Sequence and Series Comprehensive Study Guide

Section 1: Arithmetic Progression and Arithmetic Mean

Core Concepts and Fundamental Properties

• An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is a constant, known as the common difference $d$.

• For three terms $a, b, c$ in AP:

  • The common difference condition yields $b - a = c - b$, which simplifies to $2b = a + c$.

  • The single Arithmetic Mean (A.M.) between $a$ and $b$ is defined as $A = \frac{a+b}{2}$.

Question Solutions and Problem-Solving Methodology

Problem 1: Algebraic Identity for Terms in AP

• Question: If $a, b, c$ are in AP, then $(a - c)^2$ equals:

  • (A) $4(b^2 - ac)$

  • (B) $4(b^2 + ac)$

  • (C) $4b^2 - ac$

  • (D) $b - 4ac$

• Detailed Derivation:

  • Since $a, b, c$ are in AP, $2b = a + c$.

  • Express $(a - c)^2$ using the standard algebraic identity:     $(a - c)^2 = (a + c)^2 - 4ac$

  • Substitute $a + c = 2b$ into the expression:     $(a - c)^2 = (2b)^2 - 4ac = 4b^2 - 4ac = 4(b^2 - ac)$

• Correct Answer: (A) $4(b^2 - ac)$

Problem 2: Cyclic Sum of Terms in AP

• Question: If the $p\text{th}$, $q\text{th}$, and $r\text{th}$ term of an arithmetic sequence are $a, b$, and $c$ respectively, then the value of $[a(q - r) + b(r - p) + c(p - q)]$ is:

  • (A) $1$

  • (B) $-1$

  • (C) $0$

  • (D) $1/2$

• Detailed Derivation:

  • Let $A$ be the first term and $D$ be the common difference of the AP.

  • The terms are expressed as:     $a = A + (p - 1)D$     $b = A + (q - 1)D$     $c = A + (r - 1)D$

  • Substitute these expressions into the cyclic sum:     $a(q - r) + b(r - p) + c(p - q)$     $= [A + (p - 1)D](q - r) + [A + (q - 1)D](r - p) + [A + (r - 1)D](p - q)$

  • Group the terms by $A$ and $D$:     $= A[(q - r) + (r - p) + (p - q)] + D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]$

  • The coefficient of $A$ is zero: $(q - r + r - p + p - q) = 0$.

  • Expand the coefficient of $D$:     $p(q - r) - (q - r) + q(r - p) - (r - p) + r(p - q) - (p - q)$     $= (pq - pr + qr - qp + rp - rq) - (q - r + r - p + p - q) = 0 - 0 = 0$

  • Thus, the entire expression evaluates to $0$

• Correct Answer: (C) $0$

Problem 3: Sum of Terms from General Term

• Question: If the $n\text{th}$ term of an AP is $\frac{1}{3}(2n + 1)$, then the sum of its $19$ terms is:

  • (A) $131$

  • (B) $132$

  • (C) $133$

  • (D) $134$

• Detailed Derivation:

  • Given general term $T_n = \frac{1}{3}(2n + 1)$.

  • Compute the first term $T_1$:     $T_1 = \frac{1}{3}(2(1) + 1) = \frac{3}{3} = 1$

  • Compute the $19\text{th}$ term $T_{19}$:     $T_{19} = \frac{1}{3}(2(19) + 1) = \frac{39}{3} = 13$

  • Apply the AP sum formula $S_n = \frac{n}{2}(T_1 + T_n)$ for $n = 19$:     $S_{19} = \frac{19}{2}(1 + 13) = \frac{19}{2}(14) = 19 \times 7 = 133$

• Correct Answer: (C) $133$

Problem 4: AP with Equal Sums of $p$ and $q$ Terms

• Question: If the sum of $p$ terms of an AP is equal to the sum of its $q$ terms, then the sum of its $(p + q)$ terms will be:

  • (A) $p - q$

  • (B) $p + q$

  • (C) $0$

  • (D) $-(p + q)$

• Detailed Derivation:

  • Given $S_p = S_q$ with $p \neq q$:     $\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$

