Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The solution relies on the fundamental property of an arithmetic progression, where the difference between consecutive terms remains constant. If we denote the geometric progression terms as $a, ar, ar^2, ar^3$, then the modified terms $a-2, ar-7, ar^2-9, ar^3-5$ must satisfy $(ar-7) - (a-2) = (ar^2-9) - (ar-7)$ and $(ar^2-9) - (ar-7) = (ar^3-5) - (ar^2-9)$. Simplifying the first equation gives $a(r-1)^2 = -3$, while the second equation simplifies to $ar(r-1)^2 = -6$. Dividing these two equations directly reveals $r = 2$, which subsequently confirms that $a = -3$. With these values, the terms become $x_1 = -3, x_2 = -6, x_3 = -12,$ and $x_4 = -24$. The final calculation yields $\frac{1}{24} (-3 \times -6 \times -12 \times -24) = \frac{5184}{24} = 216$.

Telescoping series allow for the evaluation of complex sums by causing most intermediate terms to cancel out, leaving only the boundary values. Given $4 a_n = (n^2+5n+6)$, we determine $a_n = \frac{(n+2)(n+3)}{4}$, meaning the reciprocal $\frac{1}{a_k}$ becomes $\frac{4}{(k+2)(k+3)}$. By applying partial fraction decomposition, this expression is rewritten as $4 \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$.

When evaluating the summation from $k=1$ to $n$, the expression becomes $4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$. Expanding this series shows that each term cancels the following one, leaving only $4 \left( \frac{1}{3} - \frac{1}{n+3} \right)$. Substituting $n=2025$ leads to $4 \left( \frac{1}{3} - \frac{1}{2028} \right)$. This simplifies to $4 \left( \frac{676-1}{2028} \right)$, which is $4 \left( \frac{675}{2028} \right)$. Multiplying this by 507 results in 675, since the product of 507 and 4 equals 2028.

Geometric progressions are defined by a common ratio where each term is derived by multiplying the previous term by this constant, a property that allows us to express every term as powers of the first term and the common ratio. The product $a_3 a_5$ simplifies to $a^2 r^6$, which is equivalent to $a_4^2 = 729$, confirming that $a_4 = 27$ since the sequence consists of increasing positive terms.

Substituting this into the sum $a_2 + a_4 = \frac{111}{4}$ yields $a_2 = \frac{111}{4} - 27 = \frac{3}{4}$, and since $a_2 = ar$ and $a_4 = ar^3$, dividing these expressions provides $r^2 = \frac{ar^3}{ar} = \frac{27}{3/4} = 36$, resulting in $r = 6$. With the ratio known, the first term becomes $a = \frac{3/4}{6} = \frac{1}{8}$, and computing the final expression $24(a_1 + a_2 + a_3) = 24 \cdot \frac{1}{8}(1 + 6 + 36)$ leads to $3 \times 43 = 129$.

The total sum of the infinite series $S = \sum_{n=1}^{\infty } \frac{1}{n^4}$ is partitioned into two distinct components: $\alpha$, consisting of the reciprocal fourth powers of odd integers, and $\beta$, consisting of the reciprocal fourth powers of even integers, such that $S = \alpha + \beta$. By examining the terms within $\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots$, we can factor out $\frac{1}{2^4}$ to obtain $\beta = \frac{1}{16} \left( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots \right)$, which is equivalent to $\frac{1}{16} (\alpha + \beta)$. Rearranging the resulting equation $\beta = \frac{1}{16} (\alpha + \beta)$ leads to $16\beta = \alpha + \beta$, yielding $15\beta = \alpha$. Consequently, the ratio $\frac{\alpha}{\beta}$ evaluates to 15.

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^{\circ}$. Because these angles form an arithmetic progression, their sum can also be calculated as $\frac{n}{2}(a + l)$, where $a$ is the smallest angle and $l$ is the largest angle. Given that the largest interior angle $l$ is $219^{\circ}$ and the common difference $d$ is $6^{\circ}$, the smallest angle can be expressed using the arithmetic progression formula $l = a + (n-1)d$, which rearranges to $a = 219 - (n-1)6 = 225 - 6n$.

Equating these two expressions for the sum, we have $\frac{n}{2}(225 - 6n + 219) = 180(n-2)$. Simplifying the expression inside the parentheses results in $\frac{n}{2}(444 - 6n) = 180(n-2)$, which reduces to $n(222 - 3n) = 180n - 360$. Expanding the terms leads to $222n - 3n^2 = 180n - 360$, and rearranging the equation to one side yields $3n^2 - 42n - 360 = 0$. Dividing the entire equation by 3 gives $n^2 - 14n - 120 = 0$. Factoring this quadratic equation results in $(n-20)(n+6) = 0$, and since the number of sides must be positive, $n$ is 20.

