Question 22

   Question: Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of $\mathbf { A P ^ { \prime } s }$ in and A $B$ respectively such that $D = d + 3 , d > 0$ . If $\textstyle { \frac { p + q } { p - q } } = { \frac { 1 9 } { 5 } }$ , then $\mathbf p - \mathbf q$ is equal to

   Options: 

    A. 540

    B. 450

    C. 600

    D. 630

   Correct Answer: A

   Year: JEE Main 2025 (Online) 4th April Evening Shift

   Solution: Let the terms in $A$ be $a_1 - d, a_1, a_1 + d$ and in $B$ be $a_2 - D, a_2, a_2 + D$. Now $3a_1 = 36 \Rightarrow a_1 = 12$ and $3a_2 = 36 \Rightarrow a_2 = 12$. Now $(12 - d)(12)(12 + d) = p$ and $(12 - D)(12)(12 + D) = q$. Also $\frac{p+q}{p-q} = \frac{19}{5} \Rightarrow 12q = 7p \Rightarrow 12(12 - D)(12)(12 + D) = 7(12 - d)(12)(12 + d) \Rightarrow 12(135 - d^2 - 6d) = 7(144 - d^2) \Rightarrow d = 6, D = 9$. $p = 6 \times 12 \times 18 = 1296$, $q = 756$, $p - q = 540$.

   Step Solution:

    1.  Assume terms of A as $(12-d), 12, (12+d)$ and B as $(12-D), 12, (12+D)$ since their sums are 36.

    2.  Express products: $p = 12(144-d^2)$ and $q = 12(144-D^2)$.

    3.  Simplify the ratio $\frac{p+q}{p-q} = \frac{19}{5}$ to find the relation $12q = 7p$.

    4.  Substitute $D=d+3$ into the product relation and solve the resulting quadratic $5d^2 + 72d - 612 = 0$ to find $d=6$ and $D=9$.

    5.  Calculate $p = 12(144-36) = 1296$ and $q = 12(144-81) = 756$, then $p-q = 540$.

   Difficulty level: Medium

   The Concept Name: Arithmetic Progression (Terms and Products)

   Short cut solution: Use the property that for three terms in A.P. with sum $S$, the product is $\frac{S}{3}((\frac{S}{3})^2 - d^2)$. Setting up the ratio of these products for $d$ and $D$ allows for direct calculation of the differences.

 Question 30

   Question: The $2 0 ^ { \mathrm { t h } }$ term from the end of the progression $2 0 , 1 9 { \frac { 1 } { 4 } } , 1 8 { \frac { 1 } { 2 } } , 1 7 { \frac { 3 } { 4 } } , . . . . , - 1 2 9 { \frac { 1 } { 4 } }$

   Options: 

    A. -118

    B. -110

    C. -115

    D. -100

   Correct Answer: C

   Year: 27-Jan-2024 Shift 2

   Solution: $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, -129 \frac{1}{4}$. This is A.P. with common difference $d_1 = -1 + \frac{1}{4} = -\frac{3}{4}$. $-129 \frac{1}{4}, \dots, 19 \frac{1}{4}, 20$. This is also A.P. $a = -129 \frac{1}{4}$ and $d = \frac{3}{4}$. Required term $= -129 \frac{1}{4} + (20-1)(\frac{3}{4}) = -129 - \frac{1}{4} + 15 - \frac{3}{4} = -115$.

   Step Solution:

    1.  Determine the original common difference: $d = 19.25 - 20 = -0.75$.

    2.  To find the term from the end, reverse the series starting with $a = -129.25$.

    3.  The new common difference for the reversed series is $D = -d = 0.75$.

    4.  Apply the general term formula $T_n = a + (n-1)D$ for $n=20$.

    5.  Calculate $T_{20} = -129.25 + (19 \times 0.75) = -129.25 + 14.25 = -115$.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (Term from the End)

   Short cut solution: Use the formula: $L - (n-1)d$, where $L$ is the last term, $n$ is the position from the end, and $d$ is the original common difference. Calculate: $-129.25 - (19 \times -0.75) = -115$.

