Previous Year Problems

Let \( 7\overbrace{5\ldots5}^{r}7 \) denote the \( (r+2) \)-digit number where the first and the last digits are 7 and the remaining \( r \) digits are 5. Consider the sum \( S = 77 + 757 + 7557 + \ldots + 7\overbrace{5\ldots5}^{98}7 \). If

\[ S = \frac{7\overbrace{5\ldots5}^{99}7 + m}{n} \]

where \( m \) and \( n \) are natural numbers less than 3000, then the value of \( m + n \) is ____.

Solution:

Let \( T_r = 7\overbrace{5\ldots5}^{r}7 \) represent the \( (r+2) \)-digit number where the first and last digits are 7 and the middle \( r \) digits are 5s. We can express \( T_r \) as:

\[ T_r = 7 \cdot 10^{r+1} + 7 + 5 \left(\sum_{k=1}^{r} 10^k\right) \]

Using the formula for a geometric series:

\[ \sum_{k=1}^{r} 10^k = \frac{10(10^r - 1)}{9} \]

This gives:

\[ T_r = 7 \cdot 10^{r+1} + 7 + \frac{5(10^{r+1} - 10)}{9} \]

\[ T_r = \frac{9 \cdot 7 \cdot 10^{r+1} + 9 \cdot 7 + 5 \cdot 10^{r+1} - 50}{9} \]

\[ T_r = \frac{68 \cdot 10^{r+1} + 13}{9} \]

Summing \( T_r \) from \( r=0 \) to \( r=98 \):

\[ S = \sum_{r=0}^{98} \frac{68 \cdot 10^{r+1} + 13}{9} \]

Simplify this sum:

\[ S = \frac{68 \cdot \sum_{r=0}^{98} 10^{r+1} + 99 \cdot 13}{9} \]

\[ S = \frac{68 \cdot 10 \cdot \frac{10^{99} - 1}{9} + 99 \cdot 13}{9} \]

\[ S = \frac{680 \cdot \frac{10^{99} - 1}{9} + 1287}{9} \]

\[ S = \frac{680 \cdot 10^{99} - 680 + 11583}{81} \]

\[ S = \frac{680 \cdot 10^{99} + 10903}{81} \]

Given \( S = \frac{7\overbrace{5\ldots5}^{99}7 + m}{n} \), where \( 7\overbrace{5\ldots5}^{99}7 = \frac{68 \cdot 10^{100} + 13}{9} \), equating the two:

\[ \frac{68 \cdot 10^{100} + 13}{9} + m = \frac{680 \cdot 10^{99} + 10903}{81} \times n \]

Setting \( n = 9 \) for consistency:

\[ 68 \cdot 10^{100} + 13 + 9m = 680 \cdot 10^{99} + 10903 \]

\[ 9m = 10890\]

\[ m = 1210 \]

Therefore, \( m + n = 1210 + 9 = 1219 \).
Let \( l_1, l_2, \ldots, l_{100} \) be consecutive terms of an arithmetic progression with common difference \( d_1 \), and let \( w_1, w_2, \ldots, w_{100} \) be consecutive terms of another arithmetic progression with common difference \( d_2 \), where \( d_1d_2 = 10 \). For each \( i = 1, 2, \ldots, 100 \), let \( R_i \) be a rectangle with length \( l_i \), width \( w_i \) and area \( A_i \). If \( A_{51} - A_{50} = 1000 \), then the value of \( A_{100} - A_{90} \) is ____.

Solution:

Given the expressions for the areas of rectangles in two arithmetic progressions, \(A_i = l_i \times w_i\), where \(l_i = l_1 + (i-1)d_1\) and \(w_i = w_1 + (i-1)d_2\), and the products of the common differences \(d_1d_2 = 10\), we proceed as follows:

The area of the \(i\)-th rectangle, \(A_i\), expands to:

\[ A_i = (l_1 + (i-1)d_1)(w_1 + (i-1)d_2) = l_1w_1 + (d_1w_1 + l_1d_2)(i-1) + d_1d_2(i-1)^2 \]

Given \(A_{51} - A_{50} = 1000\), using the simplified form of \(A_i\), we have:

\[ A_{51} - A_{50} = (d_1w_1 + l_1d_2) + 99d_1d_2 = 1000 \]

