the definition of an HP.

16. (b) — (41 − 5)/4 + 1 = 10.

17. (c) — 2 × 3² = 18.

18. (d) — n(n + 1)/2; (b) is the sum of even numbers.

19. (a) — 3 / (1 − 1/3) = 9/2.

20. (b) — constant ratio 3.

Set B — Understanding (12 questions)

  1. The sum of the first 10 terms of the AP 3, 7, 11, … isa200b210c220d190
  2. Which term of the AP 3, 8, 13, … is 78?a14thb15thc17thd16th
  3. Three numbers in AP have sum 27. The middle number isa9b8c10d12
  4. Three numbers in GP have product 216. The middle number isa8b12c6d9
  5. A GP has first term 5 and fifth term 80, with positive ratio. The ratio isa3b2c4d16
  6. The sum 1+3+9+27+81+243 isa364b243c363d729
  7. The first term of an AP is 3 and the sum of its first 5 terms is 35. The common difference isa4b3c1d2
  8. The sum of all multiples of 3 between 1 and 50 isa400b425c408d416
  9. The sum 2+4+6+\cdots+100 isa2,550b5,050c2,500d5,100
  10. The sum to infinity of 0.6+0.06+0.006+\cdots isa0.6b0.7c1d\frac{2}{3}
  11. For what n is 1+2+\cdots+n=55?a9b11c10d12
  12. If three arithmetic means are inserted between 4 and 20, the second mean isa10b12c14d16
Answers and explanations — Set B

1. (b) — 5(6 + 36) = 210.

2. (d) — 3 + (n − 1)5 = 78 gives n = 16.

3. (a) — sum = 3 × middle, so 27/3 = 9.

4. (c) — (a/r)(a)(ar) = a³ = 216, a = 6.

5. (b) — r⁴ = 16, so r = 2.

6. (a) — (3⁶ − 1)/2 = 364.

7. (d) — (5/2)(6 + 4d) = 35 gives d = 2.

8. (c) — 3, 6, …, 48 is 16 terms; 16(3 + 48)/2 = 408.

9. (a) — 50 × 51 = 2,550; 5,050 is the sum of 1 to 100.

10. (d) — 0.6/(1 − 0.1) = 2/3.

11. (c) — 10 × 11/2 = 55.

12. (b) — d = 16/4 = 4, so the means are 8, 12, 16.

Set C — Application (8 questions)

  1. A man saves Rs 100 in the first month and Rs 20 more each month than the month before. His total saving in 12 months isaRs 2,400bRs 2,640cRs 2,520dRs 2,280
  2. If the sum of the first n terms of an AP is 3n^2+2n, its common difference isa6b5c3d11
  3. A GP has first term 1 and common ratio 3. Its 5th term isa27b64c243d81
  4. Three numbers in GP have sum 14 and product 64. The largest isa6b8c10d12
  5. The sum 1^2+2^2+\cdots+10^2 isa3,025b55c385d100
  6. The sum 1^3+2^3+\cdots+10^3 isa385b55c2,025d3,025
  7. The sum of the AP 25, 22, 19, …, 1 isa117b126c108d130
  8. If x, x+2, x+8 are in GP, then x isa2b1c4d3
Answers and explanations — Set C

1. (c) — (12/2)(200 + 11 × 20) = 6 × 420 = 2,520.

2. (a) — a₁ = 5, a₂ = 16 − 5 = 11, so d = 6.

3. (d) — 1 × 3⁴ = 81; (c) is the 6th term.

4. (b) — numbers 2, 4, 8: sum 14, product 64.

5. (c) — 10 × 11 × 21 / 6 = 385.

6. (d) — (55)² = 3,025.

7. (a) — 9 terms; 9 × (25 + 1)/2 = 117.

8. (b) — (x + 2)² = x(x + 8) gives 4x + 4 = 8x, x = 1.