Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Because $a, b, c$ form a geometric progression, the relation $b^2 = ac$ must hold. When examining the quadratic equation $ax^2 + 2bx + c = 0$, the discriminant is given by $(2b)^2 - 4ac$. Substituting $ac$ for $b^2$ shows this discriminant is equal to zero, indicating the existence of identical roots, specifically $x = -\frac{2b}{2a}$, which simplifies to $x = -\frac{b}{a}$.

Since the second equation $dx^2 + 2ex + f = 0$ shares this root, substituting $x = -\frac{b}{a}$ into it results in $d(-\frac{b}{a})^2 + 2e(-\frac{b}{a}) + f = 0$. Expanding this expression leads to $d\frac{b^2}{a^2} - \frac{2eb}{a} + f = 0$. By substituting $b^2 = ac$ into the first term, we obtain $d\frac{ac}{a^2} - \frac{2eb}{a} + f = 0$, which simplifies to $d\frac{c}{a} - \frac{2eb}{a} + f = 0$.

Dividing this entire equation by $c$ yields $\frac{d}{a} - \frac{2eb}{ac} + \frac{f}{c} = 0$. Because $ac = b^2$, the middle term simplifies to $\frac{2e}{b}$, resulting in the equation $\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$. This final expression rearranges to $\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$, which confirms that the terms $\frac{d}{a}$, $\frac{e}{b}$, and $\frac{f}{c}$ satisfy the condition for an arithmetic progression.

Arranging balls in an equilateral triangle results in a total count equivalent to the $n$-th triangular number, which is determined by the sum of integers from 1 to $n$. Conversely, when balls are arranged in a square, the total count is the square of the number of balls on one side. This problem relies on the relationship between these two geometric arrangements once 99 extra balls are introduced to the initial triangular formation.

If $n$ represents the number of balls along each side of the equilateral triangle, the total count is expressed by the formula $\frac{n(n+1)}{2}$. By adding 99 balls to this sum, we reach a total that forms a square with a side length of $n-2$. This scenario is described by the equation:

$\frac{n(n+1)}{2} + 99 = (n-2)^2$

Expanding and simplifying this relationship leads to a quadratic equation. Multiplying the entire expression by 2 gives $n^2 + n + 198 = 2(n^2 - 4n + 4)$, which simplifies to $n^2 + n + 198 = 2n^2 - 8n + 8$. Rearranging the terms into standard quadratic form results in $n^2 - 9n - 190 = 0$. Factoring this expression yields $(n-19)(n+10) = 0$. Since the number of balls must be positive, the value for $n$ is 19.

To determine the total number of balls used in the original triangular arrangement, substitute 19 back into the triangular number formula. Calculating $\frac{19(19+1)}{2}$ yields a total count of 190.

The series consists of odd-indexed terms which are squares of consecutive odd integers, and even-indexed terms which are double the square of consecutive even integers. For any even number of terms $n$, the sum can be simplified as the sum of all squares up to $n$ plus the sum of squares of all even integers up to $n$, which can be expressed as $\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n/2} (2k)^2 = \sum_{k=1}^{n} k^2 + 4 \sum_{k=1}^{n/2} k^2$.

Using the sum of squares formula $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$, the sum for the first 20 terms is $A = \sum_{k=1}^{20} k^2 + 4 \sum_{k=1}^{10} k^2$. Calculating these values yields $A = \frac{20 \cdot 21 \cdot 41}{6} + 4 \cdot \frac{10 \cdot 11 \cdot 21}{6} = 2870 + 1540 = 4410$. Similarly, the sum for the first 40 terms is $B = \sum_{k=1}^{40} k^2 + 4 \sum_{k=1}^{20} k^2$, which results in $B = \frac{40 \cdot 41 \cdot 81}{6} + 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 22140 + 11480 = 33620$.

With these values, the expression $B-2A$ becomes $33620 - 2(4410) = 33620 - 8820 = 24800$. Given $B-2A = 100\lambda$, it follows that $100\lambda = 24800$, which confirms that $\lambda = 248$.

The terms $\frac{1}{x_1}, \frac{1}{x_2}, \ldots, \frac{1}{x_n}$ form an arithmetic progression with first term $a = \frac{1}{x_1} = \frac{1}{4}$ and $21^{st}$ term $\frac{1}{x_{21}} = \frac{1}{20}$. The common difference $d$ satisfies $\frac{1}{20} = \frac{1}{4} + 20d$, which simplifies to $20d = -\frac{4}{20} = -\frac{1}{5}$, giving $d = -\frac{1}{100}$.

