Sequence and Series | NEB Class 11 Mathematics Notes

Chapter Overview

In mathematics, studying patterns leads to significant generalizations. A sequence is a succession of numbers arranged in a definite order according to a certain rule. When the terms of a sequence are added together, it forms a series. Sequences and series are not just abstract mathematical concepts—they appear everywhere in nature, finance, and engineering. Understanding how numbers progress and how to calculate their sums opens doors to solving real-world problems elegantly.

What this chapter covers

Sequences: Definition, Notation, and TypesSeries: Understanding the Sum of SequencesArithmetic Progression (A.P.): Finding terms and sumsGeometric Progression (G.P.): Exponential growth patternsHarmonic Progression (H.P.): Reciprocal relationshipsArithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.)Relationship between A.M., G.M., and H.M.Insertion of Means: A.M., G.M., and H.M. between two numbersSum of infinite Geometric SeriesApplication problems: Financial mathematics, population growth, bouncing ball

Learning Objectives

1

Understand the concept of sequence as a function from positive integers to real numbers.

2

Distinguish between sequences, series, and progressions, and identify which type a given pattern represents.

3

Find the $n^{th}$ term of an Arithmetic Progression using the formula $t_n = a + (n-1)d$.

4

Calculate the sum of $n$ terms of an A.P. using $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + l)$.

5

Find the $n^{th}$ term of a Geometric Progression using the formula $t_n = ar^{n-1}$.

6

Calculate the sum of $n$ terms of a G.P. using $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r \neq 1$.

7

Find the sum to infinity of a convergent Geometric Series using $S_{\infty} = \frac{a}{1-r}$ where $|r| < 1$.

8

Understand Harmonic Progressions and their relationship with Arithmetic Progressions.

9

Calculate and compare Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.) for two numbers.

10

Prove and apply the relationship $A.M. \times H.M. = G.M.^2$ and the inequality $A.M. > G.M. > H.M.$.

11

Insert Arithmetic, Geometric, and Harmonic means between two given numbers.

12

Apply sequence and series concepts to solve real-world problems in finance, physics, and biology.

Why Sequences and Series Matter: From Banking to Biology

Key Insight

Sequences and series are the mathematical language of patterns and growth. Every time a bank calculates compound interest, every time a virus spreads exponentially, every time an engineer designs stadium seating in rows—sequences and series are at work.

Real-World Scenario: Imagine you deposit ₹1,000 in a bank account that earns 5% interest annually. After 1 year, you have ₹1,050. After 2 years, ₹1,102.50. After 3 years, ₹1,157.625. Notice the amounts form a sequence: $1000, 1050, 1102.50, 1157.625, \ldots$ This is a Geometric Progression with first term $a = 1000$ and common ratio $r = 1.05$. If you want to know how much you'll have after 20 years, you use the formula $t_n = ar^{n-1}$. If you want to know the total interest earned, you sum the series!

The Big Picture: Understanding A.P., G.P., and their sums equips you to model and predict patterns in the real world. This chapter builds your mathematical intuition and problem-solving power.

Pay special attention to the condition $|r| < 1$ for infinite G.P. series convergence. This single concept appears in 90% of infinite series problems in NEB exams.

Key Definitions

Sequence

A sequence is a function whose domain is the set of positive integers $\mathbb{N} = \{1, 2, 3, \ldots\}$ and whose range is a set of real numbers. In other words, it is an ordered list of numbers where each number is called a term. The $n^{th}$ term is denoted $t_n$ or $a_n$. Example: $2, 4, 6, 8, 10, \ldots$ is a sequence where the first term $t_1 = 2$, the second term $t_2 = 4$, etc.

Series

A series is the expression formed by adding the terms of a sequence. If a sequence is $t_1, t_2, t_3, \ldots, t_n$, then the corresponding series is $t_1 + t_2 + t_3 + \cdots + t_n$. The sum of the first $n$ terms of a series is denoted $S_n$. Example: If the sequence is $2, 4, 6, 8$, the series is $2 + 4 + 6 + 8 = 20$, so $S_4 = 20$.

Progression

A progression is a special type of sequence that follows a specific, predictable pattern with a clear rule for its $n^{th}$ term. The three main progressions in this chapter are A.P. (Arithmetic Progression), G.P. (Geometric Progression), and H.P. (Harmonic Progression). Each has a defining characteristic that determines all its terms.

Arithmetic Progression (A.P.)

An Arithmetic Progression is a sequence where the difference between any term and its preceding term is a constant value called the common difference, denoted $d$. Formally, if $t_n$ and $t_{n+1}$ are consecutive terms, then $d = t_{n+1} - t_n$ for all $n \geq 1$. The general form is: $a, a+d, a+2d, a+3d, \ldots$ where $a$ is the first term. Example: $5, 8, 11, 14, \ldots$ is an A.P. with $a = 5$ and $d = 3$.

Geometric Progression (G.P.)

A Geometric Progression is a sequence where the ratio of any term to its preceding term is a constant value called the common ratio, denoted $r$. Formally, if $t_n$ and $t_{n+1}$ are consecutive terms, then $r = \frac{t_{n+1}}{t_n}$ for all $n \geq 1$ and $t_n \neq 0$. The general form is: $a, ar, ar^2, ar^3, \ldots$ where $a$ is the first term. Example: $2, 6, 18, 54, \ldots$ is a G.P. with $a = 2$ and $r = 3$.

Harmonic Progression (H.P.)

A Harmonic Progression is a sequence whose reciprocals form an Arithmetic Progression. If $t_1, t_2, t_3, \ldots$ form an H.P., then $\frac{1}{t_1}, \frac{1}{t_2}, \frac{1}{t_3}, \ldots$ form an A.P. Example: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ is an H.P. because the reciprocals $1, 2, 3, 4, \ldots$ form an A.P.

Arithmetic Mean (A.M.)

The Arithmetic Mean of two numbers $a$ and $b$ is the average of the two numbers, given by: $$A = \frac{a + b}{2}$$ Graphically, it is the middle value if $a, A, b$ are arranged in order. Example: The A.M. of 10 and 20 is $\frac{10 + 20}{2} = 15$.

Geometric Mean (G.M.)

The Geometric Mean of two positive numbers $a$ and $b$ is given by: $$G = \sqrt{ab}$$ If $a, G, b$ are in G.P., then $G$ is the geometric mean of $a$ and $b$. Example: The G.M. of 4 and 16 is $\sqrt{4 \times 16} = \sqrt{64} = 8$.

Harmonic Mean (H.M.)

The Harmonic Mean of two non-zero numbers $a$ and $b$ is given by: $$H = \frac{2ab}{a + b}$$ This is the reciprocal of the arithmetic mean of the reciprocals. Example: The H.M. of 4 and 12 is $\frac{2 \times 4 \times 12}{4 + 12} = \frac{96}{16} = 6$.

A.P. vs G.P.: The Complete Comparison

Arithmetic and Geometric progressions are the two most important progressions in NEB syllabus. Understanding their differences and when to use each formula is crucial for solving problems quickly and accurately.

Arithmetic Progression (A.P.)

$$Constant Difference: $d = t_{n+1} - t_n$$$

General Term: $t_n = a + (n-1)d$ where $a$ is first term, $d$ is common difference

Sum of n terms: $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term

Pattern: Growth is linear (constant increment). Example: $3, 5, 7, 9, 11$ (add 2 each time)

Characteristic: Differences between consecutive terms are always the same

When to use: Modeling uniform growth, stadium seating in rows, simple interest (but compound interest uses G.P.)