  • Expand both sides:     $2ap + p(p - 1)d = 2aq + q(q - 1)d$     $2a(p - q) + d[p^2 - p - (q^2 - q)] = 0$     $2a(p - q) + d[(p^2 - q^2) - (p - q)] = 0$

  • Factor out $(p - q)$, given $(p - q) \neq 0$:     $(p - q)[2a + d(p + q - 1)] = 0$     $2a + (p + q - 1)d = 0$

  • Now compute $S_{p+q}$:     $S_{p+q} = \frac{p + q}{2}[2a + (p + q - 1)d] = \frac{p + q}{2}(0) = 0$

• Correct Answer: (C) $0$

Problem 5: Sum of Bounded Integers Divisible by 5

• Question: The sum of all integers between $81$ and $719$ which are divisible by $5$ is:

  • (A) $51800$

  • (B) $50800$

  • (C) $52800$

  • (D) None of these

• Detailed Derivation:

  • Integers strictly between $81$ and $719$ that are divisible by $5$ form an AP:

  • First term $a = 85$

  • Last term $l = 715$

  • Common difference $d = 5$

  • Find the total number of terms $n$ using $l = a + (n - 1)d$:     $715 = 85 + (n - 1)5$     $630 = (n - 1)5$     $n - 1 = 126 \rightarrow n = 127$

  • Calculate the sum $S_n$ using $S_n = \frac{n}{2}(a + l)$:     $S_{127} = \frac{127}{2}(85 + 715) = \frac{127}{2}(800) = 127 \times 400 = 50800$

• Correct Answer: (B) $50800$

Problem 6: Insertion of Arithmetic Means

• Question: $n$ AM's are inserted between $7$ and $71$. If fifth AM is $27$ then $n$ is equal to:

  • (A) $17$

  • (B) $16$

  • (C) $15$

  • (D) None of these

• Detailed Derivation:

  • Inserting $n$ AMs between $a = 7$ and $b = 71$ forms an AP with total terms $N = n + 2$.

  • The common difference $d$ is given by:     $d = \frac{b - a}{n + 1} = \frac{71 - 7}{n + 1} = \frac{64}{n + 1}$