To find the sum of this sequence, we first treat the recurrence relation $2a_{n+2} = 5a_{n+1} - 3a_n$ as a linear homogeneous equation with the characteristic form $2r^2 - 5r + 3 = 0$. Solving this quadratic yields roots $r = 1$ and $r = \frac{3}{2}$, allowing us to express the general term as $a_n = A(1)^n + B(\frac{3}{2})^n$. Applying the conditions $a_0 = 0$ and $a_1 = \frac{1}{2}$ determines the coefficients, where $A + B = 0$ and $A + \frac{3}{2}B = \frac{1}{2}$, leading to $A = -1$ and $B = 1$. This simplifies the explicit formula to $a_n = (\frac{3}{2})^n - 1$. Calculating the sum from $k=1$ to $100$ involves splitting the expression into the geometric series $\sum_{k=1}^{100} (\frac{3}{2})^k$ and the constant sum $\sum_{k=1}^{100} 1$. Evaluating the geometric component results in $3[(\frac{3}{2})^{100} - 1]$, and subtracting the constant term $100$ reveals the final total as $3 a_{100} - 100$.

The number of elements in the union of two finite sets is determined by the inclusion-exclusion principle, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. Set $A$ comprises $2025$ terms $5n - 4$ for $1 \le n \le 2025$, while set $B$ comprises $2025$ terms $7m + 2$ for $1 \le m \le 2025$. The intersection $A \cap B$ contains elements $x$ such that $5n - 4 = 7m + 2$, or $5n - 7m = 6$, which yields the general form $x = 35k + 16$ with $n = 7k + 4$ and $m = 5k + 2$. Applying the constraints $1 \le n \le 2025$ and $1 \le m \le 2025$ restricts $k$ such that $0 \le k \le 288$, resulting in $289$ common terms. Substituting these values into the inclusion-exclusion formula yields $n(A \cup B) = 2025 + 2025 - 289 = 3761$.

The given series of 40 terms is composed of two alternating sub-sequences based on the position of each term. For even positions $n = 2, 4, 6, \dots, 40$, the terms follow the arithmetic progression $a_n = 2n-1$, which generates the sequence $3, 7, 11, \dots, 79$. The sum of this arithmetic progression with 20 terms is calculated using the formula $\frac{n}{2}(a+l)$, where $a=3$ and $l=79$:
$\frac{20}{2}(3+79) = 10 \times 82 = 820$

For odd positions $n = 1, 3, 5, \dots, 39$, the terms are squares of the form $a_n = (2n-1)^2$. By substituting $n = 2k-1$ for $k = 1, 2, \dots, 20$, the terms are defined by $(4k-3)^2$. The total sum for these 20 terms is calculated as follows:
$\sum_{k=1}^{20} (4k-3)^2 = \sum_{k=1}^{20} (16k^2 - 24k + 9) = 16\left(\frac{20 \cdot 21 \cdot 41}{6}\right) - 24\left(\frac{20 \cdot 21}{2}\right) + 20(9)$

Evaluating these expressions results in $16(2870) - 24(210) + 180$, which simplifies to $45920 - 5040 + 180 = 41060$. Adding the sums from the even and odd positions yields the final total of $820 + 41060 = 41880$.

The core principle for solving this problem involves the property that in an arithmetic progression with an even number of terms, $2n$, the difference between the sum of even-positioned terms and odd-positioned terms effectively captures the common difference $d$ repeated across $n$ pairs. Given that the sum of even-positioned terms is $30$ and the sum of odd-positioned terms is $24$, the total difference is $30 - 24 = 6$, which can be equated to $nd = 6$.

The problem states that the last term exceeds the first term by $\frac{21}{2}$, which is mathematically expressed as the difference between the $2n$-th term and the first term: $(2n - 1)d = \frac{21}{2}$. Expanding this expression yields $2nd - d = \frac{21}{2}$. By substituting the known value $nd = 6$ into this equation, we arrive at $12 - d = 10.5$, which simplifies to $d = 1.5$. Since $nd = 6$ and $d = 1.5$, we find $n = 4$, indicating that the total number of terms in the sequence is $2n = 8$.

The fundamental property of any arithmetic progression is that the sum of terms equidistant from the beginning and the end is constant, and this constant is always equal to the sum of the first and last terms, $a_1 + a_{2024}$. The expression $a_1 + a_5 + a_{10} + \dots + a_{2020} + a_{2024} = 2233$ comprises terms that can be paired based on this principle, where each pair sums to the same constant. Given that there are 203 such pairs in this specific grouping, we set up the equation $203(a_1 + a_{2024}) = 2233$, which simplifies to $a_1 + a_{2024} = 11$. To find the total sum of all 2024 terms, we apply the standard arithmetic progression summation formula $S_n = \frac{n}{2}(a_1 + a_n)$, substituting our known values to get $S_{2024} = \frac{2024}{2}(11)$. This calculation results in 1012 multiplied by 11, which gives a final sum of 11132.

