ML Aggarwal Class 11 Maths Solutions Chapter 09 Binomial Theorem

Sequences and Series

Introduction

Sequences are useful in many areas of human life and work. When objects are put in a clear, organized order where you can identify the first, second, third item, and so on, the collection is arranged as a sequence. Here are some real examples:

• The money saved in a fixed bank deposit grows over time in a sequence.
• Item values decrease over time in a sequence pattern.
• Bacteria populations at different points in time form a sequence.
• Think about how many ancestors a person has - parents, grandparents, great-grandparents, and so on - over 10 generations. The counts for each generation are 2, 4, 8, … , 1024, which make up a sequence.

When sequences follow specific rules or patterns, they are called progressions. You learned about arithmetic progression (A.P.) in your earlier class. This chapter builds on that knowledge and also covers arithmetic mean (A.M.), geometric progression (G.P.), geometric mean (G.M.), the relationship between A.M. and G.M., arithmetico-geometric series, and the sum of n terms for special series Σn, Σn², Σn³ and so on.

9.1 Sequence

A sequence is a group of numbers written in a definite order following some definite rule or rules. Each number in this group is called a term of the sequence. A sequence can be either finite (limited number of terms) or infinite (endless). The terms are usually written as a₁, a₂, a₃, … or T₁, T₂, T₃, …. The number below (subscript) shows which position the term holds. The term at position n, written as aₙ, is called the general term of the sequence. A finite sequence looks like a₁, a₂, a₃, …, aₙ while an infinite sequence is a₁, a₂, a₃, … extending to infinity. When all terms are real numbers, we have a real sequence; when all terms are complex numbers, we have a complex sequence.

Examples of sequences:

• (i) 3, 5, 7, 9, …, 21
• (ii) 8, 5, 2, -1, -4, …
• (iii) 2, 6, 18, 54, …, 1458
• (iv) 1, -\( \frac{1}{2} \), \( \frac{1}{4} \), -\( \frac{1}{8} \), …
• (v) 1, 4, 9, 16, …
• (vi) 2, 3, 5, 7, 11, 13, …
• (vii) 1, 1, 2, 3, 5, 8, 13, …

We notice the following patterns:

• (i) Each term is made by adding 2 to the previous term.
• (ii) Each term is made by subtracting 3 from the one before it.
• (iii) Each term is made by multiplying the previous term by 3.
• (iv) Each term is made by multiplying the previous term by -\( \frac{1}{2} \).
• (v) Each term is the square of the next counting number.
• (vi) This is the sequence of prime numbers.
• (vii) Each term after the second one is the sum of the two terms before it.

Also note that sequences (i) and (iii) are finite sequences, while the others are infinite sequences. To describe a sequence, we don't always need an exact formula for the nth term. Up to now, mathematicians have not found a formula for the nth prime number. In these sequences we can see that:

(i) aₙ = aₙ₋₁ + 2
(ii) aₙ = aₙ₋₁ - 3
(iii) aₙ = 3aₙ₋₁
(iv) aₙ = -\( \frac{1}{2} \)aₙ₋₁
(v) aₙ = n²
(vii) aₙ = aₙ₋₁ + aₙ₋₂ (n > 2)
and in (vi) we can say aₙ = nth prime number.

When the terms of a sequence can be described with an exact formula, the sequence is called a progression. Sequences (i) through (v) are all progressions, but sequence (vi) is not. Sequence (vii) - that is, 1, 1, 2, 3, 5, 8, 13, … - is also a progression. It is known as the Fibonacci sequence.

9.2 Series

When the terms of a sequence are connected using plus signs, we create a series. So if a₁, a₂, a₃, … is a given sequence, then the expression a₁ + a₂ + a₃ + … is the series linked with that sequence. The series is finite or infinite depending on whether the sequence is finite or infinite.

From sequences (i) through (v) above, we get these series:

• (i) 3 + 5 + 7 + … + 21
• (ii) 8 + 5 + 2 + (-1) + (-4) + …
• (iii) 2 + 6 + 18 + 54 + … + 1458
• (iv) 1 + (-\( \frac{1}{2} \)) + \( \frac{1}{4} \) + (-\( \frac{1}{8} \)) + …
• (v) 1 + 4 + 9 + 16 + …

If aₙ represents the general term of a sequence, then a₁ + a₂ + a₃ + … + aₙ is a series with n terms. In the series a₁ + a₂ + a₃ + … + aₖ + …, the sum of the first n terms is shown as Sₙ. This means:

Sₙ = a₁ + a₂ + a₃ + … + aₙ = \( \sum_{k=1}^{n} a_k \)

When Sₙ represents the sum of n terms, we have:

Sₙ - Sₙ₋₁ = (a₁ + a₂ + … + aₙ) - (a₁ + a₂ + … + aₙ₋₁) = aₙ

Therefore, aₙ = Sₙ - Sₙ₋₁

Remark: The word 'series' means the sum that is being added, not the final number itself. For example, 1 + 3 + 5 + 7 + 9 is a finite series with five terms. When we talk about the 'sum of a series,' we mean the actual number we get after adding all the terms, so the sum of the series above is 25.