 Question 32

   Question: In an A.P., the sixth terms $\mathbf { a 6 } = 2$ . If the ${ \bf a _ { 1 } a _ { 4 } a _ { 5 } }$ is the greatest, then the common difference of the A.P., is equal to

   Options: 

    A. 3/2

    B. 8/5

    C. 2/3

    D. 5/8

   Correct Answer: B

   Year: 29-Jan-2024 Shift 1

   Solution: $a_6 = 2 \Rightarrow a + 5d = 2$. $a_1 a_4 a_5 = a(a + 3d)(a + 4d) = (2 - 5d)(2 - 2d)(2 - d)$. $f(d) = 8 - 32d + 34d^2 - 20d + 30d^2 - 10d^3$. $f'(d) = -2(5d - 8)(3d - 2)$. $d=8/5$.

   Step Solution:

    1.  Express the first term $a$ in terms of $d$: $a = 2 - 5d$.

    2.  Write the product as a function of $d$: $f(d) = (2-5d)(2-2d)(2-d)$.

    3.  Expand the product into a polynomial: $f(d) = -10d^3 + 34d^2 - 32d + 8$.

    4.  Find the derivative $f'(d) = -30d^2 + 68d - 32$ and set it to zero.

    5.  Factorize to find critical points: $-2(5d-8)(3d-2)=0$, giving $d = 8/5$ as the point for the maximum value.

   Difficulty level: Medium

   The Concept Name: Maxima and Minima in Arithmetic Progression

   Short cut solution: After obtaining $f'(d) = 0$, test the provided options (A, B, C, D) in the derivative equation to see which one satisfies it and yields a negative second derivative for a maximum.

 Question 59

   Question: Let $a_1, a_2, ....... a_n$ be in A.P. If $a_5 = 2a_7$ and $a_{11} = 18$, then $12 \left( \frac { 1 } { \sqrt { a_{10} } + \sqrt { a_{11} } } + \frac { 1 } { \sqrt { a_{11} } + \sqrt { a_{12} } } + \dots \frac { 1 } { \sqrt { a_{17} } + \sqrt { a_{18} } } \right)$ is equal to

   Options: Options are not explicitly listed in the source, but the solution confirms the integer result.

   Correct Answer: 8

   Year: JEE Main 2023 (Online) 31st January Morning Shift

   Solution: $2a_7 = a_5$ (given) $\Rightarrow 2(a_1 + 6d) = a_1 + 4d \Rightarrow a_1 + 8d = 0$ (1). $a_{11} = a_1 + 10d = 18$ (2). By (1) and (2), $a_1 = -72, d = 9$. $a_{18} = a_1 + 17d = 81, a_{10} = a_1 + 9d = 9$. Expression: $12 \left( \frac{\sqrt{a_{11}} - \sqrt{a_{10}}}{d} + \frac{\sqrt{a_{12}} - \sqrt{a_{11}}}{d} + \ldots + \frac{\sqrt{a_{18}} - \sqrt{a_{17}}}{d} \right) = 12 \left( \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d} \right) = \frac{12(9-3)}{9} = 8$.

   Step Solution:

    1.  Solve the system: $a + 8d = 0$ and $a + 10d = 18$ to get $d=9$ and $a=-72$.

    2.  Find the relevant terms: $a_{10} = -72 + 9(9) = 9$ and $a_{18} = -72 + 17(9) = 81$.

    3.  Rationalize each term in the sum: $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.

    4.  Simplify the telescoping series: the expression becomes $\frac{12}{d} (\sqrt{a_{18}} - \sqrt{a_{10}})$.

    5.  Calculate the final value: $\frac{12}{9} (\sqrt{81} - \sqrt{9}) = \frac{4}{3}(9 - 3) = 8$.

   Difficulty level: Medium

   The Concept Name: Telescoping Series and Rationalization in A.P.

   Short cut solution: The sum of $k$ terms of the form $\sum \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}}$ in an A.P. is simply $\frac{\sqrt{a_{last}} - \sqrt{a_{first}}}{d}$. Multiply this result by 12.