VS

Geometric Progression (G.P.)

$$Constant Ratio: $r = \frac{t_{n+1}}{t_n}$$$

General Term: $t_n = ar^{n-1}$ where $a$ is first term, $r$ is common ratio

Sum of n terms: $S_n = \frac{a(r^n - 1)}{r - 1}$ if $r \neq 1$, or $S_n = na$ if $r = 1$

Sum to infinity: $S_{\infty} = \frac{a}{1-r}$ only when $|r| < 1$ (convergent)

Pattern: Growth is exponential (constant multiplication). Example: $2, 4, 8, 16, 32$ (multiply by 2 each time)

Characteristic: Ratios between consecutive terms are always the same

When to use: Modeling exponential growth/decay, compound interest, viral spread, radioactive decay, bouncing ball problems

Quick Decision Rule: If you're adding a constant each time → Use A.P. formulas. If you're multiplying by a constant each time → Use G.P. formulas. If you can see repeated multiplication by the same factor, it's definitely G.P.

Complete Formula Sheet

Arithmetic Progression - General Term

$$t_n = a + (n-1)d$$

Find any term if you know the first term $a$, common difference $d$, and position $n$. This formula is used in 80% of A.P. problems.

Arithmetic Progression - Sum of n Terms (Form 1)

S_n = \frac{n}{2}[2a + (n-1)d]

Use this form when you know the first term and common difference. It expands to include all terms efficiently.

Arithmetic Progression - Sum of n Terms (Form 2)

S_n = \frac{n}{2}(a + l)

Use this form when you know the first term $a$ and the last term $l$. It's often simpler if the last term is already calculated.

Geometric Progression - General Term

t_n = ar^{n-1}

Find any term if you know the first term $a$, common ratio $r$, and position $n$. Remember: the exponent is $(n-1)$, not $n$.

Geometric Progression - Sum of n Terms

S_n = \frac{a(r^n - 1)}{r - 1} \text{ (if } r \neq 1\text{)}

Calculate the sum of the first $n$ terms. The formula works for any $r$ except 1. If $r = 1$, all terms equal $a$, so $S_n = na$.

Geometric Progression - Sum to Infinity

S_{\infty} = \frac{a}{1-r} \text{ (only when } |r| < 1\text{)}

This critical formula only works if the absolute value of $r$ is less than 1 (convergent series). If $|r| \geq 1$, the series diverges and $S_{\infty}$ does not exist as a finite value.

Harmonic Progression - General Term

t_n = \frac{1}{a + (n-1)d}

For H.P., the reciprocals form an A.P. with first term $a$ and common difference $d$. To find the $n^{th}$ term of H.P., take the reciprocal of the $n^{th}$ term of the corresponding A.P.

Arithmetic Mean (A.M.) of Two Numbers

A = \frac{a + b}{2}

The A.M. is simply the average. If three numbers are in A.P., the middle one is the A.M. of the other two.

Geometric Mean (G.M.) of Two Numbers

G = \sqrt{ab}

Both $a$ and $b$ must be positive. If three positive numbers are in G.P., the middle one is the G.M. of the other two. Relationship: $G^2 = ab$

Harmonic Mean (H.M.) of Two Numbers

H = \frac{2ab}{a + b}

The H.M. is often used in problems involving rates and averages of reciprocals. Relationship: $\frac{1}{H} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)$

Fundamental Relationship Between Means

A \times H = G^2

This beautiful relationship always holds for any two positive numbers $a$ and $b$. It shows that the arithmetic and harmonic means, when multiplied, give the square of the geometric mean.

Inequality Between Means (for unequal positive numbers)

$$A > G > H$$

For any two unequal positive numbers, the Arithmetic Mean is always greater than the Geometric Mean, which is always greater than the Harmonic Mean. Equality holds only when $a = b$.

Number of Arithmetic Means Between Two Numbers

d = \frac{b - a}{n + 1}

To insert $n$ arithmetic means between $a$ and $b$, calculate the common difference $d$. The complete A.P. becomes: $a, a+d, a+2d, \ldots, a+nd, b$ with $(n+2)$ total terms.

Number of Geometric Means Between Two Numbers

r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}

To insert $n$ geometric means between $a$ and $b$, calculate the common ratio $r$. Both $a$ and $b$ must have the same sign. The complete G.P. becomes: $a, ar, ar^2, \ldots, ar^n, b$ with $(n+2)$ total terms.

Problem-Solving Decision Tree

When you encounter a sequence or series problem, follow this decision tree to identify which formulas and techniques to use.

Is the sequence given or identified in the problem?

Yes

Look at the differences or ratios between consecutive terms to classify it.

No

Read carefully to understand what the problem is asking you to find.

Are the differences between consecutive terms constant?

Yes

It's an ARITHMETIC PROGRESSION (A.P.). Use formulas: $t_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$

No

Check if the ratios are constant.

Are the ratios between consecutive terms constant?

Yes

It's a GEOMETRIC PROGRESSION (G.P.). Use formulas: $t_n = ar^{n-1}$ and $S_n = \frac{a(r^n-1)}{r-1}$

No

Could the reciprocals form an A.P.? (Check for H.P.)

Do the reciprocals of the terms form an A.P.?

Yes

It's a HARMONIC PROGRESSION (H.P.). Convert to A.P., solve, then convert back.

No

It may be a special sequence requiring problem-specific analysis.

Does the problem ask for a single term, sum of n terms, or sum to infinity?

Yes

For single term: Use $t_n = a + (n-1)d$ (A.P.) or $t_n = ar^{n-1}$ (G.P.)

No

For sum of n terms: Use $S_n$ formulas. For infinite sum (G.P. only): Check if $|r| < 1$, then use $S_{\infty} = \frac{a}{1-r}$

Is the problem about means (A.M., G.M., H.M.) or inserting means?

Yes

Calculate using: $A = \frac{a+b}{2}$, $G = \sqrt{ab}$, $H = \frac{2ab}{a+b}$. For inserting: Use $d = \frac{b-a}{n+1}$ (A.M.) or $r = (b/a)^{1/(n+1)}$ (G.M.)

No

Proceed to solve using the formulas identified above.

Top 8 Memory Tricks and Quick Methods

Trick: These shortcuts and mnemonics will save you valuable exam time and help prevent common calculation errors.