The summation relies on recognizing the structure of the general term $T_r = \frac{4r}{1+4r^4}$, which is a perfect candidate for a telescoping series. By factoring the denominator as $(2r^2-2r+1)(2r^2+2r+1)$, the expression can be decomposed into the difference of two fractions, specifically $\frac{1}{(r-1)^2+r^2} - \frac{1}{r^2+(r+1)^2}$.

Applying this across the series from $r=1$ to $10$ causes every intermediate term to cancel out, leaving only the initial and final values. The sum simplifies significantly as shown below:
$\frac{1}{0^2+1^2} - \frac{1}{10^2+11^2} = 1 - \frac{1}{100+121} = \frac{220}{221}$
Given that the sum is $\frac{m}{n} = \frac{220}{221}$, we identify $m=220$ and $n=221$, which results in $m+n=441$.

In any geometric progression, terms follow the form $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio. We are given the equations $a_1 a_5 = 28$ and $a_2 + a_4 = 29$, which translate to $a^2 r^4 = 28$ and $ar + ar^3 = 29$. Observing that $ar$ and $ar^3$ have a product of $a^2 r^4 = 28$ and a sum of $29$, these values act as roots of the quadratic equation $k^2 - 29k + 28 = 0$. Factoring this quadratic yields the roots $1$ and $28$. For an increasing sequence, we assign $ar = 1$ and $ar^3 = 28$, which implies $r^2 = 28$. Consequently, the sixth term $a_6 = ar^5$ is found by calculating $ar^3 \cdot r^2$, resulting in $28 \cdot 28 = 784$.

An Arithmetic Progression is governed by the relation $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference. Because the sum of the first three terms is 54, we have $3a + 3d = 54$, which simplifies to $a + d = 18$, or $a = 18 - d$. The sum of the first twenty terms is $S_{20} = 10(2a + 19d)$, and given this value falls between 1600 and 1800, the condition $160 < 2a + 19d < 180$ must hold.

Substituting $a = 18 - d$ into this inequality leads to $160 < 2(18 - d) + 19d < 180$, which resolves to $160 < 36 + 17d < 180$. Subtracting 36 from all parts gives $124 < 17d < 144$, and since $d$ must be an integer, $d = 8$. With $d = 8$, the first term becomes $a = 10$, making the $11^{\text{th}}$ term $a_{11} = a + 10d = 10 + 10(8) = 90$.

The general term of an arithmetic progression is defined as $T_r = a + (r-1)d$. Using the sum formula for the first 25 terms, $S_{25} = \frac{25}{2}(a + T_{25}) = \frac{13}{20}$, we substitute the given value $T_{25} = \frac{1}{20}$ to find the first term $a = \frac{1}{500}$. From the relation $T_{25} = a + 24d = \frac{1}{20}$, we determine that the common difference is also $d = \frac{1}{500}$.

Given $T_m = a + (m-1)d = md = \frac{m}{500} = \frac{1}{25}$, solving for $m$ yields $m = 20$. The required expression $5m \sum_{r=m}^{2m} T_r$ simplifies to $100 \sum_{r=20}^{40} T_r$. The sum of these 21 terms is calculated using the formula $\frac{21}{2}(T_{20} + T_{40}) = \frac{21}{2}(20d + 40d) = 630d$. Applying the value $d = \frac{1}{500}$, the computation $100 \times 630 \times \frac{1}{500}$ results in 126.

This series is an Arithmetico-Geometric Progression following the form $a + (a+d)r + (a+2d)r^2 + \dots$, where the infinite sum is given by the formula $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ provided that $|r| < 1$. By comparing the series to this structure, we identify the first term $a = 5$, the common ratio $r = 1/7$, the total sum $S = 7$, and the common difference $d = \alpha$. Substituting these values into the sum formula yields the following equation:

$7 = \frac{5}{1 - 1/7} + \frac{\alpha \cdot 1/7}{(1 - 1/7)^2}$

Simplifying the denominator $1 - 1/7$ results in $6/7$. The first part of the sum formula then becomes $\frac{5}{6/7} = \frac{35}{6}$, while the second part simplifies to $\frac{\alpha/7}{36/49} = \frac{7\alpha}{36}$. Setting up the equation $7 = \frac{35}{6} + \frac{7\alpha}{36}$ and subtracting $\frac{35}{6}$ from both sides leads to $\frac{42}{6} - \frac{35}{6} = \frac{7}{6} = \frac{7\alpha}{36}$. Multiplying both sides of this result by $\frac{36}{7}$ confirms that $\alpha = 6$.

