Comprehensive Study Guide to Sequences and Series

Fundamentals of Sequences

• An ordered collection of numbers or mathematical objects arranged in a definite order according to a specific rule is called a sequence. The individual elements possess well-defined positional ranks, establishing an identifiable first member, second member, third member, and continuing onward.

• Practical examples where sequences arise naturally include:

  • The growth of human or bacterial populations observed at successive regular intervals of time.

  • The accumulated amount of money on deposit in an interest-bearing bank account over a series of years.

  • The depreciated residual value of an asset or commodity calculated over successive accounting periods.

• A sequence is formally defined as a function whose domain is the set of natural numbers $\mathbb{N}$ or a finite subset of natural numbers of the form $\{1, 2, 3, \dots, k\}$.

• The notation $a(n)$ denotes the value of the function for the natural number $n$, conventionally written using subscript notation as $a_n$.

• The distinct elements comprising a sequence are called its terms:

  • Terms are designated as $a_1, a_2, a_3, \dots, a_n, \dots$.

  • Subscripts denote the ordinal position of each term within the sequence.

  • The term occupying the $n\text{th}$ position, denoted $a_n$, is called the $n\text{th}$ term or the general term of the sequence.

• Sequences are classified into two broad categories based on their cardinality:

  • Finite Sequence: A sequence containing a fixed, countable number of terms. For example, counting the number of biological ancestors across $10$ generations (assuming a $30$-year generation gap across $300$ years, yielding $\frac{300}{30} = 10$ generations) produces the sequence $2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$, which terminates at $a_{10} = 1024$.

  • Infinite Sequence: A sequence containing an unending progression of terms. For example, successive decimal approximations obtained by repeatedly dividing $10$ by $3$ generate the non-terminating sequence $3, 3.3, 3.33, 3.333, \dots$.

• Progressions are sequences whose successive terms follow specific, identifiable algebraic patterns.

• In many sequences, individual terms can be generated through an explicit algebraic formula for the general term $a_n$:

  • Even natural numbers: $2, 4, 6, 8, \dots$ has the rule $a_n = 2n$, where $a_1 = 2(1) = 2$, $a_2 = 2(2) = 4$, $a_{23} = 2(23) = 46$, and $a_{24} = 2(24) = 48$.

  • Odd natural numbers: $1, 3, 5, 7, \dots$ has the rule $a_n = 2n - 1$.

• Certain sequences lack an explicit non-recursive algebraic formula but are defined by a recurrence relation:

  • Fibonacci Sequence: An arrangement defined by initial conditions and a recursive relation:     $a_1 = 1$     $a_2 = 1$     $a_n = a_{n-2} + a_{n-1} \quad \text{for } n > 2$     This generates the terms $1, 1, 2, 3, 5, 8, 13, 21, \dots$.

• Some sequences cannot be generated by any algebraic formula or simple recurrence relation. The sequence of prime numbers $2, 3, 5, 7, 11, 13, \dots$ has no single algebraic formula for the $n\text{th}$ prime and must be characterized by a verbal description.

Concept and Notation of Series

• For any given sequence $a_1, a_2, a_3, \dots, a_n, \dots$, the formal algebraic expression representing the sum of its terms:   $a_1 + a_2 + a_3 + \dots + a_n + \dots$   is defined as the series associated with the given sequence.

• A series is classified as finite or infinite according to whether the underlying sequence is finite or infinite.

• Summation Notation (Sigma Notation): Series are compactly represented using the uppercase Greek letter $\Sigma$ (sigma):   $\sum_{k=1}^n a_k = a_1 + a_2 + a_3 + \dots + a_n$

• Conceptual Distinction Between a Series and the Sum of a Series:

  • The term series refers specifically to the formal indicated addition of terms (e.g., $1 + 3 + 5 + 7$).

  • The phrase sum of a series refers to the resulting single numerical value obtained upon evaluating the addition (e.g., the sum of $1 + 3 + 5 + 7$ is $16$).

Analysis of Illustrative Problems on Sequences and Series

• Determining Initial Terms from an Explicit Formula:

  • For the sequence defined by $a_n = 2n + 5$:     $a_1 = 2(1) + 5 = 7$     $a_2 = 2(2) + 5 = 9$     $a_3 = 2(3) + 5 = 11$     The first three terms are $7, 9, 11$.

  • For the sequence defined by $a_n = \frac{n-3}{4}$:     $a_1 = \frac{1-3}{4} = -\frac{2}{4} = -\frac{1}{2}$     $a_2 = \frac{2-3}{4} = -\frac{1}{4}$     $a_3 = \frac{3-3}{4} = 0$     The first three terms are $-\frac{1}{2}, -\frac{1}{4}, 0$.