Steps to Remember:

Trick 1 - 'Add vs Multiply' Identification: If the problem says 'increases by' or 'decreases by' a fixed amount → A.P. (use addition/subtraction). If it says 'multiplied by' or 'doubled' or 'tripled' → G.P. (use multiplication). If you see 'interest compounded' → G.P. (never A.P.!).
Trick 2 - 'Quick A.M. Check': For A.P., if middle term is $b$ and outer terms are $a, c$, then $b = \frac{a+c}{2}$ is automatic. This also means $2b = a + c$. Use this to set up quick equations.
Trick 3 - 'Three Terms Shortcut': When selecting 3 terms in A.P., always use $(p-d), p, (p+d)$ instead of $a, a+d, a+2d$. The reason: the $d$ cancels out when you add them, making calculations faster. Example: Sum of 3 terms in A.P. = $3p$ (so if sum = 15, then $p = 5$).
Trick 4 - 'Three Terms G.P. Shortcut': When selecting 3 terms in G.P., always use $\frac{p}{r}, p, pr$ instead of $a, ar, ar^2$. Reason: multiplication/division simplifies. Example: Product of 3 terms = $p^3$ (so if product = 8, then $p = 2$).
Trick 5 - 'A-G-H Alphabet Rule': The three means follow alphabetical order: $A > G > H$ (for unequal positive numbers). Alphabetically, A comes first (biggest value), G in middle, H last (smallest value). This helps you remember which is largest.
Trick 6 - 'Magic Product Rule': Always remember $A \times H = G^2$. This is the golden relationship. If you know any two of A, G, H, you can find the third instantly. Example: If $A = 10, G = 8$, then $H = \frac{G^2}{A} = \frac{64}{10} = 6.4$.
Trick 7 - 'Infinite Series Convergence': For $S_{\infty} = \frac{a}{1-r}$, the MOST CRITICAL step is checking $|r| < 1$. If $|r| \geq 1$, write 'diverges—sum does not exist' and move on. This single mistake costs marks on 30% of G.P. infinite series problems.
Trick 8 - 'Decimal to Fraction': For repeating decimals like $0.\overline{23} = 0.232323\ldots$, use G.P. series: $0.23 + 0.0023 + 0.000023 + \ldots = \frac{0.23}{1-0.01} = \frac{0.23}{0.99} = \frac{23}{99}$. The pattern is: recurring block goes in numerator, and 9 repeats in denominator.

Mnemonic: Remember: 'A.P. = Add, G.P. = Grow, H.P. = Harmonize (by reciprocals)'. When in doubt, test with simple numbers (like 2, 4, 6 for A.P. or 2, 4, 8 for G.P.) to verify your formula choice.

Common Exam Mistakes and How to Avoid Them

These errors appear in 75% of incomplete student answers. Studying these prevents costly mark loss.

Mistake 1 - Confusing $n$ and $t_n$: A student writes: 'The 5th term is 5.' This is vague! Clearly state: $n = 5$ (position) and $t_5 = $ [value]. For example, 'The 5th term is $t_5 = 21$' is correct. In A.P. $1, 4, 7, 10, 13, \ldots$, the 5th term is $t_5 = 13$, not 5.
Mistake 2 - Forgetting $|r| < 1$ for Infinite Series: A student applies $S_{\infty} = \frac{a}{1-r}$ to the series $1 + 2 + 4 + 8 + \ldots$ (where $r = 2$). This series diverges! $S_{\infty}$ does NOT exist. The formula only works if $|r| < 1$. Always check this condition first. This error loses 2-3 marks.
Mistake 3 - Wrong Sign in Common Difference: Given $2, 5, 8, 11$, a student calculates $d = 2 - 5 = -3$. Wrong! Common difference is $d = 5 - 2 = 3$ (subtract first term from second, not the other way). For a decreasing A.P. like $10, 7, 4, 1$, we get $d = 7 - 10 = -3$ (negative, which makes sense).
Mistake 4 - Exponent Error in G.P.: Finding $t_5$ in G.P. with $a = 2, r = 3$, a student writes $t_5 = 2 \times 3^5 = 2 \times 243 = 486$. But the formula is $t_n = ar^{n-1}$, so $t_5 = 2 \times 3^{5-1} = 2 \times 3^4 = 162$. The exponent is $(n-1)$, not $n$. This is the #1 calculation error in G.P. problems.
Mistake 5 - Misidentifying Progression Type: Given $2, 4, 6, 8$, a student writes 'G.P.' because the numbers are increasing. This is A.P. (constant difference $d = 2$). For G.P., the ratio must be constant. Here, $\frac{4}{2} = 2$ but $\frac{6}{4} = 1.5$ (not constant). Always check differences first, then ratios.
Mistake 6 - Harmonic Progression Confusion: A student confuses H.P. terms with their reciprocals. Remember: If $1, \frac{1}{2}, \frac{1}{3}$ are in H.P., then the reciprocals $1, 2, 3$ form an A.P. To solve H.P. problems: (1) Take reciprocals, (2) Work in A.P., (3) Convert back to H.P. Skipping step 1 leads to wrong answers.
Mistake 7 - Inserting Means Without Correct Formula: To insert 2 means between 3 and 12, a student guesses the middle numbers are 6 and 9 (halfway). Wrong! The correct method: $d = \frac{12-3}{2+1} = 3$, so the sequence is $3, 6, 9, 12$. The difference is constant at 3. If you had guessed, you'd miss the 'equal spacing' requirement.
Mistake 8 - Mixing Sum Formulas: Using $S_n = \frac{n}{2}(a+l)$ for G.P. is wrong! This formula is ONLY for A.P. For G.P., use $S_n = \frac{a(r^n-1)}{r-1}$. If you're unsure, derive from first principles instead of guessing formulas. This error costs 3-4 marks on long-answer questions.
Mistake 9 - Forgetting $r \neq 1$ Condition: A student applies $S_n = \frac{a(r^n-1)}{r-1}$ when $r = 1$. This gives division by zero! When $r = 1$, all terms are equal to $a$, so $S_n = n \times a$. Always check $r = 1$ separately before using the standard formula.
Mistake 10 - G.M. for Negative Numbers: Finding G.M. of $-4$ and $-9$, a student writes $G = \sqrt{(-4)(-9)} = \sqrt{36} = 6$. But G.M. is typically defined for positive numbers only. For negative numbers, use absolute values: $G = \sqrt{|-4| \times |-9|} = \sqrt{36} = 6$. Or state 'G.M. is undefined for negative numbers' (check your curriculum).

Solved Examples (Easy → Medium → Hard)

1

Find the 10th term of the A.P.: $5, 8, 11, 14, \ldots$

Easy

Solution

Given: A.P. with terms $5, 8, 11, 14, \ldots$

Step 1: Identify the first term and common difference

From the sequence:
- First term: $a = 5$
- Common difference: $d = 8 - 5 = 3$ (or we can verify: $11 - 8 = 3$, and $14 - 11 = 3$ ✓)

Step 2: Identify the position

We need to find the 10th term, so $n = 10$.

Step 3: Apply the formula for the $n^{th}$ term of A.P.**

$$t_n = a + (n-1)d$$

$$t_{10} = 5 + (10-1) \times 3$$

$$t_{10} = 5 + 9 \times 3$$

$$t_{10} = 5 + 27$$

$$t_{10} = 32$$

Step 4: Verify by writing out the sequence

$t_1 = 5, t_2 = 8, t_3 = 11, t_4 = 14, t_5 = 17, t_6 = 20, t_7 = 23, t_8 = 26, t_9 = 29, t_{10} = 32$ ✓

Therefore, the 10th term is $\boxed{32}$.

∴ $t_{10} = 32$

2

Find the sum of the first 8 terms of the G.P.: $2, 4, 8, 16, \ldots$
Easy

Solution

Given: G.P. with terms $2, 4, 8, 16, \ldots$
Step 1: Identify the first term and common ratio
From the sequence:
- First term: $a = 2$
- Common ratio: $r = \frac{4}{2} = 2$ (or verify: $\frac{8}{4} = 2$, and $\frac{16}{8} = 2$ ✓)
Step 2: Check that $r \neq 1$
Since $r = 2 \neq 1$, we can use the standard formula. ✓
Step 3: Identify the number of terms
We need the sum of the first 8 terms, so $n = 8$.
Step 4: Apply the formula for the sum of $n$ terms of G.P.**
$$S_n = \frac{a(r^n - 1)}{r - 1}$$
$$S_8 = \frac{2(2^8 - 1)}{2 - 1}$$
Step 5: Calculate $2^8$
$2^8 = 256$
$$S_8 = \frac{2(256 - 1)}{1}$$
$$S_8 = \frac{2 \times 255}{1}$$
$$S_8 = \frac{510}{1}$$
$$S_8 = 510$$
Step 6: Verify by adding the terms
$S_8 = 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 = 510$ ✓
Therefore, the sum of the first 8 terms is $\boxed{510}$.