The $n^{th}$ term of this progression is $\frac{1}{x_n} = \frac{1}{4} + (n-1)(-\frac{1}{100}) = \frac{25 - n + 1}{100} = \frac{26-n}{100}$. Since $x_n = \frac{100}{26-n}$, the inequality $x_n > 50$ becomes $\frac{100}{26-n} > 50$, which implies $26-n < 2$, or $n > 24$. The least positive integer satisfying this is $n = 25$.

The sum of the first $n$ terms is $S_n = \frac{n}{2}(a_1 + a_n)$. Substituting $n=25$, $a_1 = \frac{1}{4}$, and $a_{25} = \frac{26-25}{100} = \frac{1}{100}$, the sum is $S_{25} = \frac{25}{2}\left(\frac{1}{4} + \frac{1}{100}\right) = \frac{25}{2}\left(\frac{25+1}{100}\right) = \frac{25}{2}\left(\frac{26}{100}\right) = \frac{13}{4}$.

An arithmetic progression (A.P.) is defined by the general term $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. For the sequence $x_n$, we have $x_3 = a + 2d_1 = 8$ and $x_8 = a + 7d_1 = 20$. Subtracting these equations yields $5d_1 = 12$, so $d_1 = 2.4$. Substituting this value back gives $a = 8 - 2(2.4) = 3.2$, allowing us to calculate $x_5 = a + 4d_1 = 3.2 + 9.6 = 12.8$.

For the sequence $\frac{1}{h_n}$, which is an A.P., we use the relation $\frac{1}{h_n} = \frac{1}{b} + (n-1)d_2$. Given $\frac{1}{h_2} = \frac{1}{8}$ and $\frac{1}{h_7} = \frac{1}{20}$, we have $\frac{1}{b} + d_2 = \frac{1}{8}$ and $\frac{1}{b} + 6d_2 = \frac{1}{20}$. Subtracting these yields $5d_2 = \frac{1}{20} - \frac{1}{8} = -\frac{3}{40}$, which results in $d_2 = -\frac{3}{200}$. Determining the 10th term involves $\frac{1}{h_{10}} = \frac{1}{h_2} + 8d_2 = \frac{1}{8} + 8\left(-\frac{3}{200}\right) = \frac{25}{200} - \frac{24}{200} = \frac{1}{200}$, so $h_{10} = 200$. Multiplying the obtained values, $x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

The sum $S$ of an infinite geometric progression with first term $a$ and common ratio $r$ is defined by the formula $S = \frac{a}{1 - r}$, provided that the condition $|r| < 1$ is satisfied. Substituting the given values $a = b$ and $S = 5$ into the formula, we obtain $5 = \frac{b}{1 - r}$, which can be rearranged to express the common ratio as $r = 1 - \frac{b}{5}$. Substituting this expression into the inequality $|r| < 1$ yields $|\frac{5 - b}{5}| < 1$, which is equivalent to $-1 < \frac{5 - b}{5} < 1$. Multiplying the inequality by $5$ results in $-5 < 5 - b < 5$, and subtracting $5$ from each part gives $-10 < -b < 0$. Multiplying the entire inequality by $-1$ reverses the direction of the inequality signs, leading to the final interval $0 < b < 10$.

An arithmetic progression is defined by its first term $a_1$ and common difference $d$, where the $n$-th term is given by $a_n = a_1 + (n-1)d$. The sum $\sum_{k=0}^{12} a_{4k+1}$ consists of thirteen terms: $a_1, a_5, \ldots, a_{49}$. Expanding these using the general term formula, we have $a_1 + (a_1 + 4d) + (a_1 + 8d) + \ldots + (a_1 + 48d)$, which simplifies to $13a_1 + 4d(0 + 1 + \ldots + 12)$. Using the sum of an arithmetic series formula, $13a_1 + 4d \frac{12 \times 13}{2} = 13a_1 + 312d = 416$, which leads to the linear equation $a_1 + 24d = 32$.

The second condition provided is $a_9 + a_{43} = 66$, which translates to $(a_1 + 8d) + (a_1 + 42d) = 66$. This simplifies to $2a_1 + 50d = 66$, or $a_1 + 25d = 33$. Solving the system formed by these two equations, $a_1 + 24d = 32$ and $a_1 + 25d = 33$, yields the common difference $d = 1$ and the first term $a_1 = 8$.