• Calculating High-Order Specific Terms:

  • For $a_n = (n-1)(2-n)(3+n)$, calculating the $20\text{th}$ term requires substituting $n = 20$:     $a_{20} = (20 - 1)(2 - 20)(3 + 20) = (19)(-18)(23) = -7866$

• Evaluating Sequences and Constructing Associated Series from Recurrence Relations:

  • For $a_1 = 1$ with $a_n = a_{n-1} + 2$ for $n \ge 2$:     $a_2 = a_1 + 2 = 1 + 2 = 3$     $a_3 = a_2 + 2 = 3 + 2 = 5$     $a_4 = a_3 + 2 = 5 + 2 = 7$     $a_5 = a_4 + 2 = 7 + 2 = 9$     The first five terms are $1, 3, 5, 7, 9$, and the corresponding series is $1 + 3 + 5 + 7 + 9 + \dots$.

• Fundamental Problems of Exercise 8.1:

  • Finding the first five terms of sequences defined by general terms:

  • $a_n = n(n+2) \implies 3, 8, 15, 24, 35$

  • $a_n = \frac{n}{n+1} \implies \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$

  • $a_n = 2^n \implies 2, 4, 8, 16, 32$

  • $a_n = \frac{2n-3}{6} \implies -\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}$

  • $a_n = (-1)^{n-1} 5^{n+1} \implies 25, -125, 625, -3125, 15625$

  • $a_n = \frac{n(n^2+5)}{4} \implies \frac{3}{2}, \frac{9}{2}, 14, 21, \frac{75}{2}$

  • Evaluating designated terms:

  • $a_n = 4n - 3$; $a_{17} = 4(17) - 3 = 65$, $a_{24} = 4(24) - 3 = 93$

  • $a_n = \frac{n^2}{2^n}$; $a_7 = \frac{7^2}{2^7} = \frac{49}{128}$

  • $a_n = (-1)^{n-1} n^3$; $a_9 = (-1)^{9-1} (9^3) = 729$

  • $a_n = \frac{n(n-2)}{n+3}$; $a_{20} = \frac{20(20-2)}{20+3} = \frac{360}{23}$

  • Generating terms and series under given recurrence bounds:

  • $a_1 = 3$, $a_n = 3a_{n-1} + 2$ for $n > 1$ gives terms $3, 11, 35, 107, 323$ and series $3 + 11 + 35 + 107 + 323 + \dots$

  • $a_1 = -1$, $a_n = \frac{a_{n-1}}{n}$ for $n \ge 2$ gives terms $-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}$ and series $(-1) + \left(-\frac{1}{2}\right) + \left(-\frac{1}{6}\right) + \left(-\frac{1}{24}\right) + \left(-\frac{1}{120}\right) + \dots$

  • $a_1 = a_2 = 2$, $a_n = a_{n-1} - 1$ for $n > 2$ gives terms $2, 2, 1, 0, -1$ and series $2 + 2 + 1 + 0 + (-1) + \dots$

  • Evaluating successive Fibonacci ratios $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$:

  • Given $a_1 = 1$, $a_2 = 1$, $a_3 = 2$, $a_4 = 3$, $a_5 = 5$, $a_6 = 8$:

    • For $n = 1$: $\frac{a_2}{a_1} = \frac{1}{1} = 1$

    • For $n = 2$: $\frac{a_3}{a_2} = \frac{2}{1} = 2$

    • For $n = 3$: $\frac{a_4}{a_3} = \frac{3}{2}$

    • For $n = 4$: $\frac{a_5}{a_4} = \frac{5}{3}$

    • For $n = 5$: $\frac{a_6}{a_5} = \frac{8}{5}$

Theory of Geometric Progression (G.P.)

• A sequence of non-zero numbers $a_1, a_2, a_3, \dots, a_n, \dots$ is called a geometric sequence or geometric progression (abbreviated G.P.) if each term is non-zero and bears a constant ratio to the term immediately preceding it:   $\frac{a_{k+1}}{a_k} = r \quad \text{for all } k \ge 1$

• The constant multiplier $r$ is defined as the common ratio of the G.P.

• Demonstrative examples of geometric progressions and their common ratios:

  • Sequence $2, 4, 8, 16, \dots$ has common ratio $r = \frac{4}{2} = \frac{8}{4} = 2$.

  • Sequence $-\frac{1}{9}, \frac{1}{27}, -\frac{1}{81}, \frac{1}{243}, \dots$ has common ratio $r = \frac{1/27}{-1/9} = -\frac{1}{3}$.

  • Sequence $0.01, 0.0001, 0.000001, \dots$ has common ratio $r = \frac{0.0001}{0.01} = 0.01$.

• Canonical Form of a Geometric Progression:

  • Designating the initial term $a_1$ as $a$, the sequence is expressed as:     $a, ar, ar^2, ar^3, \dots, ar^{n-1}, \dots$

• Standard Mathematical Notations for Progressions:

  • $a$: The first term.