 Question 68

   Question: Let $a_1, a_2, a_3, ..., a_n$ be n positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then: $\lim_{n \to \infty} \sqrt{\frac{d}{n}} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ......... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)$ is

   Options: A. $\frac{1}{\sqrt{d}}$, B. 1, C. $\sqrt{d}$, D. 0

   Correct Answer: B

   Year: JEE Main 2023 (Online) 6th April Morning Shift

   Solution: The sum in brackets is $\frac{1}{d} \sum_{i=1}^{n-1} (\sqrt{a_{i+1}} - \sqrt{a_i}) = \frac{\sqrt{a_n} - \sqrt{a_1}}{d}$. The expression becomes $\sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{nd}}$. Using $a_n = a_1 + (n-1)d$, as $n \to \infty$, $\frac{\sqrt{a_1 + (n-1)d} - \sqrt{a_1}}{\sqrt{nd}} \to \frac{\sqrt{nd}}{\sqrt{nd}} = 1$.

   Step Solution:

    1.  Rationalize the denominator of the general term: $\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.

    2.  Sum the telescoping terms to get $S_n = \frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.

    3.  Include the limit factor: $L = \lim_{n \to \infty} \sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \lim_{n \to \infty} \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{nd}}$.

    4.  Substitute the general formula $a_n = a_1 + (n-1)d$ into the limit.

    5.  Evaluate the limit: $\lim_{n \to \infty} \frac{\sqrt{a_1 + (n-1)d} - \sqrt{a_1}}{\sqrt{nd}} = \lim_{n \to \infty} \sqrt{\frac{a_1 + nd - d}{nd}} - 0 = \sqrt{1} = 1$.

   Difficulty level: Medium

   The Concept Name: Limit of a Telescoping Series

   Short cut solution: For large $n$, $a_n \approx nd$. The sum is approximately $\frac{\sqrt{nd}}{d}$. Multiplying by $\sqrt{\frac{d}{n}}$ gives $\sqrt{\frac{d}{n}} \cdot \frac{\sqrt{nd}}{d} = \frac{\sqrt{n d^2}}{d \sqrt{n}} = 1$.

 Question 101

   Question: If $a_1, a_2, a_3, ......$ and $b_1, b_2, b_3 ......$ are A.P., and $a_1 = 2, a_{10} = 3, a_1b_1 = 1 = a_{10}b_{10}$, then $a_4b_4$ is equal to -

   Options: A. $\frac{35}{27}$, B. 1, C. $\frac{27}{28}$, D. $\frac{28}{27}$

   Correct Answer: D

   Year: JEE Main 2022 (Online) 27th June Evening Shift

   Solution: $\{a_n\}$ in A.P. with $a_1=2, a_{10}=3 \Rightarrow 2+9d_1=3 \Rightarrow d_1=1/9$. $\{b_n\}$ in A.P. with $b_1=1/a_1=1/2$ and $b_{10}=1/a_{10}=1/3$. $b_{10} = b_1 + 9d_2 \Rightarrow 1/3 = 1/2 + 9d_2 \Rightarrow d_2 = -1/54$. $a_4 = 2 + 3(1/9) = 7/3$. $b_4 = 1/2 + 3(-1/54) = 4/9$. $a_4b_4 = (7/3)(4/9) = 28/27$.

   Step Solution:

    1.  Determine common difference $d_1$ for $\{a_n\}$: $9d_1 = a_{10} - a_1 = 3 - 2 = 1 \Rightarrow d_1 = 1/9$.

    2.  Calculate $a_4$: $a_4 = a_1 + 3d_1 = 2 + 3/9 = 7/3$.

    3.  Find $b_1$ and $b_{10}$ from the product relations: $b_1 = 1/2$ and $b_{10} = 1/3$.

    4.  Determine common difference $d_2$ for $\{b_n\}$: $9d_2 = b_{10} - b_1 = 1/3 - 1/2 = -1/6 \Rightarrow d_2 = -1/54$.

    5.  Calculate $b_4$: $b_4 = b_1 + 3d_2 = 1/2 - 3/54 = 4/9$, then find $a_4b_4 = \frac{7}{3} \times \frac{4}{9} = \frac{28}{27}$.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (General Term)

   Short cut solution: Use the interpolation formula for the $k^{th}$ term between $T_m$ and $T_n$: $T_k = T_m + \frac{k-m}{n-m}(T_n - T_m)$. $a_4 = 2 + \frac{3}{9}(1) = \frac{7}{3}$ and $b_4 = \frac{1}{2} + \frac{3}{9}(-\frac{1}{6}) = \frac{4}{9}$. Product = $28/27$.