∴ $S_8 = 510$

3

Find the A.M., G.M., and H.M. of 6 and 24.
Easy

Solution

Given: Two numbers $a = 6$ and $b = 24$
Step 1: Calculate Arithmetic Mean (A.M.)**
$$A = \frac{a + b}{2}$$
$$A = \frac{6 + 24}{2}$$
$$A = \frac{30}{2}$$
$$A = 15$$
Step 2: Calculate Geometric Mean (G.M.)**
$$G = \sqrt{ab}$$
$$G = \sqrt{6 \times 24}$$
$$G = \sqrt{144}$$
$$G = 12$$
Step 3: Calculate Harmonic Mean (H.M.)**
$$H = \frac{2ab}{a + b}$$
$$H = \frac{2 \times 6 \times 24}{6 + 24}$$
$$H = \frac{2 \times 144}{30}$$
$$H = \frac{288}{30}$$
$$H = 9.6$$
Step 4: Verify the relationship $A \times H = G^2$**
$$A \times H = 15 \times 9.6 = 144$$
$$G^2 = 12^2 = 144$$ ✓
Step 5: Verify the inequality $A > G > H$**
$$15 > 12 > 9.6$$ ✓
Therefore: $A = 15$, $G = 12$, $H = 9.6$

∴ A.M. = 15, G.M. = 12, H.M. = 9.6

4

Find the sum to infinity of the series: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$
Medium

Solution

Given: Series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$
Step 1: Identify that this is a G.P. series
This is clearly a G.P. because each term is half of the previous term.
Step 2: Identify the first term and common ratio
- First term: $a = 1$
- Common ratio: $r = \frac{1/2}{1} = \frac{1}{2}$ (or verify: $\frac{1/4}{1/2} = \frac{1}{2}$ ✓)
Step 3: CRITICAL CHECK - Verify the convergence condition $|r| < 1$**
$$|r| = \left|\frac{1}{2}\right| = \frac{1}{2} < 1$$ ✓
Since $|r| < 1$, the series CONVERGES and the sum to infinity EXISTS.
Step 4: Apply the formula for sum to infinity**
$$S_{\infty} = \frac{a}{1 - r}$$
$$S_{\infty} = \frac{1}{1 - \frac{1}{2}}$$
$$S_{\infty} = \frac{1}{\frac{1}{2}}$$
$$S_{\infty} = 2$$
Step 5: Verify with partial sums
- $S_1 = 1 = 1$
- $S_2 = 1 + 0.5 = 1.5$
- $S_3 = 1.5 + 0.25 = 1.75$
- $S_4 = 1.75 + 0.125 = 1.875$
- $S_5 = 1.875 + 0.0625 = 1.9375$
As we add more terms, the sum approaches 2 ✓
Therefore, the sum to infinity is $\boxed{2}$.

∴ $S_{\infty} = 2$

5

Insert 3 arithmetic means between 2 and 14.
Medium

Solution

Given: Two numbers $a = 2$ and $b = 14$. We need to insert $n = 3$ arithmetic means between them.
Step 1: Understand what we're finding
We want to create an A.P.: $2, A_1, A_2, A_3, 14$ with 5 total terms (where $A_1, A_2, A_3$ are the three means we seek).
Step 2: Find the common difference**
For an A.P. with $(n+2)$ total terms (including the endpoints):
$$d = \frac{b - a}{n + 1}$$
$$d = \frac{14 - 2}{3 + 1}$$
$$d = \frac{12}{4}$$
$$d = 3$$
Step 3: Find each arithmetic mean
- $A_1 = a + d = 2 + 3 = 5$
- $A_2 = a + 2d = 2 + 2(3) = 2 + 6 = 8$
- $A_3 = a + 3d = 2 + 3(3) = 2 + 9 = 11$
Step 4: Verify the complete A.P.**
Complete A.P.: $2, 5, 8, 11, 14$
Check differences: $5 - 2 = 3$, $8 - 5 = 3$, $11 - 8 = 3$, $14 - 11 = 3$ ✓
Therefore, the three arithmetic means are $\boxed{5, 8, 11}$.

∴ The three A.M.s are 5, 8, and 11.

6

If three positive numbers are in G.P. and their product is 64, find the middle term.
Medium

Solution

Given: Three positive numbers in G.P. with product = 64
Step 1: Set up the three terms using the G.P. shortcut
For three terms in G.P., use the form: $\frac{a}{r}, a, ar$ (instead of $a, ar, ar^2$)
This makes calculation easier because:
- First term: $\frac{a}{r}$
- Second term (middle): $a$ ← This is what we want!
- Third term: $ar$
Step 2: Apply the constraint - product = 64
$$\frac{a}{r} \times a \times ar = 64$$
$$\frac{a \times a \times ar}{r} = 64$$
$$\frac{a^3 r}{r} = 64$$
$$a^3 = 64$$
Step 3: Solve for $a$ (the middle term)
$$a = \sqrt[3]{64}$$
$$a = \sqrt[3]{4^3}$$
$$a = 4$$
Step 4: Verify with specific values
We don't know $r$ exactly, but let's check with $r = 2$:
- First term: $\frac{4}{2} = 2$
- Second term: $4$
- Third term: $4 \times 2 = 8$
Check: $2 \times 4 \times 8 = 64$ ✓ and ratios are $\frac{4}{2} = 2$ and $\frac{8}{4} = 2$ (constant) ✓
Therefore, the middle term (G.M. of the three numbers) is $\boxed{4}$.

∴ The middle term is 4.

7

Prove that if $a, b, c$ are in A.P., then $2b = a + c$.
Medium

Solution

Given: Three numbers $a, b, c$ are in Arithmetic Progression
To Prove: $2b = a + c$
Proof:**
Step 1: Use the definition of A.P.**
Since $a, b, c$ are in A.P., there exists a common difference $d$ such that:
- $b = a + d$ (the second term is the first plus $d$)
- $c = b + d = a + 2d$ (the third term is the first plus $2d$)
Step 2: Express $2b$ in terms of $a$ and $d$**
$$2b = 2(a + d)$$
$$2b = 2a + 2d$$
Step 3: Express $a + c$ in terms of $a$ and $d$**
$$a + c = a + (a + 2d)$$
$$a + c = 2a + 2d$$
Step 4: Compare**
From Step 2: $2b = 2a + 2d$
From Step 3: $a + c = 2a + 2d$
Therefore: $2b = a + c$ ✓ (Proved)
Alternatively (using the mean concept):**
If $a, b, c$ are in A.P., then $b$ is the Arithmetic Mean of $a$ and $c$.
$$b = \frac{a + c}{2}$$
Multiply both sides by 2:
$$2b = a + c$$ ✓

∴ Proved: If $a, b, c$ are in A.P., then $2b = a + c$.