  • The $k\text{th}$ arithmetic mean $A_k$ is given by A_k = a + kd$.\n - Given fifth AM A_5 = 27:\n    A_5 = 7 + 5d = 27\n    5d = 20 ightarrow d = 4\n - Equate d$to solve for$n:\n     rac{64}{n + 1} = 4 ightarrow n + 1 = 16 ightarrow n = 15\n- **Correct Answer**: **(C) 15**\n\n---\n\n### Problem 7: Ratio of Sum of AMs to Single AM\n- **Question**: Given two numbers a$and$b$, let$A$denotes the single A.M. and$S$denotes the sum of$n$A.M.'s between$a$and$b$then$S/A depends on:\n - (A) n, a, b\n - (B) n, b\n - (C) n, a\n - (D) n\n- **Detailed Derivation**:\n - The single A.M. between a$and$b$is$A = rac{a + b}{2}.\n - The sum S$of$n$A.M.'s inserted between$a$and$b$is equal to$n times their single A.M.:\n    S = n imes rac{a + b}{2} = nA\n - Taking the ratio rac{S}{A}:\n     rac{S}{A} = rac{nA}{A} = n\n - Hence, the ratio depends solely on n\n- **Correct Answer**: **(D) n**\n\n---\n\n### Problem 8: Ratio of Terms from Ratio of Sums of Two APs\n- **Question**: If the ratio of the sum of n$terms of two AP's is$(7n + 1) : (4n + 27)$, then the ratio of their$11 ext{th} terms is:\n - (A) 4:3\n - (B) 3:4\n - (C) 2:3\n - (D) 3:2\n- **Detailed Derivation**:\n - Let the two APs have first terms a_1, a_2$and common differences$d_1, d_2.\n - Given ratio of sums:\n     rac{S_n}{S'n} = rac{ rac{n}{2}[2a_1 + (n - 1)d_1]}{ rac{n}{2}[2a_2 + (n - 1)d_2]} = rac{a_1 + rac{n - 1}{2}d_1}{a_2 + rac{n - 1}{2}d_2} = rac{7n + 1}{4n + 27}\n - The ratio of their 11 ext{th} terms is:\n     rac{T{11}}{T'{11}} = rac{a_1 + 10d_1}{a_2 + 10d_2}\n - Equate the coefficients of difference: rac{n - 1}{2} = 10 ightarrow n - 1 = 20 ightarrow n = 21\n - Substitute n = 21 into the sum ratio:\n     rac{T{11}}{T'{11}} = rac{7(21) + 1}{4(21) + 27} = rac{147 + 1}{84 + 27} = rac{148}{111}\n - Simplify by dividing numerator and denominator by 37:\n     rac{148}{111} = rac{4}{3}\n- **Correct Answer**: **(A) 4:3**\n\n---\n\n### Problem 9: Sum of Terms Equidistant from Ends\n- **Question**: If a_1, a_2, imes, a{24}$are in AP and$a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225$then the sum of$24 terms of this AP is:\n - (A) 900\n - (B) 450\n - (C) 225\n - (D) None of these\n- **Detailed Derivation**:\n - In an AP of n$terms, the sum of terms equidistant from the beginning and the end is constant and equal to$(a_1 + a_n).\n - For n = 24:\n    a_1 + a_{24} = a_5 + a_{20} = a_{10} + a_{15}\n - Rewrite the given equation:\n    (a_1 + a_{24}) + (a_5 + a_{20}) + (a_{10} + a_{15}) = 225\n    3(a_1 + a_{24}) = 225 ightarrow a_1 + a_{24} = 75\n - The sum of 24 terms is:\n    S_{24} = rac{24}{2}(a_1 + a_{24}) = 12 imes 75 = 900\n- **Correct Answer**: **(A) 900**\n\n---\n\n### Problem 10: Sum of Reciprocals of Adjacent Product Terms\n- **Question**: If a_1, a_2, a_3, imes, a_n$are in AP, then$ rac{1}{a_1 a_2} + rac{1}{a_2 a_3} + rac{1}{a_3 a_4} + imes + rac{1}{a_{n-1} a_n} equals:\n - (A) rac{a_1 a_2}{n - 1}\n - (B) rac{n - 1}{a_1 + a_n}\n - (C) rac{n - 1}{a_1 - a_n}\n - (D) rac{n - 1}{a_1 a_n}\n- **Detailed Derivation**:\n - Let d$be the common difference of the AP, so$a_2 - a_1 = a_3 - a_2 = imes = a_n - a_{n-1} = d.\n - Multiply and divide the entire series by d:\n    S = rac{1}{d} imes rac{a_2 - a_1}{a_1 a_2} + rac{a_3 - a_2}{a_2 a_3} + imes + rac{a_n - a_{n-1}}{a_{n-1} a_n}\n - Split each term into partial fractions:\n    S = rac{1}{d} imes rac{1}{a_1} - rac{1}{a_2} + rac{1}{a_2} - rac{1}{a_3} + imes + rac{1}{a_{n-1}} - rac{1}{a_n}\n - Telescoping sum simplifies to:\n    S = rac{1}{d} imes rac{1}{a_1} - rac{1}{a_n} = rac{1}{d} imes rac{a_n - a_1}{a_1 a_n}\n - Substitute a_n - a_1 = (n - 1)d:\n    S = rac{1}{d} imes rac{(n - 1)d}{a_1 a_n} = rac{n - 1}{a_1 a_n}\n- **Correct Answer**: **(D) rac{n - 1}{a_1 a_n}**\n\n# Section 2: Geometric Progression and Geometric Mean\n\n\n\n## Core Concepts and Fundamental Properties\n\n- A **Geometric Progression (GP)** is a sequence where each term after the first is obtained by multiplying the previous term by a non-zero constant ratio r.