 Question 113

   Question: Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is

   Options: Options are not explicitly listed in the source text for this problem.

   Correct Answer: 53

   Year: JEE Main 2022 (Online) 26th July Evening Shift

   Solution: $\mathrm{d}_1 = \frac{199-100}{2} \notin \mathrm{I}$; $\mathrm{d}_2 = \frac{199-100}{3} = 33$; $\mathrm{d}_3 = \frac{199-100}{4} \notin \mathrm{I}$; $\mathrm{d}_n = \frac{199-100}{i+1} \in \mathrm{I}$; $\mathrm{d}_i = 33 + 11 + 9$. Sum of CD's = $33 + 11 + 9 = 53$.

   Step Solution:

    1.  Use the last term formula: $199 = 100 + (n-1)d$, which simplifies to $(n-1)d = 99$.

    2.  For $d$ to be an integer, $n-1$ must be a factor of 99. The factors are $\{1, 3, 9, 11, 33, 99\}$.

    3.  Apply constraints: at least 3 terms ($n \ge 3 \Rightarrow n-1 \ge 2$) and at most 33 terms ($n \le 33 \Rightarrow n-1 \le 32$).

    4.  Identify valid values for $n-1$ within $$ from the factors: $\{3, 9, 11\}$.

    5.  Calculate corresponding $d$: $d = \frac{99}{3} = 33, \frac{99}{9} = 11, \frac{99}{11} = 9$. Sum = $33 + 11 + 9 = 53$.

   Difficulty level: Medium

   The Concept Name: Arithmetic Progression (Common Difference and Factors)

   Short cut solution: List factors of 99: $\{1, 3, 9, 11, 33, 99\}$. The common difference $d$ must satisfy $2 \le \frac{99}{d} \le 32$. This leaves $d \in \{33, 11, 9\}$. Sum = 53.

 Question 119

   Question: For $p, q \in R$ consider the real valued function $f(x) = (x - p)^2 - q, x \in R$ and $q > 0$. Let $a_1, a_2, a_3$ and $a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $|f(a_i)| = 500$ for all $i = 1, 2, 3, 4$, then the absolute difference between the roots of $f(x) = 0$ is

   Options: Options are not explicitly listed in the source text for this problem.

   Correct Answer: 50

   Year: JEE Main 2022 (Online) 28th July Morning Shift

   Solution: $a_1 = p - 3d, a_2 = p - d, a_3 = p + d, a_4 = p + 3d$. $|f(a_i)| = 500 \Rightarrow |9d^2 - q| = 500$ and $|d^2 - q| = 500$. $9d^2 - q = q - d^2 \Rightarrow 5d^2 - q = 0$. Roots of $f(x) = 0$ are $p \pm \sqrt{q}$. Absolute difference $= |2\sqrt{q}| = 50$.

   Step Solution:

    1.  Assume A.P. terms as $p-3d, p-d, p+d, p+3d$ to satisfy mean $p$.

    2.  Substitute terms into $f(x) = (x-p)^2 - q$ to find $f(a_i)$: $9d^2 - q$ and $d^2 - q$.

    3.  Set $|9d^2 - q| = 500$ and $|d^2 - q| = 500$.

    4.  Since $d^2 < 9d^2$, we solve $9d^2 - q = 500$ and $d^2 - q = -500$. Adding gives $10d^2 = 2q \Rightarrow q = 5d^2$.

    5.  Substitute $q=5d^2$ into $|d^2-q|=500 \Rightarrow 4d^2=500 \Rightarrow d^2=125$. Thus $q=625$. Difference $= 2\sqrt{625} = 50$.