  • $r$: The common ratio.

  • $l$: The last term.

  • $n$: The total count of terms.

  • $S_n$: The sum of the first $n$ terms.

• Derivation of the General Term ($n\text{th}$ Term):

  • First term: $a_1 = a = a r^{1-1}$

  • Second term: $a_2 = a_1 r = ar = a r^{2-1}$

  • Third term: $a_3 = a_2 r = (ar)r = ar^2 = a r^{3-1}$

  • Fourth term: $a_4 = a_3 r = (ar^2)r = ar^3 = a r^{4-1}$

  • Fifth term: $a_5 = a_4 r = (ar^3)r = ar^4 = a r^{5-1}$

  • By mathematical induction and pattern identification, the $n\text{th}$ term is:     $a_n = a r^{n-1}$

• Structural Forms of Geometric Progressions:

  • Finite G.P.: $a, ar, ar^2, ar^3, \dots, ar^{n-1}$

  • Infinite G.P.: $a, ar, ar^2, ar^3, \dots, ar^{n-1}, \dots$

  • Finite Geometric Series: $a + ar + ar^2 + \dots + ar^{n-1}$

  • Infinite Geometric Series: $a + ar + ar^2 + \dots + ar^{n-1} + \dots$

Summation Formulas for Geometric Progressions

• Let $S_n$ denote the sum of the first $n$ terms of a geometric progression:   $S_n = a + ar + ar^2 + ar^3 + \dots + ar^{n-1}$

• Case 1: When the common ratio is unity ($r = 1$):   $S_n = a + a + a + \dots + a \ (n \text{ terms}) = na$

• Case 2: When the common ratio is not unity ($r \ne 1$):

  • Multiplying the entire summation expression by $r$ yields:     $r S_n = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n$

  • Subtracting this equation from the original summation equation:     $S_n - r S_n = (a + ar + ar^2 + \dots + ar^{n-1}) - (ar + ar^2 + \dots + ar^n)$     $(1 - r) S_n = a - ar^n = a(1 - r^n)$

  • Dividing by $(1 - r)$ gives the standard summation formula:     $S_n = \frac{a(1 - r^n)}{1 - r}$

  • Equivalently, multiplying both numerator and denominator by $-1$ yields:     $S_n = \frac{a(r^n - 1)}{r - 1}$

• Exemplary Applications and Step-by-Step Solutions:

  • Finding High-Order and General Terms (Example 4):

  • In the progression $5, 25, 125, \dots$, $a = 5$ and $r = 5$.

  • The $10\text{th}$ term is $a_{10} = 5(5)^{10-1} = 5(5^9) = 5^{10}$.

  • The general term is $a_n = 5(5)^{n-1} = 5^n$.

  • Determining the Number of Terms Given the Final Term (Example 5):

  • In $2, 8, 32, \dots$, determine which term equals $131072$.

  • Here $a = 2$ and $r = 4$.

  • Setting $a_n = 131072 \implies 2(4)^{n-1} = 131072$.

  • $4^{n-1} = 65536$.

  • Since $4^8 = 65536$, equating exponents yields $n - 1 = 8 \implies n = 9$.

  • Thus, $131072$ is the $9\text{th}$ term of the progression.

  • Deducing the Progression from Non-Consecutive Terms (Example 6):

  • Given $a_3 = 24$ and $a_6 = 192$, determine $a_{10}$.

  • $a r^2 = 24$ and $a r^5 = 192$.

  • Dividing yields $\frac{a r^5}{a r^2} = r^3 = \frac{192}{24} = 8 \implies r = 2$.

  • Substituting back: $a(2^2) = 24 \implies 4a = 24 \implies a = 6$.

  • The $10\text{th}$ term is $a_{10} = a r^9 = 6(2^9) = 6(512) = 3072$.

  • Summing Fractions in a Geometric Series (Example 7):

  • For the series $1 + \frac{2}{3} + \frac{4}{9} + \dots$, $a = 1$ and $r = \frac{2}{3}$.

  • The sum of $n$ terms is:       $S_n = \frac{1\left[1 - \left(\frac{2}{3}\right)^n\right]}{1 - \frac{2}{3}} = 3\left[1 - \left(\frac{2}{3}\right)^n\right]$

  • For the first $5$ terms:       $S_5 = 3\left[1 - \left(\frac{2}{3}\right)^5\right] = 3\left[1 - \frac{32}{243}\right] = 3\left(\frac{211}{243}\right) = \frac{211}{81}$

  • Solving for the Index from a Rational Sum (Example 8):

  • For $3, \frac{3}{2}, \frac{3}{4}, \dots$, find $n$ such that $S_n = \frac{3069}{512}$.

  • Here $a = 3$ and $r = \frac{1}{2}$.