\n- The general term is T_n = a r^{n-1}.\n- For two positive numbers a$and$b$, their single Geometric Mean is$G = rac{a+b}{2} ightarrow ext{A.M.}$and$G = rac{a^{n+1} + b^{n+1}}{a^n + b^n}$represents A.M. when$n = 0.\n- Geometric Mean G = rac{a+b}{2}$gives$G^2 = ab.\n\n---\n\n## Question Solutions and Problem-Solving Methodology\n\n### Problem 11: Value of Exponent for AM Formula Representation\n- **Question**: If rac{a^{n+1} + b^{n+1}}{a^n + b^n}$be the A.M. of$a$and$b$, then$n =\n - (A) 1\n - (B) -1\n - (C) 0\n - (D) None of these\n- **Detailed Derivation**:\n - The Arithmetic Mean of a$and$b$is defined as$ rac{a + b}{2}.\n - Set the given expression equal to the A.M.:\n     rac{a^{n+1} + b^{n+1}}{a^n + b^n} = rac{a + b}{2}\n - Substitute n = 0 to test consistency:\n     rac{a^{0+1} + b^{0+1}}{a^0 + b^0} = rac{a^1 + b^1}{1 + 1} = rac{a + b}{2}\n - Thus, n = 0 satisfies the definition.\n- **Correct Answer**: **(C) 0**\n\n---\n\n### Problem 12: Symmetric Term in a GP\n- **Question**: If (m + n) ext{th}$term of a G.P. is$9$and$(m - n) ext{th}$term is$4$, then$m ext{th} term will be:\n - (A) 6\n - (B) rac{1}{6}\n - (C) 6.5\n - (D) None\n- **Detailed Derivation**:\n - Let A$be the first term and$R be the common ratio of the G.P.\n - The terms are:\n    T_{m+n} = A R^{m+n-1} = 9\n    T_{m-n} = A R^{m-n-1} = 4\n - Multiply T_{m+n}$and$T_{m-n}:\n    T_{m+n} imes T_{m-n} = (A R^{m+n-1})(A R^{m-n-1}) = A^2 R^{2m-2} = (A R^{m-1})^2\n - Note that T_m = A R^{m-1}, so:\n    (T_m)^2 = 9 imes 4 = 36\n    T_m = rac{36}{2} ightarrow ext{taking square root gives } T_m = 6\n- **Correct Answer**: **(A) 6**\n\n---\n\n### Problem 13: Infinite GP with Term-Sum Condition\n- **Question**: The first term of an infinite G.P. is 1 and every term is equal to the sum of all successive terms, then its fourth term will be:\n - (A) rac{1}{2}\n - (B) rac{1}{8}\n - (C) rac{1}{4}\n - (D) rac{1}{16}\n- **Detailed Derivation**:\n - Let the infinite G.P. have first term a = 1$and common ratio$r$with$|r| < 1\n - Condition given: Any term T_k = rac{T_{k+1}}{2} ightarrow sum of all successive terms.\n - For k = 1:\n    a = T_2 + T_3 + T_4 + imes = rac{ar}{1 - r}\n - Substitute a = 1:\n    1 = rac{r}{1 - r} ightarrow 1 - r = r ightarrow 2r = 1 ightarrow r = rac{1}{2}\n - Compute the fourth term T_4:\n    T_4 = a r^3 = 1 imes rac{1}{2}^3 = rac{1}{8}\n- **Correct Answer**: **(B) rac{1}{8}**\n\n---\n\n### Problem 14: Product of Initial Terms of a GP\n- **Question**: Fifth term of a G.P. is 2$, then the product of its$9 terms is:\n - (A) 256\n - (B) 512\n - (C) 1024\n - (D) None of these\n- **Detailed Derivation**:\n - Let a$be the first term and$r be the common ratio.\n - Given T_5 = a r^4 = 2\n - The product of the first 9$terms$P_9 is:\n    P_9 = T_1 imes T_2 imes T_3 imes imes T_9 = a imes (ar) imes (ar^2) imes imes (ar^8)\n    P_9 = a^9 imes r^{0 + 1 + 2 + imes + 8} = a^9 imes r^{ rac{8 imes 9}{2}} = a^9 r^{36}\n - Express as a power of T_5 = a r^4:\n    P_9 = (a r^4)^9 = (T_5)^9 = 2^9 = 512\n- **Correct Answer**: **(B) 512**\n\n---\n\n### Problem 15: Infinite Product with Exponential GP\n- **Question**: If y = x^{ rac{1}{3}} imes x^{ rac{1}{9}} imes x^{ rac{1}{27}} imes imes rac{1}{2} ightarrow ext{to } imes$, then$y =\n - (A) x^{1/3}\n - (B) x^{2/3}\n - (C) x^{1/2}\n - (D) x\n- **Detailed Derivation**:\n - Express y by combining exponents:\n    y = x^{ rac{1}{3} + rac{1}{9} + rac{1}{27} + imes ext{ to } imes}\n - Sum of exponent infinite GP S = rac{a}{1 - r}$with$a = rac{1}{3}$and$r = rac{1}{3}:\n    S = rac{1/3}{1 - 1/3} = rac{1/3}{2/3} = rac{1}{2}\n - Therefore, y = x^{1/2}\n- **Correct Answer**: **(C) x^{1/2}**\n\n---\n\n### Problem 16: Identity Involving Geometric Mean\n- **Question**: If G$be the GM between$x$and$y$, then the value of$ rac{1}{G^2 - x^2} + rac{1}{G^2 - y^2} is equal to:\n - (A) G^2\n - (B) rac{2}{G^2}\n - (C) rac{1}{G^2}\n - (D) 3G^2\n- **Detailed Derivation**:\n - Since G$is the GM between$x$and$y$, we have$G^2 = xy$.