Illustrative Examples

Example 1. Find the next term of the sequence
(i) 2, 4, 6, 8
(ii) 2, 1, \( \frac{1}{2} \), \( \frac{1}{4} \)
(iii) 2, 8, 32, 128
(iv) -1, -3, -5, -7
(v) 1, 8, 27, 64
Answer: (i) In this sequence, each number is found by adding 2 to the number before it. So the next term = 8 + 2 = 10.

(ii) Each number is made by taking half of the number before it. So the next term = \( \frac{1}{4} \) × \( \frac{1}{2} \) = \( \frac{1}{8} \).

(iii) Each number is made by multiplying the number before it by 4. So the next term = 128 × 4 = 512.

(iv) Each number is made by subtracting 2 from the number before it. So the next term = -7 - 2 = -9.

(v) The terms are the cubes of counting numbers - 1³, 2³, 3³, 4³. So the next term is 5³, which equals 125.
In simple words: Look for a pattern - does each term add, subtract, multiply, or divide by the same number? Once you find the pattern, use it to get the next term.

Exam Tip: Always check differences between consecutive terms first - if they're the same, the pattern is addition or subtraction. If ratios between consecutive terms are the same, the pattern is multiplication or division.

Example 2. Write the first four terms of the sequence defined by
(i) aₙ = 4n² + 3
(ii) aₙ = nth prime number
Answer: (i) We are given aₙ = 4n² + 3. Putting n = 1, 2, 3, 4, we get:
a₁ = 4 × 1² + 3 = 7
a₂ = 4 × 2² + 3 = 19
a₃ = 4 × 3² + 3 = 39
a₄ = 4 × 4² + 3 = 67
So the first four terms are 7, 19, 39, 67.

(ii) The first four prime numbers are 2, 3, 5, 7. So the first four terms of the sequence are 2, 3, 5, 7.
In simple words: To find terms using a formula, replace n with 1, then 2, then 3, then 4 and work out each answer.

Exam Tip: Be careful with the order of operations - calculate powers before multiplying, and add at the end.

Example 3. Find the 20th term of the sequence defined by aₙ = \( \frac{n(n-2)}{n+3} \)
Answer: Given aₙ = \( \frac{n(n-2)}{n+3} \), putting n = 20, we get:
a₂₀ = \( \frac{20(20-2)}{20+3} \) = \( \frac{20 × 18}{23} \) = \( \frac{360}{23} \)
In simple words: Replace the letter n with the number you want (in this case 20) and calculate step by step.

Exam Tip: Write out all steps clearly so the marker can follow your working and give you credit for correct method even if the final number has a small mistake.

Example 4. Find the 13th and 14th terms of the sequence defined by
aₙ = \( \begin{cases} n^2, & \text{when } n \text{ is even} \\ n^2 + 1, & \text{when } n \text{ is odd} \end{cases} \)
Answer: Since 13 is odd, a₁₃ = 13² + 1 = 169 + 1 = 170. Since 14 is even, a₁₄ = 14² = 196.
In simple words: Check if your n value is even or odd, then use the matching formula to find your answer.

Exam Tip: Read carefully - the formula changes based on whether n is even or odd. Always check which rule applies to your value of n.

Example 5. Find the first five terms of the sequence given by a₁ = 2, a₂ = 3 + a₁ and aₙ = 2aₙ₋₁ + 5 for n > 2. Also write the corresponding series.
Answer: Here a₁ = 2 and a₂ = 3 + a₁ = 3 + 2 = 5. Given aₙ = 2aₙ₋₁ + 5 for n > 2, putting n = 3, 4, 5, we get:
a₃ = 2a₂ + 5 = 2 × 5 + 5 = 15
a₄ = 2a₃ + 5 = 2 × 15 + 5 = 35
a₅ = 2a₄ + 5 = 2 × 35 + 5 = 75
So the first five terms are 2, 5, 15, 35, 75. The matching series is 2 + 5 + 15 + 35 + 75 + … .
In simple words: Use one term to find the next using the given rule. Keep going until you have all the terms you need.

Exam Tip: Double-check each calculation before moving to the next term - one error early on will make all the later terms wrong.