The general term of the series, $\frac{4k}{k^4 + 3k^2 + 4}$, can be decomposed using partial fractions after factorizing the denominator as $(k^2-k+2)(k^2+k+2)$, resulting in the expression $\frac{4k}{k^4 + 3k^2 + 4} = \frac{2}{k^2-k+2} - \frac{2}{k^2+k+2}$. Defining the function $f(k) = \frac{2}{k^2-k+2}$, the series becomes a telescoping sum $\sum_{k=1}^{20} (f(k) - f(k+1))$, which simplifies to the difference $f(1) - f(21)$. Substituting the values into this function yields $f(1) = \frac{2}{1^2-1+2} = 1$ and $f(21) = \frac{2}{21^2-21+2} = \frac{2}{422} = \frac{1}{211}$, which results in the total sum $\frac{210}{211}$. With $m=210$ and $n=211$ being coprime, their sum $m+n$ equals $421$.

An arithmetic progression consisting of $2k$ terms can be divided into $k$ odd-positioned terms and $k$ even-positioned terms. Since the difference between any even-positioned term and its preceding odd-positioned term is the constant common difference $d$, the sum of the even terms minus the sum of the odd terms is equal to $k$ times the common difference, or $kd$.

With the sum of the even terms given as 55 and the sum of the odd terms as 40, we find $kd = 55 - 40$, which simplifies to $d = \frac{15}{k}$. The last term of an A.P. with $2k$ terms is expressed as $a + (2k-1)d$, so the difference between the last and first terms is $(2k-1)d = 27$. Substituting $d = \frac{15}{k}$ into this equation yields $(2k-1) \frac{15}{k} = 27$, which expands to $30k - 15 = 27k$, resulting in $3k = 15$, or $k=5$.

A geometric progression is defined by a starting term $a$ and a common ratio $r$, where each term follows the form $a r^{n-1}$. We are given the sum of the second, fourth, and sixth terms as
$a r + a r^3 + a r^5 = 21$
and the sum of the eighth, tenth, and twelfth terms as
$a r^7 + a r^9 + a r^{11} = 15309$
By factoring $r^6$ out of the second expression, it becomes $r^6(a r + a r^3 + a r^5) = 15309$. Substituting the first expression into this gives $21 r^6 = 15309$, which simplifies to $r^6 = 729$. Since the terms are positive, we determine $r = 3$.

Using the first equation with $r = 3$ yields $a(3 + 3^3 + 3^5) = 21$, which is $a(3 + 27 + 243) = 21$, or $273a = 21$, resulting in $a = 1/13$. To find the sum of the first nine terms, we use the formula $S_n = a \frac{r^n - 1}{r - 1}$. Substituting our values gives
$S_9 = \frac{1}{13} \times \frac{3^9 - 1}{3 - 1}$
which simplifies to 757.

The summation of the given series relies on the method of partial fractions, where each general term is written as $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Summing these terms causes intermediate values to cancel, leaving the simplified result $S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$. Applying this to the specific case of $S_{2025}$, the expression $\sqrt{2026 S_{2025}}$ becomes $\sqrt{2026 \times \frac{2025}{2026}}$, which simplifies to $\sqrt{2025} = 45$.

Given that this value corresponds to the sum of the first 6 terms of an arithmetic progression with first term $-p$ and common difference $p$, we utilize the sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$. Substituting 6 for $n$, $-p$ for $a$, and $p$ for $d$, the sum becomes $\frac{6}{2}(2(-p) + 5p) = 3(3p) = 9p$. Equating $9p = 45$ confirms that $p = 5$. Since the absolute difference between the 20th and 15th terms of an arithmetic progression is $5d$, which equals $5p$, the final result is $5 \times 5 = 25$.

Every arithmetic progression is determined entirely by its initial term and the constant common difference added at each step, and we can isolate these variables by setting up a system of equations from the information given. Starting with the sixth term, we know that $a + 5d = 7$, and utilizing the sum formula for the first seven terms, $\frac{7}{2}(2a + 6d) = 7$, simplifies to $a + 3d = 1$. Subtracting these two linear equations reveals that $2d = 6$, which fixes the common difference at $d = 3$ and the starting term at $a = -8$.

Once these values are established, the sum of the first $n$ terms follows the equation $700 = \frac{n}{2}(2(-8) + (n-1)3)$. Expanding this expression results in the quadratic equation $3n^2 - 19n - 1400 = 0$. Solving for the positive integer $n$ provides $n = 25$. Substituting this position back into the general term formula $a_n = a + (n-1)d$ yields $-8 + 24(3)$, which evaluates to 64.