From \(d_1d_2 = 10\):

\[ (d_1w_1 + l_1d_2) + 990 = 1000 \]

\[ d_1w_1 + l_1d_2 = 10 \]

Using this relationship for \(A_{100} - A_{99}\):

\[ A_{100} - A_{99} = (d_1w_1 + l_1d_2) + 10(99^2 - 98^2) \]

Thus:

\[ A_{100} - A_{99} = 10 + 10 \times 197 = 10 + 1970 = 1980 \]

Therefore, \(A_{100} - A_{99} = 1980\).
Let \( a_1, a_2, a_3, \ldots \) be an arithmetic progression with \( a_1 = 7 \) and common difference 8. Let \( T_1, T_2, T_3, \ldots \) be such that \( T_1 = 3 \) and \( T_{n+1} - T_n = a_n \) for \( n \geq 1 \). Then, which of the following is/are TRUE?

(A) \( T_{20} = 1604 \)

(B) \( \sum_{k=1}^{20} T_k = 10510 \)

(C) \( T_{30} = 3454 \)

(D) \( \sum_{k=1}^{30} T_k = 35610 \)
Solution:

(A, B, C)

Given an arithmetic sequence \( a_n \) where \( a_1 = 7 \) and the common difference is 8, we have \( a_n = 7 + (n-1) \times 8 = 8n - 1 \). The sequence \( T_n \) is defined recursively such that \( T_1 = 3 \) and \( T_{n+1} - T_n = a_n \).

The term \( T_n \) can thus be derived from the sums of the terms of \( a_n \). The quadratic expression proposed for \( T_n \) is \( T_n = a(n-1)(n-2) + b(n-1) + c \). Using the initial terms:
- \( T_1 = 3 \)
- \( T_2 = T_1 + a_1 = 3 + 7 = 10 \)
- \( T_3 = T_2 + a_2 = 10 + (8 \times 2 - 1) = 25 \)
We insert \( n = 1, 2, 3 \) into the general form to solve for \( a \), \( b \), and \( c \):
- From \( T_1 = 3 \), we get \( c = 3 \).
- From \( T_2 = 10 \), we substitute \( n = 2 \) and solve \( a(1)(0) + b(1) + 3 = 10 \), giving \( b = 7 \).
- From \( T_3 = 25 \), we substitute \( n = 3 \) and solve \( a(2)(1) + b(2) + 3 = 25 \), giving \( 2a + 2b + 3 = 25 \), which resolves to \( a = 4 \) after substituting \( b = 7 \).
This establishes \( T_n = 4(n-1)(n-2) + 7(n-1) + 3 \). Expanding and simplifying, we find \( T_n = 4n^2 - 5n + 4 \).

To compute the sum \( S_n = \sum_{k=1}^n T_k \):

\[ S_n = \sum_{k=1}^n (4k^2 - 5k + 4) \]

\[ S_n = 4\sum_{k=1}^n k^2 - 5\sum_{k=1}^n k + 4\sum_{k=1}^n 1 \]

Using the formulas for the sum of squares and linear terms:

\[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^n k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^n 1 = n \]

\[ S_n = 4\left(\frac{n(n+1)(2n+1)}{6}\right) - 5\left(\frac{n(n+1)}{2}\right) + 4n \]

Substituting \( n = 20 \) and \( n = 30 \), we find:

\[ S_{20} = 4\left(\frac{20 \times 21 \times 41}{6}\right) - 5\left(\frac{20 \times 21}{2}\right) + 4 \times 20 = 10510 \]

\[ S_{30} = 4\left(\frac{30 \times 31 \times 61}{6}\right) - 5\left(\frac{30 \times 31}{2}\right) + 4 \times 30 = 35615 \]

Therefore, we confirm:
- \( T_{20} = 1604 \)
- \( \sum_{k=1}^{20} T_k = 10510 \)
- \( T_{30} = 3454 \)
- \( \sum_{k=1}^{30} T_k = 35615 \)
For any positive integer \( n \), let \( S_n: (0, \infty) \rightarrow \mathbb{R} \) be defined by

\[ S_n(x) = \sum_{k=1}^{n} \cot^{-1}\left( \frac{1 + k(k + 1)x^2}{x} \right), \]

where for any \( x \in \mathbb{R}, \cot^{-1}(x) \in (0, \pi) \) and \( \tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Then which of the following statements is (are) TRUE?