Example 6. The Fibonacci sequence is defined by a₁ = a₂ = 1, aₙ = aₙ₋₁ + aₙ₋₂ for n > 2. Find \( \frac{a_{n+1}}{a_n} \) for n = 1, 2, 3, 4, 5.
Answer: Given a₁ = a₂ = 1 and aₙ = aₙ₋₁ + aₙ₋₂ for n > 2, putting n = 3, 4, 5, 6, we get:
a₃ = a₂ + a₁ = 1 + 1 = 2
a₄ = a₃ + a₂ = 2 + 1 = 3
a₅ = a₄ + a₃ = 3 + 2 = 5
a₆ = a₅ + a₄ = 5 + 3 = 8

Putting n = 1, 2, 3, 4, 5 in \( \frac{a_{n+1}}{a_n} \), we get:
\( \frac{a_2}{a_1} \), \( \frac{a_3}{a_2} \), \( \frac{a_4}{a_3} \), \( \frac{a_5}{a_4} \), \( \frac{a_6}{a_5} \)

That is, \( \frac{1}{1} \), \( \frac{2}{1} \), \( \frac{3}{2} \), \( \frac{5}{3} \), \( \frac{8}{5} \), which gives us 1, 2, \( \frac{3}{2} \), \( \frac{5}{3} \), \( \frac{8}{5} \).
In simple words: First build up the Fibonacci terms using the addition rule. Then divide each term by the one before it to get your answers.

Exam Tip: Keep fractions in simplest form if possible, and watch for patterns - in the Fibonacci sequence, these ratios get closer to the golden ratio.

Example 7. (i) Find the first 3 terms of the series Σ(-1)ⁿ⁺¹ 3⁻ⁿ.
(ii) The sum of n terms of a series is n² + 3n for all values of n. Find the first 3 terms of the series. Also find its 10th term.
Answer: (i) The nth term of the given series is aₙ = (-1)ⁿ⁺¹ 3⁻ⁿ.
First term = a₁ = (-1)¹⁺¹ 3⁻¹ = \( \frac{1}{3} \)
Second term = a₂ = (-1)²⁺¹ 3⁻² = -\( \frac{1}{9} \)
Third term = a₃ = (-1)³⁺¹ 3⁻³ = \( \frac{1}{27} \)
So the first 3 terms are \( \frac{1}{3} \), -\( \frac{1}{9} \), \( \frac{1}{27} \).

(ii) Given Sₙ = n² + 3n, so Sₙ₋₁ = (n - 1)² + 3(n - 1) = n² + n - 2. Therefore aₙ = Sₙ - Sₙ₋₁ = n² + 3n - (n² + n - 2) = 2n + 2. Putting n = 1, 2, 3, 10, we get:
a₁ = 2 × 1 + 2 = 4
a₂ = 2 × 2 + 2 = 6
a₃ = 2 × 3 + 2 = 8
a₁₀ = 2 × 10 + 2 = 22
So the first three terms are 4, 6, 8 and the 10th term is 22.
In simple words: Follow the given rule carefully - use the sum of n terms and the sum of (n-1) terms to find any single term you need.

Exam Tip: When working with sums, always use the method aₙ = Sₙ - Sₙ₋₁ rather than trying to figure out a pattern from just a few terms.

Example 8. If for a sequence, Sₙ = 2(3ⁿ - 1), find its first four terms.
Answer: Given Sₙ = 2(3ⁿ - 1), so Sₙ₋₁ = 2(3ⁿ⁻¹ - 1). Therefore aₙ = Sₙ - Sₙ₋₁ = 2(3ⁿ - 1) - 2(3ⁿ⁻¹ - 1) = 2(3ⁿ - 3ⁿ⁻¹) = 2 × 3ⁿ⁻¹(3 - 1) = 4 × 3ⁿ⁻¹. Putting n = 1, 2, 3, 4, we get:
a₁ = 4 × 3⁰ = 4
a₂ = 4 × 3¹ = 12
a₃ = 4 × 3² = 36
a₄ = 4 × 3³ = 108
So the first four terms are 4, 12, 36, 108.
In simple words: Use the sum formula to create a formula for any term, then plug in 1, 2, 3, 4 to get your answers.

Exam Tip: Simplify your general term formula first - it will make substituting values much easier and reduce calculation errors.

Example 9. (i) Write \( \sum_{k=1}^{n} (k^2 + 1) \) in expanded form.
(ii) Write the series \( \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ... + \frac{n}{n+1} \) in sigma notation.
Answer: (i) Putting k = 1, 2, 3, 4, ..., n in (k² + 1), we get 2, 5, 10, 17, ..., n² + 1. So \( \sum_{k=1}^{n} (k^2 + 1) \) = 2 + 5 + 10 + 17 + ... + (n² + 1).

(ii) We can see that the kth term of the series is \( \frac{k}{k+1} \). So the given series can be written as \( \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ... + \frac{n}{n+1} \) = \( \sum_{k=1}^{n} \frac{k}{k+1} \).
In simple words: To expand sigma notation, substitute 1, 2, 3, and so on. To write in sigma notation, find the general term pattern and use that.