  • $S_n = \frac{3\left[1 - \left(\frac{1}{2}\right)^n\right]}{1 - \frac{1}{2}} = 6\left(1 - \frac{1}{2^n}\right)$.

  • Setting $6\left(1 - \frac{1}{2^n}\right) = \frac{3069}{512} \implies 1 - \frac{1}{2^n} = \frac{3069}{3072}$.

  • $\frac{1}{2^n} = 1 - \frac{3069}{3072} = \frac{3}{3072} = \frac{1}{1024}$.

  • $2^n = 1024 = 2^{10} \implies n = 10$.

  • Determining Three Consecutive Terms via System of Equations (Example 9):

  • Let the three terms be represented symmetrically as $\frac{a}{r}, a, ar$.

  • Given product: $\left(\frac{a}{r}\right)(a)(ar) = a^3 = -1 \implies a = -1$ (real root).

  • Given sum: $\frac{a}{r} + a + ar = \frac{13}{12}$.

  • Substituting $a = -1$:       $-\frac{1}{r} - 1 - r = \frac{13}{12} \implies -\left(\frac{1 + r + r^2}{r}\right) = \frac{13}{12}$       $12(1 + r + r^2) = -13r \implies 12r^2 + 25r + 12 = 0$

  • Factoring the quadratic: $(3r + 4)(4r + 3) = 0 \implies r = -\frac{3}{4} \text{ or } r = -\frac{4}{3}$.

  • If $r = -\frac{3}{4}$, the terms are $\frac{4}{3}, -1, \frac{3}{4}$.

  • If $r = -\frac{4}{3}$, the terms are $\frac{3}{4}, -1, \frac{4}{3}$.

  • Summing Repunit-Derived Composite Sequences (Example 10):

  • To find the sum of $7 + 77 + 777 + 7777 + \dots$ to $n$ terms:       $S_n = 7(1 + 11 + 111 + 1111 + \dots \text{ to } n \text{ terms})$

  • Multiplying and dividing by $9$:       $S_n = \frac{7}{9}(9 + 99 + 999 + 9999 + \dots \text{ to } n \text{ terms})$

  • Expressing each component as a power of $10$ minus $1$:       $S_n = \frac{7}{9}\left[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)\right]$       $S_n = \frac{7}{9}\left[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ to } n \text{ terms})\right]$

  • Applying the geometric series formula to the first bracket ($a = 10, r = 10$):       $S_n = \frac{7}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right] = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$

  • Generational Lineage Counting (Example 11):

  • A person has $2$ parents, $4$ grandparents, $8$ great-grandparents, continuing over $10$ preceding generations.

  • Here $a = 2, r = 2, n = 10$.

  • Applying the sum formula:       $S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2(1023) = 2046$

  • The individual has a total of $2046$ ancestors across those ten generations.

Geometric Mean (G.M.) and Insertion of Means

• Definition of the Geometric Mean of Two Positive Numbers:

  • For two positive real numbers $a$ and $b$, their geometric mean is defined as:     $G = \sqrt{ab}$

  • The sequence $a, \sqrt{ab}, b$ constitutes a geometric progression because:     $\frac{\sqrt{ab}}{a} = \sqrt{\frac{b}{a}} = \frac{b}{\sqrt{ab}}$

  • For instance, the geometric mean of $2$ and $8$ is $\sqrt{2 \times 8} = \sqrt{16} = 4$, forming the G.P. $2, 4, 8$.

• Insertion of $n$ Geometric Means Between Two Positive Numbers:

  • Let $G_1, G_2, G_3, \dots, G_n$ be $n$ positive numbers inserted between $a$ and $b$ such that:     $a, G_1, G_2, G_3, \dots, G_n, b$     forms a geometric progression.

  • The complete progression contains $n + 2$ terms.

  • The term $b$ occupies the $(n+2)\text{th}$ position:     $b = a r^{(n+2)-1} = a r^{n+1}$

  • Solving for the common ratio $r$:     $r^{n+1} = \frac{b}{a} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$

  • The explicit formulas for the inserted geometric means are:     $G_1 = ar = a \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$     $G_2 = ar^2 = a \left(\frac{b}{a}\right)^{\frac{2}{n+1}}$     $G_3 = ar^3 = a \left(\frac{b}{a}\right)^{\frac{3}{n+1}}$     $G_k = ar^k = a \left(\frac{b}{a}\right)^{\frac{k}{n+1}}$     $G_n = ar^n = a \left(\frac{b}{a}\right)^{\frac{n}{n+1}}$

• Practical Insertion of Means (Example 12):

  • Insert three geometric means between $1$ and $256$.

  • Here $a = 1$, $b = 256$, and $n = 3$.

  • The total number of terms is $3 + 2 = 5$.

  • $b = a r^4 \implies 256 = 1(r^4) \implies r^4 = 256$.