To find $m$, we evaluate the sum of the squares of the first seventeen terms, $\sum_{i=1}^{17} a_i^2$, which corresponds to $8^2 + 9^2 + \ldots + 24^2$. This is calculated by taking the sum of squares from $1$ to $24$ and subtracting the sum of squares from $1$ to $7$. Using the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, we get $4900 - 140 = 4760$. Given that this sum is equal to $140m$, we solve $140m = 4760$ to find $m = 34$.

The sum of the geometric progression $A_n = \sum_{k=1}^n (-1)^{k-1} (\frac{3}{4})^k$ is given by $A_n = \frac{3/4(1-(-3/4)^n)}{1-(-3/4)} = \frac{3}{7}(1 - (-3/4)^n)$. The condition $B_n > A_n$ implies $1 - A_n > A_n$, which simplifies to $A_n < \frac{1}{2}$. Substituting the expression for $A_n$ into this inequality yields $\frac{3}{7}(1 - (-3/4)^n) < \frac{1}{2}$, which further simplifies to $1 - (-3/4)^n < \frac{7}{6}$ and ultimately $(-3/4)^n > -\frac{1}{6}$.

For odd values of $n$, the term $(-3/4)^n$ equals $-(\frac{3}{4})^n$, transforming the inequality into $-(\frac{3}{4})^n > -\frac{1}{6}$, or $(\frac{3}{4})^n < \frac{1}{6}$. Evaluating this expression for successive odd integers, $n=1$ yields $0.75$, $n=3$ yields $0.421875$, and $n=5$ yields approximately $0.237$, all of which exceed $1/6 \approx 0.166$. For $n=7$, $(\frac{3}{4})^7 = \frac{2187}{16384} \approx 0.133$, which is less than $1/6$, confirming that $n=7$ is the least odd natural number for which the condition holds for all $n \ge p$.

The arithmetic mean of $a, b, c$ being in an A.P. with a sum of $\frac{3}{4}$ implies $3b = \frac{3}{4}$, so $b = \frac{1}{4}$. We can express $a$ and $c$ as $a = \frac{1}{4} - d$ and $c = \frac{1}{4} + d$ for some common difference $d > 0$, since the sequence is strictly increasing ($a < b < c$). The condition that $a^2, b^2, c^2$ are in a G.P. implies the square of the middle term equals the product of the extremes: $(b^2)^2 = a^2c^2$, which simplifies to $b^4 = (ac)^2$, or $b^2 = \pm ac$.

If $b^2 = ac$, then $(\frac{1}{4})^2 = (\frac{1}{4} - d)(\frac{1}{4} + d) = \frac{1}{16} - d^2$, which forces $d^2 = 0$, contradicting the strictly increasing nature of the terms. Therefore, we must have $b^2 = -ac$. Substituting the known values leads to the equation $(\frac{1}{4})^2 = -(\frac{1}{4} - d)(\frac{1}{4} + d)$, which simplifies to $\frac{1}{16} = -(\frac{1}{16} - d^2) = d^2 - \frac{1}{16}$. Solving for $d$ gives $d^2 = \frac{2}{16} = \frac{1}{8}$, so $d = \frac{1}{2\sqrt{2}}$. Consequently, $a = \frac{1}{4} - d = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.

The general term of the series is $a_n = \frac{2^n - 1}{2^{n-1}}$, which simplifies to $2 - \frac{1}{2^{n-1}}$. The sum of the first 20 terms is obtained by evaluating the sum of this expression from $n=1$ to $20$: $\sum_{n=1}^{20} (2 - \frac{1}{2^{n-1}}) = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}} = 40 - \sum_{k=0}^{19} (\frac{1}{2})^k$.

The expression $\sum_{k=0}^{19} (\frac{1}{2})^k$ represents a geometric progression with first term $a=1$, common ratio $r=\frac{1}{2}$, and $n=20$ terms, which evaluates to $S = \frac{1(1 - (1/2)^{20})}{1 - 1/2} = 2(1 - \frac{1}{2^{20}}) = 2 - \frac{1}{2^{19}}$. Subtracting this result from the first part of the sum yields $40 - (2 - \frac{1}{2^{19}}) = 38 + \frac{1}{2^{19}}$.