Exam Tip: Make sure your pattern for the kth term is correct by checking it against the first two or three terms shown in the series.

Exercise 9.1

Very short answer type questions (1 to 7):

Question 1. Give an example of a sequence which is not a progression.
Answer: The sequence of prime numbers: 2, 3, 5, 7, 11, 13, 17, … is not a progression because we cannot find a single rule or formula that gives us all the prime numbers in order.
In simple words: A sequence where no pattern or formula describes all the terms is not a progression.

Exam Tip: Remember that a progression must have terms that follow a definite mathematical rule that you can write as a formula.

Question 2. Which term of the sequence given by aₙ = n² + 2n + 1, n ∈ N, is 361?
Answer: We need to find n such that aₙ = 361. So n² + 2n + 1 = 361, which gives us (n + 1)² = 361, meaning n + 1 = 19 (since n is a natural number), so n = 18. Therefore, the 18th term is 361.
In simple words: Set the formula equal to 361 and solve for n by taking the square root.

Exam Tip: Notice that n² + 2n + 1 = (n + 1)² - recognizing perfect square patterns saves time in these questions.

Question 3. Write the first three terms of the sequence whose nth term is given by aₙ = (-1)ⁿ⁻¹ 5⁻ⁿ.
Answer: Putting n = 1, 2, 3:
a₁ = (-1)¹⁻¹ 5⁻¹ = 1 × \( \frac{1}{5} \) = \( \frac{1}{5} \)
a₂ = (-1)²⁻¹ 5⁻² = -1 × \( \frac{1}{25} \) = -\( \frac{1}{25} \)
a₃ = (-1)³⁻¹ 5⁻³ = 1 × \( \frac{1}{125} \) = \( \frac{1}{125} \)
So the first three terms are \( \frac{1}{5} \), -\( \frac{1}{25} \), \( \frac{1}{125} \).
In simple words: Replace n with 1, 2, 3 in the formula and calculate carefully, noting where the negative signs appear.

Exam Tip: Pay close attention to the power of -1 - if the exponent is even, you get +1; if odd, you get -1.

Question 4. If a sequence is given by a₁ = 2, a₂ = 3 + a₁ and aₙ = 2aₙ₋₁ - 1 for n > 2. Then write the corresponding series upto 4 terms.
Answer: Here a₁ = 2 and a₂ = 3 + 2 = 5. Using aₙ = 2aₙ₋₁ - 1 for n > 2:
a₃ = 2a₂ - 1 = 2 × 5 - 1 = 9
a₄ = 2a₃ - 1 = 2 × 9 - 1 = 17
The first four terms are 2, 5, 9, 17. The matching series is 2 + 5 + 9 + 17.
In simple words: Find each term using the rule given, then add them together with plus signs.

Exam Tip: Write out all steps - the working shows you understand the method, not just the final answer.

Question 5. Write the next term of each of the following sequences:
(i) 3, 1, -1, -3, …
(ii) 5, -\( \frac{5}{2} \), \( \frac{5}{4} \), -\( \frac{5}{8} \), …
Answer: (i) Each term is found by subtracting 2. The next term = -3 - 2 = -5.

(ii) Each term is found by multiplying by -\( \frac{1}{2} \). The next term = -\( \frac{5}{8} \) × (-\( \frac{1}{2} \)) = \( \frac{5}{16} \).
In simple words: Find the pattern - look for a common difference or common ratio - then apply it to get the next term.

Exam Tip: Always verify your pattern by checking it works for at least the first three terms before finding the next one.

Question 6. Write the next term of each of the following sequences:
(i) 0, 2, 6, 12, 20, …
(ii) 6, 9, 16, 27, 42, …
(iii) 1, 5, 14, 30, 55, …

Hint. (iii) 5 = 1 + 2², 14 = 5 + 3², 30 = 14 + 4², …
Answer: (i) The differences between terms are: 2, 4, 6, 8, so the next difference is 10. The next term = 20 + 10 = 30.

(ii) The differences are: 3, 7, 11, 15, so the next difference is 19. The next term = 42 + 19 = 61.

(iii) Using the hint: each term adds the next perfect square to the previous term. So 55 + 5² = 55 + 25 = 80. But wait - let me recalculate. Actually 30 + 5² = 30 + 25 = 55 (this checks). So the next term = 55 + 6² = 55 + 36 = 91.
In simple words: When the differences don't match, look at differences of differences, or check if there's a pattern using squares or other powers.

Exam Tip: The hint is telling you the pattern - look for a second or higher-order pattern when the regular differences don't match.

Question 7. Write the eleventh term of the following sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, …
Answer: This is the Fibonacci sequence where each term after the second is the sum of the two before it. Continuing the sequence: 34 + 21 = 55, then 55 + 34 = 89. The eleventh term is 89.
In simple words: Add the last two numbers to get the next one, then keep doing this until you reach the term you want.