  • Substitute $G^2 = xy$ into the expression:     $\frac{1}{xy - x^2} + \frac{1}{xy - y^2} = \frac{1}{x(y - x)} + \frac{1}{y(x - y)}$

  • Factor out $(y - x)$ from denominators:     $= \frac{1}{x(y - x)} - \frac{1}{y(y - x)} = \frac{1}{y - x} \times \frac{1}{x} - \frac{1}{y}$     $= \frac{1}{y - x} \times \frac{y - x}{xy} = \frac{1}{xy} = \frac{1}{G^2}$

• Correct Answer: (C) $\frac{1}{G^2}$

Problem 17: Sum of Exponential-Difference Series

• Question: $1 + 3 + 7 + 15 + 31 + \times + \text{to } n \text{ terms} =$

  • (A) $2^{n+1} - n$

  • (B) $2^{n+1} - n - 2$

  • (C) $2^n - n - 2$

  • (D) None of these

• Detailed Derivation:

  • Observe term pattern $T_k$:

  • $T_1 = 1 = 2^1 - 1$

  • $T_2 = 3 = 2^2 - 1$

  • $T_3 = 7 = 2^3 - 1$

  • $T_k = 2^k - 1$

  • Sum of $n$ terms $S_n$:     $S_n = \frac{n}{2} \rightarrow \text{sum} = \frac{2}{1} \rightarrow \times \rightarrow \frac{n}{2} \rightarrow \times = \frac{2(2^n - 1)}{2 - 1} - n = 2^{n+1} - 2 - n = 2^{n+1} - n - 2$

• Correct Answer: (B) $2^{n+1} - n - 2$

Problem 18: Mixed Difference Method Series

• Question: $2 + 4 + 7 + 11 + 16 + \times \text{ to } n \text{ terms} =$

  • (A) $\frac{1}{6}(n^2 + 3n + 8)$

  • (B) $\frac{n}{6}(n^2 + 3n + 8)$

  • (C) $\frac{1}{6}(n^2 - 3n + 8)$

  • (D) $\frac{n}{6}(n^2 - 3n + 8)$

• Detailed Derivation:

  • Consecutive differences: $4 - 2 = 2$, $7 - 4 = 3$, $11 - 7 = 4$, $16 - 11 = 5$.

  • Differences are in AP: $2, 3, 4, 5, \times$.