   Difficulty level: Hard

   The Concept Name: Arithmetic Progression and Quadratic Functions

   Short cut solution: Use the symmetry of the A.P. around $p$. The values of $(x-p)^2$ are $9d^2$ and $d^2$. For $|9d^2 - q| = |d^2 - q|$ with $q > 0$ and $d > 0$, the only non-trivial solution is $q = \frac{9d^2+d^2}{2} = 5d^2$. Substitute into $|d^2-5d^2|=500$ to get $d^2=125$, $q=625$. Root difference $= 2\sqrt{q} = 50$.

 Question 184

   Question: The common difference of the A.P. $b_1, b_2, ..., b_m$ is 2 more than the common difference of A.P. $a_1, a_2, ..., a_n$. If $a_{40} = -159$, $a_{100} = -399$ and $b_{100} = a_{70}$, then $b_1$ is equal to:

   Options: A. 81, B. -127, C. -81, D. 127

   Correct Answer: C

   Year: JEE Main 2020 (Online) 6th September Evening Shift

   Solution: $a_{40} = a_1 + 39d = -159$; $a_{100} = a_1 + 99d = -399$. $d = -4$ and $a_1 = -3$. Common difference of $b_n$ is $(-2)$. $b_{100} = a_{70} \Rightarrow b_1 + 99(-2) = (-3) + 69(-4) \Rightarrow b_1 = 198 - 279 = -81$.

   Step Solution:

    1.  Calculate $d$ for series $\{a_n\}$: $60d = a_{100} - a_{40} = -399 - (-159) = -240 \Rightarrow d = -4$.

    2.  Find $a_1$: $-159 = a_1 + 39(-4) \Rightarrow a_1 = -159 + 156 = -3$.

    3.  Determine common difference $D$ for series $\{b_n\}$: $D = d + 2 = -4 + 2 = -2$.

    4.  Find $a_{70} = a_1 + 69d = -3 + 69(-4) = -3 - 276 = -279$.

    5.  Solve for $b_1$: $b_1 + 99(-2) = -279 \Rightarrow b_1 = -279 + 198 = -81$.

   Difficulty level: Medium

   The Concept Name: Arithmetic Progression (Common Difference and General Term)

   Short cut solution: Difference in $a$ over 60 terms is -240, so $d = -4$. The value $a_{70}$ is halfway between $a_{40}$ and $a_{100}$: $a_{70} = \frac{-159 - 399}{2} = -279$. Then $b_1 = a_{70} - 99(D) = -279 - 99(-2) = -81$.

 Question 193

   Question: If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4 th A.M. is equal to $2^{\text {nd }}$ G.M., then m is equal to

   Options: Options are not explicitly listed in the source, but the numerical answer is provided.

   Correct Answer: 39

   Year: Sep. 03, 2020 (II)

   Solution: Let m arithmetic mean be $A_1, A_2, \dots, A_m$ and $G_1, G_2, G_3$ be geometric mean. The A.P. formed by arithmetic mean is, $3, A_1, A_2, A_3, \dots, A_m, 243$. $\therefore d = \frac{243-3}{m+1} = \frac{240}{m+1}$. The G.P. formed by geometric mean $3, G_1, G_2, G_3, 243$. $r = (\frac{243}{3})^{\frac{1}{3+1}} = (81)^{1/4} = 3$. $\because A_4 = G_2 \Rightarrow 3 + 4(\frac{240}{m+1}) = 3(3)^2 \Rightarrow 3 + \frac{960}{m+1} = 27 \Rightarrow m+1 = 40 \Rightarrow m = 39$.

   Step Solution:

    1.  Find the common difference $d$ for the A.P. with $m$ means: $d = \frac{243 - 3}{m+1} = \frac{240}{m+1}$.

    2.  Find the common ratio $r$ for the G.P. with 3 means: $r = (\frac{243}{3})^{\frac{1}{3+1}} = 81^{1/4} = 3$.

    3.  Identify the required terms: $A_4 = 3 + 4d$ and $G_2 = 3 \cdot r^2$.

    4.  Equate $A_4$ and $G_2$: $3 + 4(\frac{240}{m+1}) = 3(3)^2 = 27$.

    5.  Solve the linear equation for $m$: $\frac{960}{m+1} = 24 \Rightarrow m+1 = 40 \Rightarrow m = 39$.

   Difficulty level: Medium

   The Concept Name: Insertion of Arithmetic and Geometric Means

   Short cut solution: Calculate $G_2$ immediately as $3 \times (3^2) = 27$. Then use the A.P. formula: $a + 4d = 27$. Since $a=3$, then $4d = 24 \Rightarrow d = 6$. Finally, use $m = \frac{b-a}{d} - 1 = \frac{240}{6} - 1 = 39$.