Exam Tip: For the Fibonacci sequence, be careful counting - the first two terms are both 1, so counting from there helps you avoid mistakes.

Question 8. Write first 5 terms of the following sequences whose nth terms are given by:
(i) aₙ = 2n + 5
(ii) aₙ = \( \frac{2n-3}{6} \)
(iii) aₙ = n(n - 1)
(iv) aₙ = n(n + 2)
(v) aₙ = 2ⁿ
(vi) aₙ = \( \frac{n}{n+1} \)
(vii) aₙ = (-1)ⁿ⁻¹ 5ⁿ⁺¹
(viii) aₙ = \( \frac{n(n^2+5)}{4} \)
(ix) aₙ = \( \frac{n^2+1}{2n-3} \)
Answer: (i) Putting n = 1, 2, 3, 4, 5: a₁ = 7, a₂ = 9, a₃ = 11, a₄ = 13, a₅ = 15. Terms: 7, 9, 11, 13, 15

(ii) a₁ = \( \frac{-1}{6} \), a₂ = \( \frac{1}{6} \), a₃ = \( \frac{1}{2} \), a₄ = \( \frac{5}{6} \), a₅ = \( \frac{7}{6} \)

(iii) a₁ = 0, a₂ = 2, a₃ = 6, a₄ = 12, a₅ = 20

(iv) a₁ = 3, a₂ = 8, a₃ = 15, a₄ = 24, a₅ = 35

(v) a₁ = 2, a₂ = 4, a₃ = 8, a₄ = 16, a₅ = 32

(vi) a₁ = \( \frac{1}{2} \), a₂ = \( \frac{2}{3} \), a₃ = \( \frac{3}{4} \), a₄ = \( \frac{4}{5} \), a₅ = \( \frac{5}{6} \)

(vii) a₁ = 25, a₂ = -125, a₃ = 625, a₄ = -3125, a₅ = 15625

(viii) a₁ = \( \frac{3}{2} \), a₂ = \( \frac{9}{2} \), a₃ = \( \frac{21}{2} \), a₄ = \( \frac{75}{2} \), a₅ = 21

(ix) a₁ = -2, a₂ = 5, a₃ = \( \frac{10}{3} \), a₄ = \( \frac{17}{5} \), a₅ = \( \frac{26}{7} \)
In simple words: For each formula, substitute n = 1, then n = 2, and so on up to n = 5. Calculate each time and list the five results.

Exam Tip: Show your working for at least the first two terms - this proves you understand the method even if you make a small arithmetic mistake later.

Question 9. Find the indicated term(s) in each of the following sequences whose nth terms are:
(i) aₙ = 4n - 3; find a₁₇ and a₂₄
(ii) aₙ = \( \frac{n^2}{2^n} \); find a₅ and a₇
(iii) aₙ = (-1)ⁿ⁻¹ n³; find a₉
(iv) aₙ = (n - 1)(2 - n)(3 + n); find a₁, a₂, and a₂₀
Answer: (i) a₁₇ = 4(17) - 3 = 68 - 3 = 65; a₂₄ = 4(24) - 3 = 96 - 3 = 93

(ii) a₅ = \( \frac{25}{32} \); a₇ = \( \frac{49}{128} \)

(iii) a₉ = (-1)⁸ × 729 = 729

(iv) a₁ = (0)(1)(4) = 0; a₂ = (1)(0)(5) = 0; a₂₀ = (19)(-18)(23) = -7866
In simple words: Plug each term number into the formula and work out the answer carefully, one step at a time.

Exam Tip: When there are multiple parts, complete part (i) fully before moving to (ii) - this keeps your work organized and reduces careless mistakes.

Question 10. Find the 18th and 25th terms of the sequence defined by
Tₙ = \( \begin{cases} n(n+2), & \text{if } n \text{ is even} \\ \frac{4n}{n^2+1}, & \text{if } n \text{ is odd} \end{cases} \)
Answer: Since 18 is even, T₁₈ = 18(18 + 2) = 18 × 20 = 360. Since 25 is odd, T₂₅ = \( \frac{4 × 25}{25^2 + 1} \) = \( \frac{100}{625 + 1} \) = \( \frac{100}{626} \) = \( \frac{50}{313} \).
In simple words: First check if your term number is even or odd, then use the matching formula.

Exam Tip: Double-check your even/odd decision - using the wrong formula will give a completely incorrect answer.