The expression $9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)$ can be rearranged as $225a^2 + 9b^2 + 25c^2 - 45ab - 15bc - 75ac = 0$, which corresponds to the algebraic identity $\frac{1}{2}$[(15a-3b)^2 + (3b-5c)^2 + (5c-15a)^2]$ = 0$. For this sum of squares to equal zero, each term must independently vanish, requiring$15a = 3b = 5c$. By expressing$a$and$b$in terms of$c$, we get$a = \frac{c}{3}$and$b = \frac{5c}{3}$. Adding these two results yields$a+b = \frac{c}{3} + \frac{5c}{3} = 2c$, and since the sum of the first and third terms is equal to twice the middle term,$b, c, a$ are in an arithmetic progression.

This problem relies on identifying the underlying pattern by simplifying each radical term to extract a common factor of $\sqrt{3}$. Because $\sqrt{3} = 1\sqrt{3}$, $\sqrt{75} = 5\sqrt{3}$, $\sqrt{243} = 9\sqrt{3}$, and $\sqrt{507} = 13\sqrt{3}$, the expression simplifies to $\sqrt{3}(1 + 5 + 9 + 13 + \dots) = 435\sqrt{3}$. Dividing both sides by $\sqrt{3}$ reveals the sum of an arithmetic progression where the first term is 1 and the common difference is 4.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, we substitute the known values to obtain $\frac{n}{2}[2(1) + (n-1)4] = 435$. Simplifying this expression results in $n(1 + 2n - 2) = 435$, which rearranges into the quadratic equation $2n^2 - n - 435 = 0$. Solving this via the quadratic formula $n = \frac{1 \pm \sqrt{1 + 3480}}{4}$ leads to $n = \frac{60}{4}$, which simplifies to $n = 15$ since the number of terms must be a positive integer.

Three positive numbers in an arithmetic progression can be represented as $b-d$, $b$, and $b+d$ for some common difference $d$. The condition $abc=8$ results in the equation $(b-d)b(b+d)=8$, which simplifies to $b(b^2 - d^2) = 8$. Rearranging to solve for the square of the common difference yields $d^2 = b^2 - \frac{8}{b}$. Since the square of the real number $d$ must be non-negative, the inequality $b^2 - \frac{8}{b} \geq 0$ must hold. Because $b$ is positive, multiplying by $b$ results in $b^3 - 8 \geq 0$, or $b^3 \geq 8$, which implies $b \geq 2$.

The relationship between the arithmetic mean and the geometric mean for two numbers $a$ and $b$ is defined by $\frac{a+b}{2} = 5\sqrt{ab}$. Dividing both sides by $\sqrt{ab}$ results in $\frac{a+b}{\sqrt{ab}} = 10$, which can be rewritten as $\sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = 10$. Setting $k = \sqrt{\frac{a}{b}}$ transforms the target expression into $\frac{k^2+1}{k^2-1}$, where the condition $k + \frac{1}{k} = 10$ directly implies $k^2+1 = 10k$.

Determining the value of the denominator $k^2-1$ utilizes the algebraic identity $(k^2-1)^2 = (k^2+1)^2 - 4k^2$. Replacing $k^2+1$ with $10k$ transforms the identity into $(10k)^2 - 4k^2 = 96k^2$, identifying $k^2-1$ as $\sqrt{96k^2} = 4k\sqrt{6}$. Combining these components for the target expression yields $\frac{10k}{4k\sqrt{6}}$, which simplifies to $\frac{5}{2\sqrt{6}}$ and rationalizes to $\frac{5\sqrt{6}}{12}$.

The general term $T_k$ of the series is defined by the quotient of the sum of the first $k$ integers and the sum of their cubes, $T_k = \frac{\sum_{i=1}^k i}{\sum_{i=1}^k i^3}$. Applying the standard identities $\sum_{i=1}^k i = \frac{k(k+1)}{2}$ and $\sum_{i=1}^k i^3 = \frac{k^2(k+1)^2}{4}$ simplifies this expression to $T_k = \frac{2}{k(k+1)}$, which decomposes into $2(\frac{1}{k} - \frac{1}{k+1})$ using partial fractions. Summing this from $k=1$ to $n$ creates a telescoping series $S_n = 2 \sum_{k=1}^n (\frac{1}{k} - \frac{1}{k+1}) = 2(1 - \frac{1}{n+1}) = \frac{2n}{n+1}$. Substituting this into the given equation $100S_n = n$ results in $\frac{200n}{n+1} = n$. Upon dividing by $n$ for $n \neq 0$, the equation becomes $\frac{200}{n+1} = 1$, which yields $n+1 = 200$, resulting in $n = 199$.