  • Taking real roots yields $r = \pm 4$.

  • For the positive ratio $r = 4$:     $G_1 = 1(4) = 4$     $G_2 = 1(4^2) = 16$     $G_3 = 1(4^3) = 64$     The resulting progression is $1, 4, 16, 64, 256$.

  • For the alternating negative ratio $r = -4$:     $G_1 = -4$     $G_2 = 16$     $G_3 = -64$     The resulting progression is $1, -4, 16, -64, 256$.

Relationship Between Arithmetic Mean and Geometric Mean

• Formal Definitions for Two Positive Numbers $a$ and $b$:

  • Arithmetic Mean (A.M.): $A = \frac{a+b}{2}$

  • Geometric Mean (G.M.): $G = \sqrt{ab}$

• Proof of the Fundamental Inequality $A \ge G$:

  • Subtracting the geometric mean from the arithmetic mean:     $A - G = \frac{a+b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2}$

  • Recognizing the numerator as a perfect square:     $A - G = \frac{(\sqrt{a} - \sqrt{b})^2}{2}$

  • Since the square of any real number is non-negative, $\frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$.

  • Consequently:     $A - G \ge 0 \implies A \ge G$

  • The equality condition $A = G$ holds if and only if $\sqrt{a} - \sqrt{b} = 0$, which implies $a = b$.

• Determining Unknown Numbers from Their Means (Example 13):

  • Given that the A.M. and G.M. of two positive numbers $a$ and $b$ are $10$ and $8$, respectively:     $A = \frac{a+b}{2} = 10 \implies a + b = 20$     $G = \sqrt{ab} = 8 \implies ab = 64$

  • Using the algebraic identity for the difference of two numbers:     $(a - b)^2 = (a + b)^2 - 4ab$     $(a - b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$     $a - b = \pm 12$

  • Solving simultaneously:

  • If $a - b = 12$ and $a + b = 20$:       $2a = 32 \implies a = 16, \quad b = 4$

  • If $a - b = -12$ and $a + b = 20$:       $2a = 8 \implies a = 4, \quad b = 16$

  • Thus, the two numbers are $4$ and $16$.

Detailed Problem Analysis from Exercise 8.2

• Basic Term Evaluation and Progression Properties:

  • Problem 1: For the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$:

  • Initial term $a = \frac{5}{2}$, common ratio $r = \frac{5/4}{5/2} = \frac{1}{2}$.

  • $20\text{th}$ term: $a_{20} = a r^{19} = \left(\frac{5}{2}\right)\left(\frac{1}{2}\right)^{19} = \frac{5}{2^{20}}$.

  • General term: $a_n = a r^{n-1} = \left(\frac{5}{2}\right)\left(\frac{1}{2}\right)^{n-1} = \frac{5}{2^n}$.

  • Problem 2: Find $a_{12}$ given $a_8 = 192$ and $r = 2$:

  • $a_8 = a r^7 = 192 \implies a(2^7) = 192 \implies 128a = 192 \implies a = \frac{3}{2}$.

  • $a_{12} = a r^{11} = \left(\frac{3}{2}\right)(2^{11}) = 3 \times 2^{10} = 3072$.

  • Problem 3: Let the $5\text{th}$, $8\text{th}$, and $11\text{th}$ terms be $p$, $q$, and $s$:

  • $p = a r^4$, $q = a r^7$, $s = a r^{10}$.

  • Squaring the middle term: $q^2 = (a r^7)^2 = a^2 r^{14}$.

  • Multiplying the outer terms: $ps = (a r^4)(a r^{10}) = a^2 r^{14}$.

  • Therefore, $q^2 = ps$.

  • Problem 4: Given $a_4 = (a_2)^2$ and $a = -3$:

  • $a r^3 = (a r)^2 = a^2 r^2 \implies r = a = -3$.

  • Determination of the $7\text{th}$ term:       $a_7 = a r^6 = (-3)(-3)^6 = -3^7 = -2187$

  • Problem 5: Identifying the index of designated values:

  • (a) For $2, 2\sqrt{2}, 4, \dots$, find $n$ when $a_n = 128$:       $a = 2, r = \sqrt{2} \implies 2(\sqrt{2})^{n-1} = 128 \implies (\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12}$       $n - 1 = 12 \implies n = 13$

  • (b) For $\sqrt{3}, 3, 3\sqrt{3}, \dots$, find $n$ when $a_n = 729$:       $a = \sqrt{3}, r = \sqrt{3} \implies (\sqrt{3})^n = 729 = 3^6 = (\sqrt{3})^{12} \implies n = 12$

  • (c) For $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$, find $n$ when $a_n = \frac{1}{19683}$:       $a = \frac{1}{3}, r = \frac{1}{3} \implies \left(\frac{1}{3}\right)^n = \frac{1}{19683} = \frac{1}{3^9} \implies n = 9$

  • Problem 6: Value of $x$ ensuring $-\frac{2}{7}, x, -\frac{7}{2}$ form a G.P.:

  • Setting equal ratios: $\frac{x}{-2/7} = \frac{-7/2}{x} \implies x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1 \implies x = \pm 1$.