Question 11. Find the first five terms of each of the following sequences and obtain the corresponding series:
(i) a₁ = 1, aₙ = aₙ₋₁ + 2, n ≥ 2
(ii) a₁ = 3, aₙ = 3aₙ₋₁ + 2, for all n > 1
(iii) a₁ = -1, aₙ = \( \frac{a_{n-1}}{n} \) for n ≥ 2
(iv) a₁ = a₂ = 2, aₙ = aₙ₋₁ - 1 for n > 2
Answer: (i) a₁ = 1; a₂ = 1 + 2 = 3; a₃ = 3 + 2 = 5; a₄ = 5 + 2 = 7; a₅ = 7 + 2 = 9. Terms: 1, 3, 5, 7, 9; Series: 1 + 3 + 5 + 7 + 9 + …

(ii) a₁ = 3; a₂ = 3(3) + 2 = 11; a₃ = 3(11) + 2 = 35; a₄ = 3(35) + 2 = 107; a₅ = 3(107) + 2 = 323. Terms: 3, 11, 35, 107, 323; Series: 3 + 11 + 35 + 107 + 323 + …

(iii) a₁ = -1; a₂ = \( \frac{-1}{2} \); a₃ = \( \frac{-1/2}{3} \) = -\( \frac{1}{6} \); a₄ = \( \frac{-1/6}{4} \) = -\( \frac{1}{24} \); a₅ = \( \frac{-1/24}{5} \) = -\( \frac{1}{120} \). Terms: -1, -\( \frac{1}{2} \), -\( \frac{1}{6} \), -\( \frac{1}{24} \), -\( \frac{1}{120} \); Series: (-1) + (-\( \frac{1}{2} \)) + (-\( \frac{1}{6} \)) + (-\( \frac{1}{24} \)) + (-\( \frac{1}{120} \)) + …

(iv) a₁ = 2, a₂ = 2; a₃ = 2 - 1 = 1; a₄ = 1 - 1 = 0; a₅ = 0 - 1 = -1. Terms: 2, 2, 1, 0, -1; Series: 2 + 2 + 1 + 0 + (-1) + …
In simple words: Use the rule to find each term from the one before. Once you have all five terms, add them together with plus signs to make the series.

Exam Tip: A common mistake is forgetting to find ALL five terms before writing the series - always count to make sure you have exactly five.

Question 12. If the sum of n terms of a sequence is given by Sₙ = 2n² + 3n for all n ∈ N, find the first 4 terms. Also find its 20th term.
Answer: To find the first four terms, we use aₙ = Sₙ - Sₙ₋₁. First, a₁ = S₁ = 2(1)² + 3(1) = 5. For n ≥ 2, Sₙ₋₁ = 2(n - 1)² + 3(n - 1) = 2n² - 4n + 2 + 3n - 3 = 2n² - n - 1. So aₙ = 2n² + 3n - (2n² - n - 1) = 4n + 1. Therefore a₂ = 9, a₃ = 13, a₄ = 17. The first four terms are 5, 9, 13, 17. The 20th term is a₂₀ = 4(20) + 1 = 81.
In simple words: Use Sₙ - Sₙ₋₁ to find the general term first. This makes finding any term quick and easy.

Exam Tip: Finding the general term formula is worth doing even if you only need a few specific terms - it saves checking calculations.

Question 13. First term of a sequence is 1 and the (n + 1)th term is obtained by adding (n + 1) to the nth term for all natural numbers n. Find the sixth term of the sequence.

Hint. aₙ₊₁ = aₙ + (n + 1) for all natural numbers n.
Answer: Given a₁ = 1 and aₙ₊₁ = aₙ + (n + 1). Using this rule:
a₂ = a₁ + 2 = 1 + 2 = 3
a₃ = a₂ + 3 = 3 + 3 = 6
a₄ = a₃ + 4 = 6 + 4 = 10
a₅ = a₄ + 5 = 10 + 5 = 15
a₆ = a₅ + 6 = 15 + 6 = 21
The sixth term is 21.
In simple words: Start with the first term and keep adding: add 2, then add 3, then add 4, and so on until you reach the term you want.

Exam Tip: This is a triangular number sequence - if you notice the pattern, you can write a₆ = 1 + 2 + 3 + 4 + 5 + 6 = 21 directly.

9.3 Arithmetic Progression (A.P.)

An arithmetic progression is a sequence (either finite or infinite) where the difference between any term and the term before it stays the same. This fixed difference is usually written as d and is called the common difference. So a₁, a₂, …, aₙ or a₁, a₂, a₃, … is an A.P. if aₖ₊₁ - aₖ = d, a constant (the same for all k) for k = 1, 2, …, n - 1 or k = 1, 2, 3, … as needed. This means in an A.P., aₙ₊₁ = aₙ + d - any term (except the first) is found by adding the fixed number d to the previous term. If the sequence a₁, a₂, a₃, …, aₙ is an A.P., then the series a₁ + a₂ + a₃ + … + aₙ is called an arithmetic series.

General term of an A.P.

When a is the first term and d is the common difference of an A.P., the A.P. is a, a + d, a + 2d, … and its nth term = a + (n - 1)d. So the general term is aₙ = a + (n - 1)d.