(A) \( S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1 + 11x^2}{10x}\right), \) for all \( x > 0 \)

(B) \( \lim_{n \to \infty} cot\left(S_n(x)\right) = x, \) for all \( x > 0 \)

(C) The equation \( S_3(x) = \frac{\pi}{4} \) has a root in \( (0, \infty) \)

(D) \( \tan(S_n(x)) \leq \frac{1}{2}, \) for all \( n \geq 1 \) and \( x > 0 \)

Solution:

Given \( S_n(x) \), we have:

\[ \begin{align*} S_n(x) &= \sum_{k=1}^{n} \cot^{-1}\left( \frac{1 + k(k + 1)x^2}{x} \right) \\ &= \sum_{k=1}^{n} \tan^{-1}\left( \frac{x}{1 + k(k + 1)x^2} \right) \quad \left[ \text{Since} \, \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}(x) \, \text{for} \, x > 0 \right] \\ &= \sum_{k=1}^{n} \tan^{-1}\left( \frac{(k+1)x - kx}{1 + kx(k+1)x} \right) \\ &= \sum_{k=1}^{n} \left[ \tan^{-1}(k+1)x - \tan^{-1}(kx) \right] \quad \ \left[ \text{Using the identity } \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left( \frac{a-b}{1+ab} \right) \text{ for } a > 0, b > 0 \right] \\ &= \left[ \tan^{-1}(2x) - \tan^{-1}(x) \right] + \left[ \tan^{-1}(3x) - \tan^{-1}(2x) \right] + \ldots + \left[ \tan^{-1}((n+1)x) - \tan^{-1}(nx) \right] \\ &= -\tan^{-1}(x) + \tan^{-1}((n+1)x) \quad \left[ \text{Telescoping series} \right]\\ &= \tan^{-1}\left(\frac{(n+1)x - x}{1 + (n+1)x \cdot x}\right) \\ &= \tan^{-1}\left(\frac{nx}{1 + (n+1)x^2}\right). \end{align*} \]
Let \( m \) be the minimum possible value of \( \log_3(3^{y_1} + 3^{y_2} + 3^{y_3}) \), where \( y_1, y_2, y_3 \) are real numbers for which \( y_1 + y_2 + y_3 = 9 \). Let \( M \) be the maximum possible value of \( \log_3(x_1 + \log_3x_2 + \log_3x_3) \), where \( x_1, x_2, x_3 \) are positive real numbers for which \( x_1 + x_2 + x_3 = 9 \). Then the value of \( \log_2(m^3) + \log_3(M^2) \) is ____.
Let the function \( f: [0, 1] \rightarrow \mathbb{R} \) be defined by

\[ f(x) = \frac{4^x}{4^x + 2} \]

Then the value of

\[ f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \ldots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right) \]

is ____.
Let \( \alpha \) and \( \beta \) be the roots of \( x^2 - x - 1 = 0 \), with \( \alpha > \beta \). For all positive integers \( n \), define

\[ a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, \quad n \geq 1, \]

\[ b_1 = 1 \quad \text{and} \quad b_n = a_{n-1} + a_{n+1}, \quad n \geq 2. \]

Then which of the following options is/are correct?

(A) \( a_1 + a_2 + a_3 + \ldots + a_n = a_{n+2} - 1 \) for all \( n \geq 1 \)

(B) \( \sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{10}{89} \)

(C) \( b_n = \alpha^n + \beta^n \) for all \( n \geq 1 \)

(D) \( \sum_{n=1}^{\infty} \frac{b_n}{10^n} = \frac{8}{89} \)
Let \( AP(a; d) \) denote the set of all the terms of an infinite arithmetic progression with first term \( a \) and common difference \( d > 0 \). If

\[ AP(1; 3) \cap AP(2; 5) \cap AP(3; 7) = AP(a; d) \]

then \( a + d \) equals ____.
For non-negative integers \( n \), let

\[ f(n) = \frac{\sum_{k=0}^{n} \sin\left(\frac{k+1}{n+2}\pi\right)\sin\left(\frac{k+2}{n+2}\pi\right)}{\sum_{k=0}^{n} \sin^2\left(\frac{k+1}{n+2}\pi\right)} \]

Assuming \( \cos^{-1} x\) takes values in \( [0,\pi] \), which of the following options is/are correct?