8

Find the sum of the A.P.: $2 + 5 + 8 + 11 + \ldots$ up to 20 terms.
Hard

Solution

Given: A.P. $2 + 5 + 8 + 11 + \ldots$ and we need to find the sum of the first 20 terms.
Step 1: Identify the first term and common difference
- First term: $a = 2$
- Common difference: $d = 5 - 2 = 3$
- Number of terms: $n = 20$
Step 2: Apply the formula for sum of $n$ terms of A.P.**
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
$$S_{20} = \frac{20}{2}[2(2) + (20-1)(3)]$$
Step 3: Calculate inside the bracket
$$2a = 2(2) = 4$$
$$(n-1)d = (20-1)(3) = 19 \times 3 = 57$$
$$2a + (n-1)d = 4 + 57 = 61$$
Step 4: Calculate the sum
$$S_{20} = \frac{20}{2} \times 61$$
$$S_{20} = 10 \times 61$$
$$S_{20} = 610$$
Verification using the alternative formula:**
First, find the 20th term: $t_{20} = a + (n-1)d = 2 + 19(3) = 2 + 57 = 59$
Using $S_n = \frac{n}{2}(a + l)$ where $l = t_{20} = 59$:
$$S_{20} = \frac{20}{2}(2 + 59) = 10 \times 61 = 610$$ ✓
Therefore, the sum of the first 20 terms is $\boxed{610}$.

∴ $S_{20} = 610$

9

Convert the recurring decimal $0.\overline{27}$ into a fraction using G.P. series.
Hard

Solution

Given: Recurring decimal $0.\overline{27} = 0.272727\ldots$
Step 1: Write the decimal as an infinite series
$$0.\overline{27} = 0.27 + 0.0027 + 0.000027 + \ldots$$
$$= 0.27 + 0.27 \times 0.01 + 0.27 \times (0.01)^2 + \ldots$$
Step 2: Identify this as a G.P. series
This is a G.P. with:
- First term: $a = 0.27 = \frac{27}{100}$
- Common ratio: $r = 0.01 = \frac{1}{100}$
$$0.\overline{27} = a + ar + ar^2 + ar^3 + \ldots$$
Step 3: Check convergence**
$$|r| = \left|\frac{1}{100}\right| = 0.01 < 1$$ ✓
The series converges, so we can use $S_{\infty} = \frac{a}{1-r}$
Step 4: Apply the sum to infinity formula**
$$S_{\infty} = \frac{a}{1 - r}$$
$$S_{\infty} = \frac{\frac{27}{100}}{1 - \frac{1}{100}}$$
$$S_{\infty} = \frac{\frac{27}{100}}{\frac{99}{100}}$$
$$S_{\infty} = \frac{27}{100} \times \frac{100}{99}$$
$$S_{\infty} = \frac{27}{99}$$
Step 5: Simplify the fraction**
Find GCD of 27 and 99:
- $27 = 3^3$
- $99 = 9 \times 11 = 3^2 \times 11$
- GCD = $3^2 = 9$
$$\frac{27}{99} = \frac{27 \div 9}{99 \div 9} = \frac{3}{11}$$
Verification:**
$$\frac{3}{11} = 3 \div 11 = 0.272727\ldots = 0.\overline{27}$$ ✓
Therefore, $0.\overline{27} = \boxed{\frac{3}{11}}$

∴ $0.\overline{27} = \frac{3}{11}$

10

Three numbers in A.P. have a sum of 21 and a product of 280. Find the numbers.
Hard

Solution

Given: Three numbers in A.P. with sum = 21 and product = 280
Step 1: Set up the three numbers using the A.P. shortcut
For three numbers in A.P., use: $(a-d), a, (a+d)$ where $a$ is the middle term and $d$ is the common difference.
This form is useful because it reduces the number of unknowns.
Step 2: Apply the sum condition
$$(a - d) + a + (a + d) = 21$$
$$3a = 21$$
$$a = 7$$
So the middle term is 7.
Step 3: Apply the product condition
$$(a - d) \times a \times (a + d) = 280$$
Using the difference of squares identity: $(a-d)(a+d) = a^2 - d^2$
$$(a^2 - d^2) \times a = 280$$
Substitute $a = 7$:
$$(7^2 - d^2) \times 7 = 280$$
$$(49 - d^2) \times 7 = 280$$
$$49 - d^2 = \frac{280}{7}$$
$$49 - d^2 = 40$$
$$d^2 = 49 - 40$$
$$d^2 = 9$$
$$d = \pm 3$$
Step 4: Find the three numbers
Case 1: If $d = 3$
- First number: $a - d = 7 - 3 = 4$
- Second number: $a = 7$
- Third number: $a + d = 7 + 3 = 10$
Case 2: If $d = -3$
- First number: $a - d = 7 - (-3) = 10$
- Second number: $a = 7$
- Third number: $a + d = 7 + (-3) = 4$
Both cases give the same set of numbers (just in different order): $4, 7, 10$
Step 5: Verify both conditions
Sum: $4 + 7 + 10 = 21$ ✓
Product: $4 \times 7 \times 10 = 280$ ✓
Arithmetic Progression check: $7 - 4 = 3$ and $10 - 7 = 3$ (constant difference) ✓
Therefore, the three numbers are $\boxed{4, 7, 10}$

∴ The three numbers are 4, 7, and 10.

11

Prove that $A.M. > G.M. > H.M.$ for two unequal positive numbers $a$ and $b$.
Hard

Solution

Given: Two unequal positive numbers $a$ and $b$ where $a \neq b$
To Prove:
1. $A.M. > G.M.$
2. $G.M. > H.M.$
Proof of Part 1: $A.M. > G.M.$**
$$A = \frac{a + b}{2}, \quad G = \sqrt{ab}$$
We need to show: $\frac{a + b}{2} > \sqrt{ab}$
Square both sides (valid since both are positive):
$$\left(\frac{a + b}{2}\right)^2 > ab$$
$$\frac{(a + b)^2}{4} > ab$$
$$\frac{a^2 + 2ab + b^2}{4} > ab$$
Multiply both sides by 4:
$$a^2 + 2ab + b^2 > 4ab$$
$$a^2 - 2ab + b^2 > 0$$
$$(a - b)^2 > 0$$
This is TRUE because $a \neq b$ (so $(a-b)^2 > 0$).
Therefore: $A > G$ ✓
Proof of Part 2: $G.M. > H.M.$**
$$G = \sqrt{ab}, \quad H = \frac{2ab}{a + b}$$
We need to show: $\sqrt{ab} > \frac{2ab}{a + b}$
Divide both sides by $\sqrt{ab}$ (positive, so inequality direction unchanged):
$$1 > \frac{2ab}{(a + b)\sqrt{ab}}$$
$$1 > \frac{2\sqrt{ab}}{a + b}$$
Multiply both sides by $(a + b)$ (positive):
$$a + b > 2\sqrt{ab}$$
Square both sides:
$$(a + b)^2 > 4ab$$
$$a^2 + 2ab + b^2 > 4ab$$
$$a^2 - 2ab + b^2 > 0$$
$$(a - b)^2 > 0$$
This is TRUE because $a \neq b$.
Therefore: $G > H$ ✓
Conclusion:**
Combining both parts: $A > G > H$ ✓ (Proved)
Note:** Equality holds only when $a = b$. If $a = b$, then $A = G = H = a = b$.

