Comprehensive Study Notes on Arithmetic Progressions

Introduction to Patterns and Arithmetic Progressions

  • Patterns in Nature: Many phenomena in nature follow specific mathematical patterns, including the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple, and the spirals on a pine cone.

  • Examples of Real-Life Patterns:

    • Salary Growth: Reena's starting monthly salary is 8000INR8000\,\text{INR} with an annual increment of 500INR500\,\text{INR}. Her salary sequence for the first few years is: 8000,8500,9000,...8000, 8500, 9000, . . .
    • Ladder Rungs: A ladder's rungs decrease uniformly by 2cm2\,\text{cm} from bottom to top. If the bottom rung is 45cm45\,\text{cm}, the lengths are: 45,43,41,39,37,35,33,3145, 43, 41, 39, 37, 35, 33, 31
    • Savings Scheme: An investment of 8000INR8000\,\text{INR} grows to 54\frac{5}{4} times its value every 3years3\,\text{years}. The maturity amounts after 3,6,9 and 12years3, 6, 9\text{ and } 12\,\text{years} are: 10000,12500,15625,19531.2510000, 12500, 15625, 19531.25
    • Unit Squares: The number of unit squares in larger squares with sides 1,2,3,...units1, 2, 3, . . .\,\text{units} is: 12,22,32,...1^2, 2^2, 3^2, . . .
    • Money Box Savings: Shakila puts 100INR100\,\text{INR} into a money box on her daughter's 1st birthday and increases the amount by 50INR50\,\text{INR} every year: 100,150,200,250,...100, 150, 200, 250, . . .
    • Rabbit Population (Fibonacci pattern): Pairs of rabbits that do not produce in the first month but produce a new pair every month thereafter result in the sequence: 1,1,2,3,5,81, 1, 2, 3, 5, 8

Definition and General Form of an Arithmetic Progression (AP)

  • Term: Each number in a list follows a certain rule and is called a term.

  • Definition of Arithmetic Progression (AP): An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except for the first term.

  • Common Difference (dd): The fixed number added to each term is called the common difference. It can be positive, negative, or zero.

  • Notation:

    • First term: a1a_1 (or simply aa)
    • Second term: a2a_2
    • nth term: ana_n
    • Common Difference: dd
    • Relation: a2a1=a3a2=...=anan1=da_2 - a_1 = a_3 - a_2 = . . . = a_n - a_{n-1} = d
  • General Form of an AP: The sequence is represented as:     a,a+d,a+2d,a+3d,...a, a + d, a + 2d, a + 3d, . . .

  • Types of APs:

    • Finite AP: Contains a finite number of terms and always has a last term. Examples include student heights in a queue or temperatures for a single week.
    • Infinite AP: Does not have a last term and continues indefinitely. Examples include the set of natural numbers 1,2,3,...1, 2, 3, . . .

The nth Term of an Arithmetic Progression

  • Derivation of the Formula:

    • Let the first term be aa and the common difference be dd.
    • a1=aa_1 = a
    • a2=a+d=a+(21)da_2 = a + d = a + (2 - 1)d
    • a3=a+2d=a+(31)da_3 = a + 2d = a + (3 - 1)d
    • a4=a+3d=a+(41)da_4 = a + 3d = a + (4 - 1)d
  • General Term Formula: The nth term ana_n of an AP is given by:     an=a+(n1)da_n = a + (n - 1)d

  • Last Term (ll): If there are mm terms in an AP, ama_m represents the last term, sometimes denoted as ll.

  • Identifying an AP: To determine if a list is an AP, calculate d=ak+1akd = a_{k+1} - a_k for various values of kk. If the difference is not constant, it is not an AP.

Worked Examples: nth Term

  • Example 3: For the AP 2,7,12,...2, 7, 12, . . ., find the 10th term.

    • a=2a = 2
    • d=72=5d = 7 - 2 = 5
    • n=10n = 10
    • a10=2+(101)×5=2+45=47a_{10} = 2 + (10 - 1) \times 5 = 2 + 45 = 47
  • Example 4: Which term of the AP 21,18,15,...21, 18, 15, . . . is 81-81? Is any term 00?

    • 81=21+(n1)(3)102=3(n1)34=n1n=35-81 = 21 + (n - 1)(-3) \rightarrow -102 = -3(n - 1) \rightarrow 34 = n - 1 \rightarrow n = 35
    • For an=0a_n = 0: 0=21+(n1)(3)21=3(n1)7=n1n=80 = 21 + (n - 1)(-3) \rightarrow -21 = -3(n - 1) \rightarrow 7 = n - 1 \rightarrow n = 8
    • Answer: The 35th term is 81-81 and the 8th term is 00.
  • Example 7: How many two-digit numbers are divisible by 33?