• Geometric Progression Summations:

  • Problem 7: Sum of $20$ terms of $0.15, 0.015, 0.0015, \dots$:

  • $a = 0.15$, $r = 0.1$.

  • $S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1} = \frac{0.15}{0.9}(1 - (0.1)^{20}) = \frac{1}{6}\left[1 - (0.1)^{20}\right]$.

  • Problem 8: Sum of $n$ terms of $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots$:

  • $a = \sqrt{7}$, $r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3}$.

  • $S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)}{\sqrt{3} - 1} = \frac{\sqrt{7}(\sqrt{3} + 1)\left(3^{n/2} - 1\right)}{2}$.

  • Problem 9: Sum of $n$ terms of $1, -a, a^2, -a^3, \dots$ ($a \ne -1$):

  • First term is $1$, common ratio is $-a$.

  • $S_n = \frac{1(1 - (-a)^n)}{1 - (-a)} = \frac{1 - (-a)^n}{1 + a}$.

• Problem 10: Sum of $n$ terms of $x^3, x^5, x^7, \dots$ ($x \ne \pm 1$):

  • First term is $x^3$, common ratio is $x^2$.

  • $S_n = \frac{x^3(1 - (x^2)^n)}{1 - x^2} = \frac{x^3(1 - x^{2n})}{1 - x^2}$.

• Problem 11: Evaluate $\sum_{k=1}^{11} (2 + 3^k)$:

  • Splitting the sum:     $\sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k = 2(11) + \frac{3(3^{11} - 1)}{3 - 1}$     $22 + \frac{3(177147 - 1)}{2} = 22 + \frac{3(177146)}{2} = 22 + 265719 = 265741$

• Problem 12: Sum of first three terms is $\frac{39}{10}$ and their product is $1$:

  • Let the terms be $\frac{a}{r}, a, ar$.

  • Product gives $a^3 = 1 \implies a = 1$.

  • Sum gives $\frac{1}{r} + 1 + r = \frac{39}{10} \implies \frac{r^2 + r + 1}{r} = \frac{39}{10}$.

  • $10r^2 + 10r + 10 = 39r \implies 10r^2 - 29r + 10 = 0$.

  • Factoring: $(2r - 5)(5r - 2) = 0 \implies r = \frac{5}{2} \text{ or } r = \frac{2}{5}$.

  • The terms are $\frac{2}{5}, 1, \frac{5}{2}$ or $\frac{5}{2}, 1, \frac{2}{5}$.

• Problem 13: Terms of $3, 3^2, 3^3, \dots$ needed to yield sum $120$:

  • $a = 3, r = 3$.

  • $S_n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2} = 120 \implies 3^n - 1 = 80 \implies 3^n = 81 \implies n = 4$.

• Problem 14: Sum of first three terms is $16$, and sum of next three terms is $128$:

  • $a(1 + r + r^2) = 16$

  • $a r^3 + a r^4 + a r^5 = a r^3(1 + r + r^2) = 128$

  • Dividing yields $r^3 = \frac{128}{16} = 8 \implies r = 2$.

  • Substituting gives $a(1 + 2 + 4) = 16 \implies 7a = 16 \implies a = \frac{16}{7}$.

  • Sum to $n$ terms:     $S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1} = \frac{16}{7}(2^n - 1)$

• Problem 15: Given $a = 729$ and $a_7 = 64$, find $S_7$:

  • $a r^6 = 64 \implies 729 r^6 = 64 \implies r^6 = \frac{64}{729} = \left(\frac{2}{3}\right)^6 \implies r = \frac{2}{3}$.

  • $S_7 = \frac{729\left[1 - \left(\frac{2}{3}\right)^7\right]}{1 - \frac{2}{3}} = 3(729)\left(1 - \frac{128}{2187}\right) = 2187\left(\frac{2059}{2187}\right) = 2059$.

• Problem 16: Sum of first two terms is $-4$, and fifth term is four times third term:

  • $a_5 = 4 a_3 \implies a r^4 = 4 a r^2 \implies r^2 = 4 \implies r = \pm 2$.

  • Condition on sum: $a + ar = a(1 + r) = -4$.

  • If $r = 2$, $a(1 + 2) = -4 \implies a = -\frac{4}{3}$. G.P.: $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$

  • If $r = -2$, $a(1 - 2) = -4 \implies -a = -4 \implies a = 4$. G.P.: $4, -8, 16, -32, \dots$

• Problem 17: If $a_4 = x, a_{10} = y, a_{16} = z$, prove $x, y, z$ are in G.P.:

  • $x = a r^3$, $y = a r^9$, $z = a r^{15}$.