∴ Proved: For unequal positive numbers, $A.M. > G.M. > H.M.$

MCQ Practice (10 Questions)

Mixed Level — MCQ

1

What is the common difference of the A.P.: $-3, -1, 1, 3, \ldots$?

Correct!

Incorrect. Correct: $2$

Solution

For the A.P. $-3, -1, 1, 3, \ldots$:
Common difference $d = t_2 - t_1 = (-1) - (-3) = 2$
Therefore, $d = 2$

∴ $2$

2

Find the 5th term of the G.P. with first term $a = 2$ and common ratio $r = -2$.

Correct!

Incorrect. Correct: $-32$

Solution

For G.P. with $a = 2, r = -2, n = 5$:
$$t_n = ar^{n-1}$$
$$t_5 = 2 \times (-2)^{5-1} = 2 \times (-2)^4 = 2 \times 16 = 32$$

∴ $32$

3

If the A.M. of two numbers is 10 and their G.M. is 8, what is their H.M.?

Correct!

Incorrect. Correct: $6.4$

Solution

Given: $A = 10, G = 8$
Using the relationship $A \times H = G^2$:
$$10 \times H = 8^2$$
$$10 \times H = 64$$
$$H = \frac{64}{10} = 6.4$$

∴ $6.4$

4

What is the sum to infinity of the series $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots$?

Correct!

Incorrect. Correct: $\frac{3}{2}$

Solution

This is a G.P. with $a = 1, r = \frac{1}{3}$
Check convergence: $|r| = \frac{1}{3} < 1$ ✓
$$S_{\infty} = \frac{a}{1-r} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$$

∴ $\frac{3}{2}$

5

If the sum of the first $n$ terms of an A.P. is $S_n = 3n^2 + 2n$, what is the first term?

Correct!

Incorrect. Correct: $5$

Solution

The first term of the A.P. is: $t_1 = S_1$
$$t_1 = S_1 = 3(1)^2 + 2(1) = 3 + 2 = 5$$

∴ $5$

6

Which of the following is a Harmonic Progression?

Correct!

Incorrect. Correct: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

Solution

For H.P., the reciprocals must form an A.P.
For $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$:
Reciprocals: $1, 2, 3, 4$ (clearly an A.P. with $d=1$) ✓
Therefore, it's an H.P.

∴ $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

7

What is the condition for an infinite G.P. series to have a finite sum?

Correct!

Incorrect. Correct: $|r| < 1$

Solution

For the infinite G.P. series to converge to a finite sum, we need:
$$|r| < 1$$
This ensures that the terms approach zero and $S_{\infty} = \frac{a}{1-r}$ exists as a finite value.

∴ $|r| < 1$

8

How many terms are in the A.P.: $5, 10, 15, \ldots, 100$?

Correct!

Incorrect. Correct: $20$

Solution

Using the formula $t_n = a + (n-1)d$:
Given: $a = 5, d = 5, t_n = 100$
$$100 = 5 + (n-1) \times 5$$
$$95 = 5(n-1)$$
$$19 = n - 1$$
$$n = 20$$

∴ $20$

9

If $a, b, c$ are in G.P., then which of the following is true?

Correct!

Incorrect. Correct: $b^2 = ac$

Solution

If $a, b, c$ are in G.P.:
Then $b = ar$ and $c = ar^2$ for some ratio $r$
$$b^2 = (ar)^2 = a^2r^2$$
$$ac = a \times ar^2 = a^2r^2$$
Therefore: $b^2 = ac$ ✓

∴ $b^2 = ac$

10

Convert $0.\overline{3}$ (repeating decimal) to a fraction.

Correct!

Incorrect. Correct: $\frac{1}{3}$

Solution

The repeating decimal $0.\overline{3} = 0.333\ldots$ can be written as:
$$0.\overline{3} = 0.3 + 0.03 + 0.003 + \ldots = \sum_{n=1}^{\infty} 3 \times 10^{-n}$$
This is a G.P. with $a = 0.3, r = 0.1$
$$S_{\infty} = \frac{0.3}{1-0.1} = \frac{0.3}{0.9} = \frac{1}{3}$$

∴ $\frac{1}{3}$

Short Answer Questions (10 Questions)

1

Find the sum of the first 10 terms of the A.P.: $3, 7, 11, 15, \ldots$
[2 marks]

Solution

Given: A.P. with $a = 3, d = 7 - 3 = 4, n = 10$
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
$$S_{10} = \frac{10}{2}[2(3) + (10-1)(4)]$$
$$S_{10} = 5[6 + 36]$$
$$S_{10} = 5 \times 42 = 210$$

∴ $S_{10} = 210$

2

If the 5th term of an A.P. is 18 and the 9th term is 34, find the common difference.
[2 marks]

Solution

Given: $t_5 = 18, t_9 = 34$
Using $t_n = a + (n-1)d$:
$$t_5 = a + 4d = 18 \quad \text{...(1)}$$
$$t_9 = a + 8d = 34 \quad \text{...(2)}$$
Subtract (1) from (2):
$$(a + 8d) - (a + 4d) = 34 - 18$$
$$4d = 16$$
$$d = 4$$

∴ Common difference $d = 4$

3

Show that $2, 4, 8, 16$ is a Geometric Progression and find its common ratio.
[2 marks]

Solution

To show it's a G.P., check that the ratio between consecutive terms is constant:
$$r = \frac{t_2}{t_1} = \frac{4}{2} = 2$$
$$r = \frac{t_3}{t_2} = \frac{8}{4} = 2$$
$$r = \frac{t_4}{t_3} = \frac{16}{8} = 2$$
Since the ratio is constant ($r = 2$), the sequence is a G.P. with common ratio $r = 2$ ✓

∴ Common ratio $r = 2$

4

Find the G.M. between 9 and 25.
[2 marks]

Solution

The Geometric Mean of two numbers $a$ and $b$ is:
$$G = \sqrt{ab} = \sqrt{9 \times 25} = \sqrt{225} = 15$$

∴ G.M. = 15

5

Determine whether $0.5, 1, 2, 4, 8$ is a G.P. If yes, find the sum of 5 terms.
[3 marks]

Solution

Check if it's a G.P.:
$$r = \frac{1}{0.5} = 2, \quad r = \frac{2}{1} = 2, \quad r = \frac{4}{2} = 2, \quad r = \frac{8}{4} = 2$$
Yes, it's a G.P. with $a = 0.5, r = 2, n = 5$ ✓
Find the sum:
$$S_5 = \frac{a(r^n - 1)}{r - 1} = \frac{0.5(2^5 - 1)}{2 - 1} = \frac{0.5(32 - 1)}{1} = \frac{0.5 \times 31}{1} = 15.5$$

∴ Yes, it's a G.P. with $r = 2$ and $S_5 = 15.5$

6

The sum to infinity of a G.P. is 10 and the first term is 5. Find the common ratio.
[2 marks]

Solution

Given: $S_{\infty} = 10, a = 5$
Using $S_{\infty} = \frac{a}{1-r}$:
$$10 = \frac{5}{1-r}$$
$$10(1-r) = 5$$
$$10 - 10r = 5$$
$$10r = 5$$
$$r = \frac{1}{2}$$
Check: $|r| = 0.5 < 1$ ✓ (series converges)

∴ Common ratio $r = \frac{1}{2}$

7

Insert one Arithmetic Mean between 7 and 15.
[2 marks]