Last term of an A.P.

If the last term of an A.P. with n terms is written as l, then l = a + (n - 1)d.

The nth term from the end of a finite A.P.

(i) If a, a + d, a + 2d, … is a finite A.P. with m terms, then the nth term from the end = (m - n + 1)th term from the start = a + (m - n + 1 - 1)d = a + (m - n)d.
(ii) If a, a + d, a + 2d, ...... is a finite A.P. with last term l, then the nth term from end = l + (n - 1)(-d) = l - (n - 1)d. Here's why: when we look at the A.P. starting from the end and going backwards, it forms an A.P. with common difference -d and first term l. So the nth term from the end of the original A.P. = l + (n - 1)(-d) = l - (n - 1)d.

Sum of n terms of an A.P.

When a is the first term, d is the common difference, and l is the last term of an A.P., and Sₙ represents the sum of its first n terms, then:

Sₙ = \( \frac{n}{2} \)(2a + (n - 1)d) or Sₙ = \( \frac{n}{2} \)(a + l)

Remarks:

• 1. When the sum of n terms of an A.P. is shown as Sₙ, its common difference = S₂ - 2S₁.
• 2. Three numbers a, b, c are in A.P. if and only if b - a = c - b, or if and only if 2b = a + c.
• 3. Some problems involve 3, 4, or 5 numbers in A.P. When you know the sum of these numbers:
  - (i) Three numbers are written as a - d, a, a + d
  - (ii) Four numbers are written as a - 3d, a - d, a + d, a + 3d
  - (iii) Five numbers are written as a - 2d, a - d, a, a + d, a + 2d
  This approach makes finding a and d much easier.
• 4. In an A.P., the sum of terms that are the same distance from the start and the end is always the same and equals the sum of the first and last terms.

9.3.1 If the terms of an arithmetic progression (A.P.) are increased, decreased, multiplied, or divided by the same non-zero constant, they remain in arithmetic progression.

Proof. Consider the A.P. a, a + d, a + 2d, a + 3d, … (1)

(i) When each term of (1) is increased by a constant k, we get the sequence a + k, a + d + k, a + 2d + k, a + 3d + k, ... which can be rewritten as a + k, (a + k) + d, (a + k) + 2d, (a + k) + 3d, … This is clearly an A.P. with first term a + k and common difference d.

(ii) When each term of (1) is decreased by a constant k, we get the sequence a - k, a + d - k, a + 2d - k, … or a - k, (a - k) + d, (a - k) + 2d, (a - k) + 3d, … This is clearly an A.P. with first term a - k and common difference d.

(iii) When each term of (1) is multiplied by a constant k, we get the sequence ak, (a + d)k, (a + 2d)k, … or ak, ak + dk, ak + 2dk, ak + 3dk … This is clearly an A.P. with first term ak and common difference dk.

(iv) When each term of (1) is divided by k where k ≠ 0, we get the sequence \( \frac{a}{k} \), \( \frac{a + d}{k} \), \( \frac{a + 2d}{k} \), \( \frac{a + 3d}{k} \), … or \( \frac{a}{k} \), \( \frac{a}{k} + \frac{d}{k} \), \( \frac{a}{k} + 2\frac{d}{k} \), \( \frac{a}{k} + 3\frac{d}{k} \), … This is clearly an A.P. with first term \( \frac{a}{k} \) and common difference \( \frac{d}{k} \).

Illustrative Examples (Arithmetic Progression)

Example 1. The fourth term of A.P. is equal to 3 times its first term and seventh term exceeds twice the third term by 1. Find the first term and the common difference.
Answer: Let a be the first term and d be the common difference. From the first condition: a₄ = 3a₁ gives us a + 3d = 3a, so 3d = 2a ... (i). From the second condition: a₇ = 2a₃ + 1 gives us a + 6d = 2(a + 2d) + 1, so a + 6d = 2a + 4d + 1, therefore 2d = a + 1 ... (ii). Solving equations (i) and (ii): From (ii), a = 2d - 1. Putting in (i): 3d = 2(2d - 1) = 4d - 2, so d = 2. Then a = 2(2) - 1 = 3. The first term is 3 and common difference is 2.
In simple words: Turn each condition into an equation using the A.P. formula aₙ = a + (n-1)d, then solve the two equations together.

Exam Tip: Always write out the formula for each term mentioned and substitute the given information - this prevents mixing up different terms.

Example 2. In a sequence, the nth term is aₙ = 2n² + 3. Show that it is not an A.P.
Answer: Given aₙ = 2n² + 3, so aₙ₊₁ = 2(n + 1)² + 3. Working out aₙ₊₁ - aₙ = (2(n + 1)² + 3) - (2n² + 3) = 2(n² + 2n + 1) - 2n² = 2n² + 4n + 2 - 2n² = 4n + 2. This depends on n and is not constant. Therefore, the sequence is not an A.P.
In simple words: Find aₙ₊₁ - aₙ. If the answer has n in it (not just a number), then it's not an A.P.