    • List: 12,15,18,...,9912, 15, 18, . . . , 99
    • a=12,d=3,an=99a = 12, d = 3, a_n = 99
    • 99=12+(n1)387=3(n1)29=n1n=3099 = 12 + (n - 1)3 \rightarrow 87 = 3(n - 1) \rightarrow 29 = n - 1 \rightarrow n = 30
  • Example 8: Find the 11th term from the last term of the AP: 10,7,4,...,6210, 7, 4, . . . , -62.

    • Method 1: Find total terms (n=25n=25), then calculate the 15th term from the start.
    • Method 2: Reverse the AP: 62,59,56,...-62, -59, -56, . . .
    • a=62,d=3,n=11a = -62, d = 3, n = 11
    • a11=62+(111)×3=62+30=32a_{11} = -62 + (11 - 1) \times 3 = -62 + 30 = -32

Sum of First n Terms of an AP

  • Gauss's Method: To sum numbers from 11 to 100100, Gauss wrote the sum twice (once in reverse) and added them to find constant sums of 101101.     S=1+2+...+99+100S = 1 + 2 + . . . + 99 + 100S=100+99+...+2+1S = 100 + 99 + . . . + 2 + 12S=100×101S=50502S = 100 \times 101 \rightarrow S = 5050

  • General Sum Formulas:

    • Standard formula: S=n2[2a+(n1)d]S = \frac{n}{2} [2a + (n - 1)d]
    • Using last term (ana_n): S=n2(a+an)S = \frac{n}{2} (a + a_n)
    • Using terminal term (ll): S=n2(a+l)S = \frac{n}{2} (a + l)
  • Relationship between ana_n and SnS_n:an=SnSn1a_n = S_n - S_{n-1}

Worked Examples: Summation

  • Example 11: Sum of first 2222 terms of 8,3,2,...8, 3, -2, . . .

    • a=8,d=5,n=22a = 8, d = -5, n = 22
    • S22=222[2(8)+(221)(5)]=11[16105]=11(89)=979S_{22} = \frac{22}{2} [2(8) + (22 - 1)(-5)] = 11 [16 - 105] = 11(-89) = -979
  • Example 13: How many terms of 24,21,18,...24, 21, 18, . . . must be taken for a sum of 7878?

    • 78=n2[48+(n1)(3)]156=n(513n)3n251n+156=078 = \frac{n}{2} [48 + (n - 1)(-3)] \rightarrow 156 = n(51 - 3n) \rightarrow 3n^2 - 51n + 156 = 0
    • n217n+52=0(n4)(n13)=0n^2 - 17n + 52 = 0 \rightarrow (n - 4)(n - 13) = 0
    • Two answers: n=4n = 4 or n=13n = 13. This happens because terms from the 5th to the 13th cancel each other out to zero.
  • Sum of first n positive integers:Sn=n(n+1)2S_n = \frac{n(n + 1)}{2}

Applications and Exercise Highlights

  • Divisibility: To find the sum of the first 4040 positive integers divisible by 66, use a=6,d=6,n=40a=6, d=6, n=40.
  • Compound Interest vs. Simple Interest: Simple interest sequences (80,160,240,...80, 160, 240, . . .) form an AP, whereas compound interest sequences do not.
  • Coordinate/Geometry Applications:
    • Spiral Length: A spiral made of 13 semicircles with increasing radii (0.5,1.0,1.5,...0.5, 1.0, 1.5, . . .). Total length is the sum of circumferences l=π(r1+r2+...+r13)l = \pi(r_1 + r_2 + . . . + r_{13}).
    • Log Stacking: 2020 logs in the bottom row, 1919 in next, etc. Total logs 200200. Solve for nn in 200=n2[2(20)+(n1)(1)]200 = \frac{n}{2} [2(20) + (n - 1)(-1)].
  • Potato Race: Distance for first potato is 2×52 \times 5, second is 2×(5+3)2 \times (5 + 3). The sequence is 10,16,22,...10, 16, 22, . . . for 1010 terms.

Summary of Key Points

  • Definition: Each term is obtained by adding dd to the previous one.
  • General Term: an=a+(n1)da_n = a + (n - 1)d
  • Sum of n Terms: S=n2[2a+(n1)d]S = \frac{n}{2} [2a + (n - 1)d] or S=n2(a+l)S = \frac{n}{2} (a + l)
  • Arithmetic Mean: If a,b,ca, b, c are in AP, then b=a+c2b = \frac{a + c}{2}, where bb is the arithmetic mean of aa and cc.