Solution

For one A.M. between two numbers, use:
$$A = \frac{a + b}{2} = \frac{7 + 15}{2} = \frac{22}{2} = 11$$
Verification: The sequence $7, 11, 15$ has constant differences: $11 - 7 = 4$ and $15 - 11 = 4$ ✓

∴ The A.M. is 11

8

If $a, b, c$ are in A.P., prove that $b = \frac{a+c}{2}$.
[3 marks]

Solution

Since $a, b, c$ are in A.P., there is a common difference $d$ such that:
$$b = a + d$$
$$c = a + 2d$$
From the first equation: $d = b - a$
From the second equation: $c = a + 2(b - a) = a + 2b - 2a = 2b - a$
Therefore: $c + a = 2b$
$$b = \frac{a + c}{2}$$ (Proved) ✓

∴ Proved: $b = \frac{a+c}{2}$

9

Find $n$ if $t_n = 19$ in the A.P. $1, 4, 7, 10, \ldots$
[2 marks]

Solution

Given: $a = 1, d = 3, t_n = 19$
Using $t_n = a + (n-1)d$:
$$19 = 1 + (n-1) \times 3$$
$$18 = 3(n-1)$$
$$6 = n - 1$$
$$n = 7$$
The 7th term is 19.

∴ $n = 7$

10

Verify that the H.M. of 6 and 12 satisfies $A \times H = G^2$.
[3 marks]

Solution

Calculate A.M., G.M., H.M.:
$$A = \frac{6 + 12}{2} = 9$$
$$G = \sqrt{6 \times 12} = \sqrt{72} = 6\sqrt{2} \approx 8.49$$
$$H = \frac{2 \times 6 \times 12}{6 + 12} = \frac{144}{18} = 8$$
Verify $A \times H = G^2$:
$$A \times H = 9 \times 8 = 72$$
$$G^2 = (6\sqrt{2})^2 = 36 \times 2 = 72$$ ✓
The relationship is verified!

∴ Verified: $A \times H = 72 = G^2$

Long Answer Questions (5 Questions)

1

Three numbers are in A.P. If their sum is 24 and the sum of their squares is 200, find the three numbers.
[5 marks]

Solution

Let the three numbers in A.P. be: $(a-d), a, (a+d)$ where $a$ is the middle term and $d$ is the common difference.
Step 1: Use the sum condition**
$$(a-d) + a + (a+d) = 24$$
$$3a = 24$$
$$a = 8$$
So the middle term is 8.
Step 2: Use the sum of squares condition**
$$(a-d)^2 + a^2 + (a+d)^2 = 200$$
Expand each square:
$$(a-d)^2 = a^2 - 2ad + d^2$$
$$a^2 = a^2$$
$$(a+d)^2 = a^2 + 2ad + d^2$$
Sum:
$$a^2 - 2ad + d^2 + a^2 + a^2 + 2ad + d^2 = 200$$
$$3a^2 + 2d^2 = 200$$
Step 3: Substitute $a = 8$**
$$3(8)^2 + 2d^2 = 200$$
$$3(64) + 2d^2 = 200$$
$$192 + 2d^2 = 200$$
$$2d^2 = 8$$
$$d^2 = 4$$
$$d = \pm 2$$
Step 4: Find the three numbers**
If $d = 2$:
- First: $8 - 2 = 6$
- Second: $8$
- Third: $8 + 2 = 10$
If $d = -2$:
- First: $8 - (-2) = 10$
- Second: $8$
- Third: $8 + (-2) = 6$
Both give the same set: $\{6, 8, 10\}$
Verification:
Sum: $6 + 8 + 10 = 24$ ✓
Sum of squares: $36 + 64 + 100 = 200$ ✓
A.P.: $6, 8, 10$ with $d = 2$ ✓

∴ The three numbers are 6, 8, and 10

2

The first term of a G.P. is 3. If the sum of the first three terms is 13, find the common ratio and the sum to infinity (if it exists).
[6 marks]

Solution

Given: First term $a = 3$, sum of first three terms $S_3 = 13$
Step 1: Set up the equation for $S_3$**
$$S_3 = a + ar + ar^2 = 13$$
$$3 + 3r + 3r^2 = 13$$
$$3r^2 + 3r + 3 = 13$$
$$3r^2 + 3r - 10 = 0$$
Step 2: Solve the quadratic equation**
Using the quadratic formula:
$$r = \frac{-3 \pm \sqrt{9 + 120}}{6} = \frac{-3 \pm \sqrt{129}}{6}$$
Wait, let me recalculate: $3r^2 + 3r - 10 = 0$
For $ax^2 + bx + c = 0$: $a = 3, b = 3, c = -10$
$$r = \frac{-3 \pm \sqrt{9 - 4(3)(-10)}}{2(3)} = \frac{-3 \pm \sqrt{9 + 120}}{6} = \frac{-3 \pm \sqrt{129}}{6}$$
Hmm, this gives non-simple values. Let me recheck the problem... Actually, let me try factoring or using alternative methods.
Let me try simple values: If $r = 1$: $3 + 3 + 3 = 9$ (not 13)
If $r = 2$: $3 + 6 + 12 = 21$ (not 13)
If $r = 1/2$: $3 + 1.5 + 0.75 = 5.25$ (not 13)
Using the quadratic $3r^2 + 3r - 10 = 0$, we divide by 3:
$$r^2 + r - \frac{10}{3} = 0$$
Multiply by 3: $3r^2 + 3r - 10 = 0$
Using the quadratic formula:
$$r = \frac{-3 \pm \sqrt{9 + 120}}{6} = \frac{-3 \pm \sqrt{129}}{6}$$
$\sqrt{129} \approx 11.36$
$$r \approx \frac{-3 + 11.36}{6} \approx \frac{8.36}{6} \approx 1.39 \quad \text{or} \quad r \approx \frac{-3 - 11.36}{6} \approx -2.39$$
For practical NEB purposes, let me assume cleaner values. Perhaps I made an error. Let me try: If the three terms are $3, 3r, 3r^2$ with sum 13:
Let's try $r = 2/3$: $3 + 2 + 4/3 = 5 + 4/3 \approx 6.33$ (not 13)
Actually, using $3r^2 + 3r - 10 = 0$ is correct. For an exam, students would either:
1. Use the quadratic formula
2. Check if it factors (it doesn't easily)r>3. Solve approximately
The two solutions are:
$$r_1 = \frac{-3 + \sqrt{129}}{6} \approx 1.39$$
$$r_2 = \frac{-3 - \sqrt{129}}{6} \approx -2.39$$
Step 3: Determine which value to use and find $S_{\infty}$**
For $S_{\infty}$ to exist, we need $|r| < 1$.
- For $r_1 \approx 1.39$: $|r_1| > 1$, so $S_{\infty}$ does NOT exist (series diverges)
- For $r_2 \approx -2.39$: $|r_2| > 1$, so $S_{\infty}$ does NOT exist (series diverges)
Since both values of $r$ have $|r| > 1$, the infinite series DIVERGES and $S_{\infty}$ does not exist as a finite value.
Alternative: Check if cleaner values were intended**
If we're looking for a scenario where $S_{\infty}$ exists, the problem would need $|r| < 1$. Perhaps the sum condition was different, or the first term different. For the given conditions, the series diverges.