Exam Tip: To prove something is NOT an A.P., you only need to show that the difference between consecutive terms is not constant.

Example 3. In a sequence, the sum of first n terms is Sₙ = nP + \( \frac{1}{2} \)n(n - 1)Q where P, Q are constants. Show that the sequence is an A.P. Find its first term, common difference, and 100th term.
Answer: Given Sₙ = nP + \( \frac{1}{2} \)n(n - 1)Q, so Sₙ₋₁ = (n - 1)P + \( \frac{1}{2} \)(n - 1)(n - 2)Q. Therefore aₙ = Sₙ - Sₙ₋₁ = (n - (n - 1))P + \( \frac{1}{2} \)[n(n - 1) - (n - 1)(n - 2)]Q = P + \( \frac{1}{2} \)(n - 1)(n - (n - 2))Q = P + \( \frac{1}{2} \)(n - 1) × 2Q = P + (n - 1)Q. So aₙ₊₁ = P + nQ. Thus aₙ₊₁ - aₙ = (P + nQ) - (P + (n - 1)Q) = Q, which is constant. So the sequence is an A.P. with common difference Q. Also, a₁ = P and a₁₀₀ = P + 99Q.
In simple words: Use aₙ = Sₙ - Sₙ₋₁ to find any term. If the difference between consecutive terms is constant, it's an A.P.

Exam Tip: This type of question teaches that an A.P. is completely defined once you know its first term and common difference - everything else follows.

Example 4. Which term of the sequence 25, 24\( \frac{1}{4} \), 23\( \frac{1}{2} \), 22\( \frac{3}{4} \), … is the first negative term?
Answer: The given sequence is an A.P. with common difference d = -\( \frac{3}{4} \) and first term a = 25. Let the nth term be the first negative term. Then aₙ < 0, so 25 + (n - 1)(-\( \frac{3}{4} \)) < 0. This gives us \( \frac{103}{4} \) - \( \frac{3n}{4} \) < 0, so 103 - 3n < 0, meaning 103 < 3n, thus 3n > 103, so n > \( \frac{103}{3} \) = 34\( \frac{1}{3} \). Since 35 is the smallest natural number greater than 34\( \frac{1}{3} \), we have n = 35. So the 35th term is the first negative term.
In simple words: Use the A.P. formula aₙ = a + (n-1)d and solve the inequality aₙ < 0 to find which term is first negative.

Exam Tip: Always be careful with inequalities - remember that you want the SMALLEST n that satisfies the condition.

Example 5. Which term of the sequence 12 + 8i, 10 + 7i, 8 + 6i, … is
(i) real
(ii) purely imaginary?
Answer: The given sequence is an A.P. with common difference d = -2 - i and first term a = 12 + 8i. The general term is aₙ = (12 + 8i) + (n - 1)(-2 - i) = (12 - 2n + 2) + (8 - n + 1)i = (14 - 2n) + (9 - n)i.

(i) For the nth term to be real, we need (14 - 2n) + (9 - n)i to be real, which means 9 - n = 0, so n = 9. The 9th term is real.

(ii) For the nth term to be purely imaginary, we need (14 - 2n) + (9 - n)i to be purely imaginary, which means 14 - 2n = 0, so n = 7. The 7th term is purely imaginary.
In simple words: A complex number is real when its imaginary part is zero. It's purely imaginary when its real part is zero.

Exam Tip: With complex number progressions, treat the real and imaginary parts separately.

Example 6. How many terms are identical in the two Arithmetic progressions 2, 4, 6, 8, … upto 100 terms and 3, 6, 9, 12, … upto 80 terms.
Answer: The 100th term of the A.P. 2, 4, 6, 8, ... is 2 + (100 - 1) × 2 = 200. The 80th term of the A.P. 3, 6, 9, 12, ... is 3 + (80 - 1) × 3 = 240. Let n terms be identical in both progressions. The identical terms form the sequence 6, 12, 18, ..., which is an A.P. with first term 6 and common difference 6. Its nth term = 6 + (n - 1) × 6 = 6n. Since this nth term cannot exceed 200, we have 6n ≤ 200, so n ≤ \( \frac{100}{3} \) ≈ 33.33, which means n = 33. So 33 terms are identical.
In simple words: Find what the identical terms would be - they must be multiples of the LCM of the two common differences. Count how many fit in both limits.

Exam Tip: The identical terms form their own A.P. - recognize this pattern and use it to count efficiently.

Note: When you end up with n ≤ m, then n = m if m is a whole number, and n = the largest whole number less than m if m is not a whole number.