Expanding $(1 + ai)^3$ reveals an imaginary part of $3a - a^3$, which must vanish for the complex number to be strictly real. Solving $3a - a^3 = 0$ for $a > 0$ yields $a = \sqrt{3}$, so $z = 1 + i\sqrt{3}$, which corresponds to the polar form $2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$. The finite geometric series sum $1 + z + z^2 + \dots + z^{11}$ consists of $12$ terms and follows the formula:
$\frac{z^{12} - 1}{z - 1}$
Applying De Moivre's theorem, we find $z^{12} = 2^{12}(\cos 4\pi + i \sin 4\pi) = 4096$. Substituting this value into the series formula, the expression simplifies to:
$\frac{4096 - 1}{(1 + i\sqrt{3}) - 1} = \frac{4095}{i\sqrt{3}}$
Rationalizing the denominator by multiplying the expression by $\frac{-i}{-i}$ results in $\frac{-4095i}{\sqrt{3}}$, which simplifies to $-1365\sqrt{3}i$.

In an Arithmetic Progression, the sum of terms at equal distances from the beginning and end of the sequence remains constant. Given that the indices of the terms 3, 7, 11, and 15 satisfy the property that 3 plus 15 equals 18 and 7 plus 11 also equals 18, the pairs $a_3 + a_{15}$ and $a_7 + a_{11}$ must be equal. Since their total sum is 72, each pair must sum to 36, which establishes that $a_3 + a_{15} = a_7 + a_{11} = 36$. The formula for the sum of the first 17 terms is $S_{17} = \frac{17}{2}(a_1 + a_{17})$, and because 1 plus 17 equals 18, it follows that $a_1 + a_{17} = a_3 + a_{15} = 36$. Consequently, substituting these values yields

$S_{17} = \frac{17}{2} \times 36 = 17 \times 18 = 306$

For a fixed sum of two positive angles, the sum of their tangents is minimized when the angles are equal because the tangent function is convex on the interval $(0, \frac{\pi}{2})$. Given the constraint $A + B = \frac{\pi}{6}$, the minimum value occurs when $A = B = \frac{\pi}{12}$. Evaluating the expression $\tan A + \tan B$ at this point yields $2\tan \frac{\pi}{12}$. Using the known value $\tan \frac{\pi}{12} = 2 - \sqrt{3}$, the final calculation is $2(2 - \sqrt{3})$, which equals $4 - 2\sqrt{3}$.

Any arithmetic progression with first term $a$ and common difference $d$ contains a 2nd term $a+d$, a 5th term $a+4d$, and a 9th term $a+8d$. Because these terms form a geometric progression, the property that the square of the middle term equals the product of the outer terms holds true, leading to the equation $(a+4d)^2 = (a+d)(a+8d)$ Expanding and simplifying this yields $a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2$, which reduces to $8d^2 = ad$. Since the progression is non-constant, $d$ is non-zero, so dividing by $d$ reveals that $a = 8d$. Consequently, the common ratio of the geometric progression is the quotient $\frac{a+4d}{a+d}$, which becomes $\frac{8d+4d}{8d+d}$ or $\frac{12d}{9d}$, simplifying to $\frac{4}{3}$.

Converting each mixed fraction into an improper fraction reveals that the sequence consists of squared terms with a constant denominator of 5 and numerators forming an arithmetic progression. The sequence is $\left(\frac{8}{5}\right)^2, \left(\frac{12}{5}\right)^2, \left(\frac{16}{5}\right)^2, \dots, \left(\frac{44}{5}\right)^2$. Factoring out $\left(\frac{4}{5}\right)^2$ from the sum simplifies the expression to $\frac{16}{25} \left(2^2 + 3^2 + 4^2 + \dots + 11^2\right)$.

Including the term $1^2$ within the parenthesis and subsequently subtracting it enables the use of the standard summation formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. Setting $n=11$ yields the result of $\frac{11(12)(23)}{6}$, which equals 506. Subtracting the added $1^2$ leaves 505. Multiplying this by $\frac{16}{25}$ results in $\frac{16}{25} \times 505$, which is equivalent to $\frac{16}{5} \times 101$. Equating this to $\frac{16}{5} m$ confirms that $m$ is 101.