∴ The common ratio is $r = \frac{-3 + \sqrt{129}}{6} \approx 1.39$ or $r = \frac{-3 - \sqrt{129}}{6} \approx -2.39$. In both cases, $|r| > 1$, so the sum to infinity does NOT exist (series diverges).

3

Prove that for any two positive numbers $a$ and $b$, the relationship $A \cdot H = G^2$ holds, where $A$, $G$, $H$ are the Arithmetic, Geometric, and Harmonic Means respectively.
[5 marks]

Solution

Given: Two positive numbers $a$ and $b$
To Prove: $A \cdot H = G^2$
Step 1: Express the three means**
$$A = \frac{a + b}{2}$$
$$G = \sqrt{ab}$$
$$H = \frac{2ab}{a + b}$$
Step 2: Calculate $A \cdot H$**
$$A \cdot H = \frac{a + b}{2} \cdot \frac{2ab}{a + b}$$
$$= \frac{(a + b) \cdot 2ab}{2(a + b)}$$
$$= \frac{2ab(a + b)}{2(a + b)}$$
$$= ab$$
Step 3: Calculate $G^2$**
$$G^2 = (\sqrt{ab})^2 = ab$$
Step 4: Compare**
$$A \cdot H = ab = G^2$$
Therefore: $A \cdot H = G^2$ ✓ (Proved)
Conclusion:
This beautiful relationship shows that the product of the Arithmetic and Harmonic Means equals the square of the Geometric Mean. It holds for ANY two positive numbers!

∴ Proved: $A \cdot H = G^2$

4

A ball is dropped from a height of 10 meters. After each bounce, it rises to 80% of its previous height. What is the total vertical distance traveled by the ball until it comes to rest?
[6 marks]

Solution

Given: Initial height $h = 10$ m, bounce ratio $r = 0.8$ (80%)
Step 1: Understand the motion**
- Fall 1: $10$ m (downward)
- Rise 1: $10 \times 0.8 = 8$ m (upward)
- Fall 2: $8$ m (downward)
- Rise 2: $8 \times 0.8 = 6.4$ m (upward)
- Fall 3: $6.4$ m (downward)
- Rise 3: $6.4 \times 0.8 = 5.12$ m (upward)
- ... and so on
Step 2: Separate upward and downward distances**
Downward distances: $10 + 8 + 6.4 + 5.12 + \ldots$
This is a G.P.: $10 + 10(0.8) + 10(0.8)^2 + \ldots$
- First term: $a = 10$
- Common ratio: $r = 0.8$
Upward distances: $8 + 6.4 + 5.12 + \ldots$
This is: $10(0.8) + 10(0.8)^2 + 10(0.8)^3 + \ldots$
- First term: $a = 8$
- Common ratio: $r = 0.8$
Step 3: Calculate sum of downward distances**
$$S_{\text{down}} = \frac{10}{1 - 0.8} = \frac{10}{0.2} = 50 \text{ m}$$
Step 4: Calculate sum of upward distances**
$$S_{\text{up}} = \frac{8}{1 - 0.8} = \frac{8}{0.2} = 40 \text{ m}$$
Step 5: Calculate total distance**
$$S_{\text{total}} = S_{\text{down}} + S_{\text{up}} = 50 + 40 = 90 \text{ m}$$
Verification:**
We can also write:
$$S_{\text{total}} = 10 + 2(8) + 2(6.4) + 2(5.12) + \ldots$$
$$= 10 + 2[8 + 6.4 + 5.12 + \ldots]$$
$$= 10 + 2 \times 40 = 10 + 80 = 90 \text{ m}$$ ✓

∴ The total vertical distance traveled is 90 meters

5

If the sum of an A.P. is $S_n = 2n^2 + 3n$, find the $n^{th}$ term and the common difference.
[5 marks]

Solution

Given: $S_n = 2n^2 + 3n$
Step 1: Find the first term**
$$t_1 = S_1 = 2(1)^2 + 3(1) = 2 + 3 = 5$$
Step 2: Find the second term**
$$t_1 + t_2 = S_2 = 2(2)^2 + 3(2) = 8 + 6 = 14$$
$$t_2 = 14 - 5 = 9$$
Step 3: Find the common difference**
$$d = t_2 - t_1 = 9 - 5 = 4$$
Step 4: Find the general $n^{th}$ term**
For $n \geq 2$:
$$t_n = S_n - S_{n-1}$$
$$= [2n^2 + 3n] - [2(n-1)^2 + 3(n-1)]$$
$$= 2n^2 + 3n - [2(n^2 - 2n + 1) + 3n - 3]$$
$$= 2n^2 + 3n - [2n^2 - 4n + 2 + 3n - 3]$$
$$= 2n^2 + 3n - 2n^2 + 4n - 2 - 3n + 3$$
$$= (3n + 4n - 3n) + (- 2 + 3)$$
$$= 4n + 1$$
Step 5: Verify**
Check if $t_1 = 4(1) + 1 = 5$ ✓
Check if $t_2 = 4(2) + 1 = 9$ ✓
Check if $d = 4$ (constant) ✓
Also verify $S_n$ using the A.P. formula:
$$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(5) + (n-1)(4)]$$
$$= \frac{n}{2}[10 + 4n - 4] = \frac{n}{2}[4n + 6]$$
$$= \frac{n(4n + 6)}{2} = \frac{4n^2 + 6n}{2} = 2n^2 + 3n$$ ✓

∴ The $n^{th}$ term is $t_n = 4n + 1$ and the common difference is $d = 4$

One-Page Revision Cheat Sheet

1

Arithmetic Progression (A.P.): $t_n = a + (n-1)d$ where $a$ = first term, $d$ = common difference. Sum: $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + l)$ where $l$ = last term.

2

Geometric Progression (G.P.): $t_n = ar^{n-1}$ where $a$ = first term, $r$ = common ratio. Sum: $S_n = \frac{a(r^n-1)}{r-1}$ (if $r \neq 1$). For infinite series: $S_{\infty} = \frac{a}{1-r}$ (only if $|r| < 1$).

3

Harmonic Progression (H.P.): Reciprocals form an A.P. To solve H.P. problems: convert to A.P., solve, convert back. General term: $t_n = \frac{1}{a + (n-1)d}$ where the reciprocals have first term $a$ and common difference $d$.

4

Three Means: $A = \frac{a+b}{2}$ (Arithmetic), $G = \sqrt{ab}$ (Geometric), $H = \frac{2ab}{a+b}$ (Harmonic).

5

Golden Relationship: $A \times H = G^2$ and $A > G > H$ (for unequal positive numbers).

6

Inserting Means: For $n$ A.M.s between $a$ and $b$: $d = \frac{b-a}{n+1}$. For $n$ G.M.s: $r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$.

7

Convergence Condition: Infinite G.P. converges only if $|r| < 1$. If $|r| \geq 1$, the series diverges.

8

Common Shortcuts: For 3 terms in A.P. use $(p-d), p, (p+d)$ [sum = $3p$]. For 3 terms in G.P. use $\frac{p}{r}, p, pr$ [product = $p^3$].

9

Repeating Decimals: Convert to fraction using G.P. series. Example: $0.\overline{27} = \frac{27}{99} = \frac{3}{11}$.

10

CRITICAL CHECKS: (1) In A.P.: Does $d$ stay constant? (2) In G.P.: Does $r$ stay constant? (3) For $S_{\infty}$: Is $|r| < 1$? Missing one check loses 2-3 marks!