  • $y^2 = (a r^9)^2 = a^2 r^{18}$.

  • $xz = (a r^3)(a r^{15}) = a^2 r^{18}$.

  • Since $y^2 = xz$, $x, y, z$ are confirmed to be in G.P.

• Problem 18: Sum to $n$ terms of $8, 88, 888, 8888, \dots$:   $S_n = \frac{8}{9}\left[(10-1) + (10^2-1) + \dots + (10^n-1)\right] = \frac{8}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$

• Problem 19: Sum of products of corresponding terms of $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \frac{1}{2}$:

  • The resulting terms are:     $2 \times 128 = 256$     $4 \times 32 = 128$     $8 \times 8 = 64$     $16 \times 2 = 32$     $32 \times \frac{1}{2} = 16$

  • This generates a G.P. with $a = 256, r = \frac{1}{2}, n = 5$.

  • Sum:     $S_5 = \frac{256\left(1 - (1/2)^5\right)}{1 - 1/2} = 512\left(1 - \frac{1}{32}\right) = 512\left(\frac{31}{32}\right) = 496$

• Problem 20: Products of corresponding terms of two general G.P.s:

  • Given $a, ar, ar^2, \dots, ar^{n-1}$ and $A, AR, AR^2, \dots, AR^{n-1}$.

  • Products form the sequence $aA, (ar)(AR), (ar^2)(AR^2), \dots, (ar^{n-1})(AR^{n-1})$.

  • Ratio of any term to its immediate predecessor:     $\frac{(a r^k)(A R^k)}{(a r^{k-1})(A R^{k-1})} = rR$

  • Since this ratio is constant for all terms, the product sequence is a G.P. with common ratio $rR$.

• Problem 21: Four numbers in G.P. where $a_3 = a_1 + 9$ and $a_2 = a_4 + 18$:

  • $a r^2 = a + 9 \implies a(r^2 - 1) = 9$

  • $ar = a r^3 + 18 \implies ar(1 - r^2) = 18 \implies -ar(r^2 - 1) = 18$

  • Dividing the second equation by the first: $-r = 2 \implies r = -2$.

  • Substituting back: $a((-2)^2 - 1) = 9 \implies 3a = 9 \implies a = 3$.

  • The four numbers are $3, -6, 12, -24$.

• Problem 22: Given $a_p = a$, $a_q = b$, $a_r = c$, prove $a^{q-r} b^{r-p} c^{p-q} = 1$:

  • Let the first term of the G.P. be $A_0$ and common ratio be $R$.

  • $a = A_0 R^{p-1}, \quad b = A_0 R^{q-1}, \quad c = A_0 R^{r-1}$.

  • Substituting into the expression:     $a^{q-r} b^{r-p} c^{p-q} = (A_0 R^{p-1})^{q-r} (A_0 R^{q-1})^{r-p} (A_0 R^{r-1})^{p-q}$     $= A_0^{(q-r) + (r-p) + (p-q)} R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$     $= A_0^0 R^{(pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q)} = A_0^0 R^0 = 1$

• Problem 23: First term is $a$, $n\text{th}$ term is $b$, and product of $n$ terms is $P$. Prove $P^2 = (ab)^n$:

  • $b = a r^{n-1}$.

  • $P = a \cdot (ar) \cdot (ar^2) \dots (ar^{n-1}) = a^n r^{1 + 2 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}}$.

  • Squaring $P$:     $P^2 = a^{2n} r^{n(n-1)} = \left(a^2 r^{n-1}\right)^n = \left(a \cdot ar^{n-1}\right)^n = (ab)^n$

• Problem 24: Ratio of sum of first $n$ terms to sum of terms from $(n+1)\text{th}$ to $(2n)\text{th}$:

  • Let $S_1 = a + ar + \dots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r}$.

  • Sum of the next $n$ terms:     $S_2 = a_{n+1} + a_{n+2} + \dots + a_{2n} = ar^n + ar^{n+1} + \dots + ar^{2n-1}$     $S_2 = r^n(a + ar + \dots + ar^{n-1}) = r^n S_1$

  • Ratio:     $\frac{S_1}{S_2} = \frac{S_1}{r^n S_1} = \frac{1}{r^n}$

• Problem 25: If $a, b, c, d$ are in G.P., prove $(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$:

  • Let $b = ar, c = ar^2, d = ar^3$.