(A) \( f(4) = \frac{\sqrt{3}}{2} \)

(B) \( \lim_{n \to \infty} f(n) = \frac{1}{2} \)

(C) If \( a = \tan(\cos^{-1}(f(6))) \), then \( a^2 + 2a - 1 = 0 \)

(D) \( \sin(7\cos^{-1}(f(5))) = 0 \)
The value of

\[ \sec^{-1}\left(\frac{1}{4} \sum_{k=0}^{10} \sec\left(\frac{7\pi}{12} + \frac{k\pi}{2}\right) \sec\left(\frac{7\pi}{12} + \frac{(k+1)\pi}{2}\right)\right) \]

in the interval \(\left[ -\frac{\pi}{4}, \frac{3\pi}{4} \right]\) equals ____.
The number of real solutions of the equation

\[ \sin^{-1}\left( \sum_{i=1}^{\infty} x^{i+1} - x\sum_{i=1}^{\infty} \left(\frac{x}{2}\right)^i \right) = \frac{\pi}{2} - \cos^{-1}\left( \sum_{i=1}^{\infty} \left(-\frac{x}{2}\right)^i - \sum_{i=1}^{\infty} (-x)^i \right) \]

lying in the interval \( \left(-\frac{1}{2}, \frac{1}{2}\right) \) is ____.

(Here, the inverse trigonometric functions \( \sin^{-1}x \) and \( \cos^{-1}x \) assume values in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) and \( \left[0, \pi\right] \), respectively.)
For any positive integer \( n \), define \( f_n: (0, \infty) \rightarrow \mathbb{R} \) as

\[ f_n(x) = \sum_{j=1}^{n} \tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \text{ for all } x \in (0, \infty). \]

(Here, the inverse trigonometric function \( \tan^{-1}x \) assumes values in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).)

Then, which of the following statement(s) is (are) TRUE?

(A) \( \sum_{j=1}^{5} \tan^2(f_j(0)) = 55 \)

(B) \( \sum_{j=1}^{10} (1 + f_j'(0)) \sec^2(f_j(0)) = 10 \)

(C) For any fixed positive integer \( n \), \( \lim_{x \to \infty} \tan(f_n(x)) = \frac{1}{n} \)

(D) For any fixed positive integer \( n \), \( \lim_{x \to \infty} \sec^2(f_n(x)) = 1 \)
Let \( b_i > 1 \) for \( i = 1, 2, \ldots, 101 \). Suppose \( \log_e b_1, \log_e b_2, \ldots, \log_e b_{101} \) are in Arithmetic Progression (A.P.) with the common difference \( \log_e 2 \). Suppose \( a_1, a_2, \ldots, a_{101} \) are in A.P. such that \( a_1 = b_1 \) and \( a_{51} = b_{51} \). If \( t = b_1 + b_2 + \ldots + b_{51} \) and \( s = a_1 + a_2 + \ldots + a_{51} \), then

(A) \( s > t \) and \( a_{101} > b_{101} \)

(B) \( s > t \) and \( a_{101} < b_{101} \)

(C) \( s < t \) and \( a_{101} > b_{101} \)

(D) \( s < t \) and \( a_{101} < b_{101} \)
The value of

\[ \sum_{k=1}^{13} \frac{1}{\sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right)} \]

is equal to

(A) \( 3 - \sqrt{3} \)

(B) \( 2(3 - \sqrt{3}) \)

(C) \( 2(\sqrt{3} - 1) \)