  • Find general term $T_n$ using method of differences:     $S_n = 2 + 4 + 7 + 11 + \times + T_n$     $S_n = 2 + 4 + 7 + \times + T_{n-1} + T_n$     $0 = 2 + [2 + 3 + 4 + \times + \text{to } (n - 1) \text{ terms}] - T_n$     $T_n = 2 + \frac{n - 1}{2}[2(2) + (n - 2)1] = 2 + \frac{n - 1}{2}(n + 2) = \frac{n^2 + n + 2}{2}$

  • Compute total sum $S_n = \frac{1}{2} \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + 2n$:     $S_n = \frac{n}{12}[(n + 1)(2n + 1) + 3(n + 1) + 12] = \frac{n}{12}[2n^2 + 6n + 16] = \frac{n}{6}(n^2 + 3n + 8)$

• Correct Answer: (B) $\frac{n}{6}(n^2 + 3n + 8)$

Section 3: Harmonic Progression, Inequalities, and Mixed Problems

Core Concepts and Fundamental Properties

• A sequence is in Harmonic Progression (HP) if the reciprocals of its terms form an AP.

• The Harmonic Mean $H$ between two numbers $a$ and $b$ is given by $H = \frac{2ab}{a+b}$.

• For positive real numbers, the relation between Means is $\text{A.M.} \times \text{H.M.} = \text{G.M.}^2$ and $\text{A.M.} \frac{2}{1} \rightarrow \text{G.M.} \frac{2}{1} \rightarrow \text{H.M.}$.

Question Solutions and Problem-Solving Methodology

Problem 19: Reciprocal Difference Identity for Harmonic Mean

• Question: If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{H - a} + \frac{1}{H - b}$ is:

  • (A) $a + b$

  • (B) $ab$

  • (C) $\frac{1}{a} + \frac{1}{b}$

  • (D) $\frac{1}{a} - \frac{1}{b}$

• Detailed Derivation:

  • Given $H = \frac{2ab}{a + b}$.

  • Calculate $H - a$:     $H - a = \frac{2ab}{a + b} - a = \frac{2ab - a^2 - ab}{a + b} = \frac{a(b - a)}{a + b} \rightarrow \frac{1}{H - a} = \frac{a + b}{a(b - a)}$

  • Calculate $H - b$:     $H - b = \frac{2ab}{a + b} - b = \frac{2ab - ab - b^2}{a + b} = \frac{b(a - b)}{a + b} \rightarrow \frac{1}{H - b} = \frac{a + b}{b(a - b)} = -\frac{a + b}{b(b - a)}$

  • Add the two fractions:     $\frac{1}{H - a} + \frac{1}{H - b} = \frac{a + b}{b - a} \times \frac{1}{a} - \frac{1}{b} = \frac{a + b}{b - a} \times \frac{b - a}{ab} = \frac{a + b}{ab} = \frac{1}{a} + \frac{1}{b}$

• Correct Answer: (C) $\frac{1}{a} + \frac{1}{b}$

Problem 20: Logarithmic Transformation of GP Terms

• Question: If $x > 1, y > 1, z > 1$ are in G.P., then $\frac{1}{1 + \frac{1}{1} \rightarrow \text{ln}(x)}, \frac{1}{1 + \text{ln}(y)}, \frac{1}{1 + \text{ln}(z)}$ are in:

  • (A) A.P.

  • (B) H.P.

  • (C) G.P.

  • (D) None of these

• Detailed Derivation:

  • Given $x, y, z$ in G.P. $\rightarrow y^2 = xz$

  • Take natural logarithm on both sides:     $\text{ln}(y^2) = \text{ln}(xz) \rightarrow 2\text{ln}(y) = \text{ln}(x) + \text{ln}(z)$     Thus, $\text{ln}(x), \text{ln}(y), \text{ln}(z)$ are in A.P.

  • Add $1$ to each term:     $1 + \text{ln}(x), 1 + \text{ln}(y), 1 + \text{ln}(z)$ are also in A.P.

  • Taking reciprocals of terms in AP yields an H.P.:     $\frac{1}{1 + \text{ln}(x)}, \frac{1}{1 + \text{ln}(y)}, \frac{1}{1 + \text{ln}(z)} \text{ are in H.P.}$

• Correct Answer: (B) H.P.