  • First factor:     $a^2 + b^2 + c^2 = a^2 + a^2 r^2 + a^2 r^4 = a^2(1 + r^2 + r^4)$

  • Second factor:     $b^2 + c^2 + d^2 = a^2 r^2 + a^2 r^4 + a^2 r^6 = a^2 r^2(1 + r^2 + r^4)$

  • Product of factors (L.H.S.):     $\text{L.H.S.} = a^4 r^2 (1 + r^2 + r^4)^2$

  • Evaluating R.H.S.:     $ab + bc + cd = a(ar) + (ar)(ar^2) + (ar^2)(ar^3) = a^2 r + a^2 r^3 + a^2 r^5 = a^2 r(1 + r^2 + r^4)$     $\text{R.H.S.} = [a^2 r(1 + r^2 + r^4)]^2 = a^4 r^2 (1 + r^2 + r^4)^2$

  • L.H.S. equals R.H.S.

• Problem 26: Insert two numbers between $3$ and $81$ to form a G.P.:

  • $a = 3, b = 81, n = 2$. Total terms = $4$.

  • $b = a r^3 \implies 81 = 3 r^3 \implies r^3 = 27 \implies r = 3$.

  • $G_1 = ar = 3(3) = 9$, $G_2 = ar^2 = 3(3^2) = 27$.

  • The inserted numbers are $9$ and $27$.

• Problem 27: Find $n$ such that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$ is the geometric mean between $a$ and $b$:

  • $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab} = a^{1/2} b^{1/2}$.

  • $a^{n+1} + b^{n+1} = (a^n + b^n) a^{1/2} b^{1/2} = a^{n + 1/2} b^{1/2} + a^{1/2} b^{n + 1/2}$.

  • $a^{n+1} - a^{n+1/2} b^{1/2} = b^{n+1/2} a^{1/2} - b^{n+1}$.

  • $a^{n+1/2}(a^{1/2} - b^{1/2}) = b^{n+1/2}(a^{1/2} - b^{1/2})$.

  • Dividing by $(a^{1/2} - b^{1/2})$ gives $a^{n+1/2} = b^{n+1/2} \implies \left(\frac{a}{b}\right)^{n+1/2} = 1 = \left(\frac{a}{b}\right)^0$.

  • Equating exponents: $n + \frac{1}{2} = 0 \implies n = -\frac{1}{2}$.

• Problem 28: Given the sum of two positive numbers is $6$ times their geometric mean, prove their ratio is $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$:

  • Given $a + b = 6\sqrt{ab} \implies \frac{a+b}{2\sqrt{ab}} = 3$.

  • Applying Componendo and Dividendo:     $\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2$     $\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 2 \implies \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{2}}{1}$

  • Applying Componendo and Dividendo again:     $\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$     $\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

  • Squaring both sides:     $\frac{a}{b} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{2 + 1 + 2\sqrt{2}}{2 + 1 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$

• Problem 29: Prove that two numbers with A.M. $A$ and G.M. $G$ are $A \pm \sqrt{(A+G)(A-G)}$:

  • $a + b = 2A$ and $ab = G^2$.

  • $(a - b)^2 = (a + b)^2 - 4ab = 4A^2 - 4G^2 = 4(A^2 - G^2)$.

  • Taking the square root: $a - b = 2\sqrt{A^2 - G^2} = 2\sqrt{(A+G)(A-G)}$.

  • Adding and subtracting the expressions for $(a + b)$ and $(a - b)$:     $2a = 2A + 2\sqrt{(A+G)(A-G)} \implies a = A + \sqrt{(A+G)(A-G)}$     $2b = 2A - 2\sqrt{(A+G)(A-G)} \implies b = A - \sqrt{(A+G)(A-G)}$

  • Thus, the numbers are $A \pm \sqrt{(A+G)(A-G)}$.

• Problem 30: Population of bacteria doubling every hour with initial population $30$:

  • Initial population: $a_0 = 30$.

  • Population at end of $1\text{st}$ hour: $30(2) = 60$.

  • Population at end of $2\text{nd}$ hour: $30(2^2) = 120$.

  • Population at end of $4\text{th}$ hour: $30(2^4) = 480$.

  • Population at end of $n\text{th}$ hour: $30(2^n)$.

• Problem 31: Value of $\text{Rs } 500$ after $10$ years at $10\%$ compounded annually:

  • The compound interest formula yields $A = P(1 + i)^n$.

  • With $P = 500$, $i = 0.1$, and $n = 10$:     $A = 500(1 + 0.1)^{10} = 500(1.1)^{10}$

• Problem 32: Reconstructing a quadratic equation whose roots have A.M. $8$ and G.M. $5$:

  • Let the roots be $\alpha$ and $\beta$.

  • $A = \frac{\alpha + \beta}{2} = 8 \implies \alpha + \beta = 16$.

  • $G = \sqrt{\alpha \beta} = 5 \implies \alpha \beta = 25$.

  • The quadratic equation is $x^2 - (\alpha + \beta)x + \alpha \beta = 0$:     $x^2 - 16x + 25 = 0$