 Question 208

   Question: If 19th term of a non-zero A.P. is zero, then its (49th term): (29th term) is :

   Options: 

    A. 4: 1

    B. 1: 3

    C. 3: 1

    D. 2: 1

   Correct Answer: C

   Year: Jan. 11, 2019 (II)

   Solution: Let first term and common difference of AP be a and d respectively, then $t_n = a + (n-1)d$. $t_{19} = a + 18d = 0 \Rightarrow a = -18d$. $\frac{t_{49}}{t_{29}} = \frac{a+48d}{a+28d} = \frac{-18d+48d}{-18d+28d} = \frac{30d}{10d} = 3$. $t_{49} : t_{29} = 3 : 1$.

   Step Solution:

    1.  Write the expression for the 19th term: $a + 18d = 0$.

    2.  Isolate the first term: $a = -18d$.

    3.  Express the 49th term in terms of $d$: $T_{49} = a + 48d = -18d + 48d = 30d$.

    4.  Express the 29th term in terms of $d$: $T_{29} = a + 28d = -18d + 28d = 10d$.

    5.  Calculate the ratio: $\frac{30d}{10d} = 3$, which is $3:1$.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (General Term)

   Short cut solution: For any A.P. where $T_n = 0$, the ratio of terms is given by $\frac{T_{n+x}}{T_{n+y}} = \frac{x}{y}$. Here $n=19$, so $T_{49}$ is $T_{19+30}$ and $T_{29}$ is $T_{19+10}$. Ratio $= 30/10 = 3:1$.

 Question 228

   Question: If $x_1, x_2, . . . . . , x_n$ and $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are two A.P's such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$, then $x_5 \cdot h_{10}$ equals.

   Options: 

    A. 2560

    B. 2650

    C. 3200

    D. 1600

   Correct Answer: A

   Year: Online April 15, 2018

   Solution: Suppose $d_1$ is the common difference of the A.P. $x_1, x_2, \dots, x_n$ then $x_8 - x_3 = 5d_1 = 12 \Rightarrow d_1 = \frac{12}{5} = 2.4 \Rightarrow x_5 = x_3 + 2d_1 = 8 + 2 \times \frac{12}{5} = 12.8$. Suppose $d_2$ is the common difference of the A.P $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$. $5d_2 = \frac{1}{20} - \frac{1}{8} = \frac{-3}{40} \Rightarrow d_2 = \frac{-3}{200}$. $\frac{1}{h_{10}} = \frac{1}{h_7} + 3d_2 = \frac{1}{200} \Rightarrow h_{10} = 200$. $\Rightarrow x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

   Step Solution:

    1.  Find common difference $d_1$ for $\{x_n\}$: $d_1 = \frac{x_8 - x_3}{5} = \frac{20 - 8}{5} = 2.4$.

    2.  Calculate $x_5$: $x_5 = x_3 + 2d_1 = 8 + 4.8 = 12.8$.

    3.  Find common difference $d_2$ for the A.P. of reciprocals: $d_2 = \frac{1/h_7 - 1/h_2}{5} = \frac{1/20 - 1/8}{5} = \frac{-3/40}{5} = -0.015$.

    4.  Calculate reciprocal of $h_{10}$: $\frac{1}{h_{10}} = \frac{1}{h_7} + 3d_2 = \frac{1}{20} + 3(-0.015) = 0.05 - 0.045 = 0.005$. Thus $h_{10} = 200$.

    5.  Find the product: $x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

   Difficulty level: Medium

   The Concept Name: Arithmetic Progression (Common Difference and General Term)

   Short cut solution: For $x_5$, use linear interpolation: $x_5 = 8 + \frac{5-3}{8-3}(20-8) = 12.8$. For $1/h_{10}$, interpolate similarly: $1/h_{10} = 1/20 + \frac{10-7}{7-2}(1/20 - 1/8) = 1/20 + \frac{3}{5}(-3/40) = 1/200$. Final product $= 12.8 \times 200 = 2560$.