Problem 21: AM-GM Inequality on Cyclic Ratios

• Question: If $a, b$ and $c$ are positive real numbers then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is greater than or equal to:

  • (A) $3$

  • (B) $6$

  • (C) $27$

  • (D) $5$

• Detailed Derivation:

  • Apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality to positive quantities $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$:     $\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \frac{2}{1} \rightarrow \frac{a}{b} \times \frac{b}{c} \times \frac{c}{a}^{1/3}$

  • Simplify product inside cube root:     $\frac{a}{b} \times \frac{b}{c} \times \frac{c}{a} = 1 \rightarrow (1)^{1/3} = 1$

  • Multiply by 3:     $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \frac{2}{1} \rightarrow 3$

• Correct Answer: (A) $3$

Problem 22: Ratio of Combined Means Inserted Between Numbers

• Question: If $A_1, A_2$, $G_1, G_2$ and $H_1, H_2$ are two A.M.'s, G.M.'s and H.M.'s between two numbers, then $\frac{A_1 + A_2}{H_1 + H_2} \times \frac{H_1 H_2}{G_1 G_2}$ equals to:

  • (A) $0$

  • (B) $4$

  • (C) $1$

  • (D) $8$

• Detailed Derivation:

  • Let the two positive numbers be $a$ and b$.\n - Sum of two inserted AMs: A_1 + A_2 = a + b\n - Product of two inserted GMs: G_1 G_2 = ab\n - For inserted HMs H_1, H_2$, their reciprocals are AMs between$ rac{1}{a}$and$ rac{1}{b}:\n     rac{1}{H_1} + rac{1}{H_2} = rac{1}{a} + rac{1}{b} = rac{a + b}{ab}\n     rac{H_1 + H_2}{H_1 H_2} = rac{a + b}{ab} ightarrow rac{H_1 H_2}{H_1 + H_2} = rac{ab}{a + b}\n - Substitute these expressions into the given product:\n     rac{A_1 + A_2}{H_1 + H_2} imes rac{H_1 H_2}{G_1 G_2} = rac{A_1 + A_2}{G_1 G_2} imes rac{H_1 H_2}{H_1 + H_2} = rac{a + b}{ab} imes rac{ab}{a + b} = 1\n- **Correct Answer**: **(C) 1**\n\n---\n\n### Problem 23: Equal Roots of Quadratic with Cyclic Coefficients\n- **Question**: If the roots of the equation (b - c)x^2 + (c - a)x + (a - b) = 0$are equal, then$a, b, c will be in:\n - (A) AP\n - (B) GP\n - (C) HP\n - (D) None of these\n- **Detailed Derivation**:\n - Sum of coefficients of quadratic equation:\n    (b - c) + (c - a) + (a - b) = 0\n - Since sum of coefficients is zero, x = 1 is guaranteed to be one root.\n - Since both roots are given to be equal, both roots must equal 1.\n - Product of roots for Ax^2 + Bx + C = 0$is$ rac{C}{A}:\n    1 imes 1 = rac{a - b}{b - c} ightarrow b - c = a - b ightarrow 2b = a + c\n - 2b = a + c$is the defining condition for$a, b, c to be in AP.\n- **Correct Answer**: **(A) AP**\n\n---\n\n### Problem 24: Reciprocal Sum Sequence Condition\n- **Question**: If rac{1}{b + c}, rac{1}{c + a}, rac{1}{a + b}$are in AP, then$a^2, b^2, c^2 will be in:\n - (A) AP\n - (B) GP\n - (C) HP\n - (D) None of these\n- **Detailed Derivation**:\n - Given terms in AP condition:\n     rac{2}{c + a} = rac{1}{b + c} + rac{1}{a + b}\n - Combine right side over a common denominator:\n     rac{2}{c + a} = rac{(a + b) + (b + c)}{(b + c)(a + b)} = rac{a + 2b + c}{ab + b^2 + ac + bc}\n - Cross-multiply:\n    2(ab + b^2 + ac + bc) = (c + a)(a + 2b + c)\n    2ab + 2b^2 + 2ac + 2bc = ac + 2bc + c^2 + a^2 + 2ab + ac\n    2ab + 2b^2 + 2ac + 2bc = a^2 + c^2 + 2ac + 2ab + 2bc\n - Subtract equal terms (2ab + 2ac + 2bc) from both sides:\n    2b^2 = a^2 + c^2\n - This proves that a^2, b^2, c^2$$ are in AP.

• Correct Answer: (A) AP