(D) \( 2(2 + \sqrt{3}) \)
Let the harmonic mean of two positive real numbers \( a \) and \( b \) be 4. If \( q \) is a positive real number such that \( a, 5, q, b \) is an arithmetic progression, then the value(s) of \( |q - a| \) is (are) ____.
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is \( 6 : 11 \) and the seventh term lies between 130 and 140, then the common difference of this A.P. is ____.
Let \( a, b, c \) be positive integers such that \( \frac{b}{a} \) is an integer. If \( a, b, c \) are in geometric progression and the arithmetic mean of \( a, b, c \) is \( b + 2 \), then the value of

\[ \frac{a^2 + a - 14}{a + 1} \]

is ____.
The value of \( \cot \left( \sum_{n=1}^{23} \cot^{-1}\left( 1 + \sum_{k=1}^{n} 2k \right) \right) \) is

(A) \( \frac{23}{25} \)

(B) \( \frac{25}{23} \)

(C) \( \frac{23}{24} \)

(D) \( \frac{24}{23} \)
Let \( S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2 \). Then \( S_n \) can take value(s)

(A) 1056

(B) 1088

(C) 1120

(D) 1332
Let \( a_1, a_2, a_3, \ldots \) be in harmonic progression with \( a_1 = 5 \) and \( a_{20} = 25 \). The least positive integer \( n \) for which \( a_n < 0 \) is

(A) 22

(B) 23

(C) 24

(D) 25
Let \( a_1, a_2, a_3, ..., a_{100} \) be an arithmetic progression with \( a_1 = 3 \) and \( S_p = \sum_{i=1}^{p} a_i \), \( 1 \leq p \leq 100 \). For any integer \( n \) with \( 1 \leq n \leq 20 \), let \( m = 5n \). If \( \frac{S_m}{S_n} \) does not depend on \( n \), then \( a_2 \) is _____.
The image contains a mathematical problem involving an infinite geometric series and a summation:

Let \( S_k \), \( k = 1, 2, ..., 100 \), denote the sum of the infinite geometric series whose first term is \( \frac{k-1}{k!} \) and the common ratio is \( \frac{1}{k} \). Then the value of \( \frac{100^2}{100!} + \sum_{k=1}^{100} \left| (k^2 - 3k + 1)S_k \right| \) is ...
If the sum of first \( n \) terms of an A.P. is \( cn^2 \), then the sum of squares of these \( n \) terms is

(A) \( \frac{n(4n^2 - 1)c^2}{6} \)

(B) \( \frac{n(4n^2 + 1)c^2}{3} \)

(C) \( \frac{n(4n^2 - 1)c^2}{3} \)

(D) \( \frac{n(4n^2 + 1)c^2}{6} \)
For \( 0 < \theta < \frac{\pi}{2} \), the solution(s) of

\[ \sum_{m=1}^{6} \csc\left(\theta + \frac{(m-1)\pi}{4}\right) \csc\left(\theta + \frac{m\pi}{4}\right) = 4\sqrt{2} \]

is(are)

(A) \( \frac{\pi}{4} \)

(B) \( \frac{\pi}{6} \)

(C) \( \frac{\pi}{12} \)

(D) \( \frac{5\pi}{12} \)
PARAGRAPH ( 1-3)

Let \( V_r \) denote the sum of the first \( r \) terms of an arithmetic progression (A.P.) whose first term is \( r \) and the common difference is \( 2r - 1 \). Let \( T_r = V_{r+1} - V_r \), and \( Q_r = T_{r+1} - T_r \), for \( r = 1, 2, \ldots \)

Question 1:

The sum \( V_1 + V_2 + \ldots + V_n \) is

(A) \( \frac{1}{12}n(n + 1)(3n^2 - n + 1) \)

(B) \( \frac{1}{12}n(n + 1)(3n^2 + n + 2) \)

(C) \( \frac{1}{2}n(2n^2 - n + 1) \)

(D) \( \frac{1}{3}(2n^3 - 2n + 3) \)

Question 2:

\( T_r \) is always

(A) an odd number

(B) an even number

(C) a prime number

(D) a composite number

Question 3:

Which one of the following is a correct statement?

(A) \( Q_1, Q_2, Q_3, \ldots \) are in A.P. with common difference 5

(B) \( Q_1, Q_2, Q_3, \ldots \) are in A.P. with common difference 6

(C) \( Q_1, Q_2, Q_3, \ldots \) are in A.P. with common difference 11

(D) \( Q_1 = Q_2 = Q_3 = \ldots \)