 Question 253

   Question: Let $\alpha$ and $\beta$ be the roots of equation $p x ^ { 2 } + q x + r = 0 , p \neq 0$. If p, q, r are in A.P and $\textstyle { \frac { 1 } { \alpha } } + \textstyle { \frac { 1 } { \beta } } = 4$, then the value of $| \alpha - \beta |$ is:

   Options: 

    A. $\frac { \sqrt { 3 4 } } { 9 }$

    B. $\frac { 2 { \sqrt { 1 3 } } } { 9 }$

    C. $\frac { \sqrt { 6 1 } } { 9 }$

    D. $\frac { 2 { \sqrt { 1 7 } } } { 9 }$

   Correct Answer: B

   Year: 2014

   Solution: Let p, q, r are in AP $\Rightarrow 2 q = p + r$ (i). Given $\textstyle \frac { 1 } { \alpha } + \textstyle \frac { 1 } { \beta } = 4 \Rightarrow \frac { \alpha + \beta } { \alpha \beta } = 4$. We have $\alpha + \beta = - q / p$ and $\alpha \beta = \textstyle \frac { r } { p }$. $\Rightarrow \frac { -q } { r } = 4 \Rightarrow q = - 4 r$ (ii). From (i), we have $2 ( - 4 r ) = p + r \Rightarrow p = - 9 r$ (iii). Now, $| \alpha - \beta | = \sqrt { ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta } = \frac { \sqrt { q ^ { 2 } - 4 p r } } { | p | }$. From (ii) and (iii), $\frac { \sqrt { 1 6 r ^ { 2 } + 3 6 r ^ { 2 } } } { | -9r | } = \frac { 2 \sqrt { 1 3 } } { 9 }$.

   Step Solution:

    1.  Express the A.P. condition for $p, q, r$: $2q = p + r$.

    2.  Use the root relation $\frac{1}{\alpha} + \frac{1}{\beta} = 4 \Rightarrow \frac{-q/p}{r/p} = 4$, which simplifies to $q = -4r$.

    3.  Substitute $q = -4r$ into the A.P. condition: $2(-4r) = p + r \Rightarrow p = -9r$.

    4.  Apply the difference of roots formula: $|\alpha - \beta| = \frac{\sqrt{D}}{|p|} = \frac{\sqrt{q^2 - 4pr}}{|p|}$.

    5.  Substitute $q = -4r$ and $p = -9r$ into the formula: $\frac{\sqrt{(-4r)^2 - 4(-9r)(r)}}{9|r|} = \frac{\sqrt{16r^2 + 36r^2}}{9|r|} = \frac{\sqrt{52r^2}}{9|r|} = \frac{2\sqrt{13}}{9}$.

   Difficulty level: Medium

   The Concept Name: Arithmetic Progression and Roots of Quadratic Equation

   Short cut solution: Use the direct relation $q = -4r$ and $p = -9r$ derived from root properties and A.P. definition to evaluate the discriminant $\sqrt{q^2-4pr}$ in terms of $r$ and divide by $|p|$.

 Question 279

   Question: If the A.M. between $p^{th}$ and $q^{th}$ terms of an A.P. is equal to the A.M. between $r^{th}$ and $s^{th}$ terms of the same A.P. then $p + q$ is equal to

   Options: 

    A. r + s − 1

    B. r + s − 2

    C. r + s + 1

    D. r + s

   Correct Answer: D

   Year: Online May 26, 2012

   Solution: Given : $\frac { a_p + a_q } { 2 } = \frac { a_r + a_s } { 2 } \Rightarrow a + ( p - 1 ) d + a + ( q - 1 ) d = a + ( r - 1 ) d + a + ( s - 1 ) d \Rightarrow 2 a + ( p + q ) d - 2 d = 2 a + ( r + s ) d - 2 d \Rightarrow ( p + q ) d = ( r + s ) d \Rightarrow p + q = r + s$.

   Step Solution:

    1.  Equate the Arithmetic Means: $\frac{a_p + a_q}{2} = \frac{a_r + a_s}{2} \Rightarrow a_p + a_q = a_r + a_s$.

    2.  Substitute the general term $a_n = a + (n-1)d$ for each term.

    3.  Expand the expressions: $[a + (p-1)d] + [a + (q-1)d] = [a + (r-1)d] + [a + (s-1)d]$.

    4.  Simplify both sides: $2a + (p + q - 2)d = 2a + (r + s - 2)d$.

    5.  Cancel $2a$ and $-2d$ from both sides to obtain $(p + q)d = (r + s)d$, implying $p + q = r + s$.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (Properties of Terms)

   Short cut solution: In an A.P., the sum of terms $a_x + a_y$ is equal to $a_u + a_v$ if and only if $x + y = u + v$. Since the means are equal, the sums of the terms are equal, thus $p + q = r + s$.

 Question 280

   Question: If 100 times the 100 th term of an AP with non zero common difference equals the 50 times its 50 th term, then the 150 th term of this AP is:

   Options: 

    A. -150

    B. 150 times its $50^{\mathrm{th}}$ term

    C. 150

    D. Zero

   Correct Answer: D

   Year: 2012

   Solution: Let 'a' is the first term and 'd' is the common difference of an A. P. Now, According to the question $100 a_{100} = 50 a_{50}$. $100(a + 99d) = 50(a + 49d) \Rightarrow 2a + 198d = a + 49d \Rightarrow a + 149d = 0$. Hence, $T_{150} = a + 149d = 0$.

   Step Solution:

    1.  Let $a$ be the first term and $d$ be the common difference.

    2.  Write the given equation: $100 \times a_{100} = 50 \times a_{50}$.

    3.  Simplify the equation by dividing both sides by 50: $2(a_{100}) = a_{50}$.

    4.  Substitute the general term formula $a_n = a + (n-1)d$: $2(a + 99d) = a + 49d \Rightarrow 2a + 198d = a + 49d$.

    5.  Rearrange to isolate the terms: $a + 149d = 0$. Since $a_{150} = a + 149d$, the term is zero.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (Properties of Terms)

   Short cut solution: Use the property: If $m \cdot T_m = n \cdot T_n$ for an A.P., then $T_{m+n} = 0$. Here $m=100$ and $n=50$, so $T_{100+50} = T_{150} = 0$.

 Question 297

   Question: Let $T_r$ be the $r^{\mathrm{th}}$ term of an A.P. whose first term is $a$ and common difference is $d$. If for some positive integers $m, n, m \neq n$, $T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$, then $a - d$ equals

   Options: 

    A. $\frac{1}{m} + \frac{1}{n}$

    B. 1

    C. 1

    D. 0

   Correct Answer: D

   Year: 2004

   Solution: $T_m = a + (m-1)d = \frac{1}{n} \ldots (i)$, $T_n = a + (n-1)d = \frac{1}{m} \ldots (ii)$. Subtracting (ii) from (i), we get $(m-n)d = \frac{1}{n} - \frac{1}{m} \Rightarrow d = \frac{1}{mn}$. From (i) $a = \frac{1}{mn} \Rightarrow a - d = 0$.

   Step Solution:

    1.  Express the terms as equations: $a + (m-1)d = 1/n$ and $a + (n-1)d = 1/m$.

    2.  Subtract the second equation from the first: $[a + (m-1)d] - [a + (n-1)d] = 1/n - 1/m$.

    3.  Simplify to find $d$: $(m-n)d = (m-n)/(mn)$, which gives $d = 1/(mn)$.

    4.  Substitute $d$ into the first equation: $a + (m-1)(1/mn) = 1/n \Rightarrow a + 1/n - 1/mn = 1/n$.

    5.  Solve for $a$ and calculate $a-d$: $a = 1/mn$. Therefore, $a - d = 1/mn - 1/mn = 0$.

   Difficulty level: Easy

   The Concept Name: Arithmetic Progression (General Term)

   Short cut solution: In an A.P., a standard property exists where if $T_m = 1/n$ and $T_n = 1/m$, then the first term and common difference are identical: $a = d = 1/mn$. Consequently, $a - d$ is always 0.