Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Identifying the general terms for the given sequences allows us to apply standard summation identities. The sequence $a_n$ is an arithmetic progression with a first term of 1 and a common difference of 2, which simplifies to the general term $a_n = 2n - 1$. The recursive relationship $b_n = a_n + b_{n-1}$ indicates that $b_n$ represents the sum of the first $n$ odd integers, which is recognized as $b_n = n^2$.

Evaluating the summation of the product of these sequences involves calculating $\sum_{n=1}^{15} (2n - 1)n^2$, which expands to the difference of two separate sums: $2 \sum_{n=1}^{15} n^3 - \sum_{n=1}^{15} n^2$. Using the standard formulas for the sum of the first $n$ cubes and squares, we calculate the values for $n=15$ as:

$2 \left( \frac{15 \cdot 16}{2} \right)^2 - \frac{15 \cdot 16 \cdot 31}{6}$

This expression simplifies to $2 \cdot 120^2 - 1240$, which results in $28800 - 1240 = 27560$.

The number of terms in each set follows the sequence of odd integers, where the $n^{\text{th}}$ set contains $2n - 1$ elements. Since the sum of the first $n$ odd numbers is $n^2$, we can determine the exact position of the multiples of 3 by calculating how many terms appear before the desired set.

In the first 10 sets, there are a total of $10^2 = 100$ terms, meaning the 11th set begins with the 101st multiple of 3. This set contains $2(11) - 1 = 21$ terms, extending from the 101st to the 121st multiple of 3. The sum of these elements is $3 \times (101 + 102 + \dots + 121)$. Calculating this using the arithmetic series formula $\frac{N}{2}(a + l)$, where the number of terms $N = 21$, the first term $a = 101$, and the last term $l = 121$, we find the sum as $\frac{21}{2}(101 + 121) = \frac{21 \times 222}{2}$. Simplifying this results in $21 \times 111 = 2331$, and multiplying by 3 yields a final value of 6993.

The intersection of two arithmetic progressions generates a new arithmetic progression where the common difference is the least common multiple of the original differences. Determining the upper bound for these common terms involves finding the last term of each sequence: the first series reaches $3+(77)3 = 234$ and the second reaches $5+(58)4 = 237$. Since a common term cannot exceed the smaller limit of 234, we restrict our analysis to values up to 234. The first term appearing in both series is 9, and the common difference for the new progression is $\text{LCM}(3,4) = 12$. The general term is defined by the equation:
$t_n = 9 + (n-1)12$
Setting $t_n \le 234$ to find the total count of common terms leads to the inequality $9 + (n-1)12 \le 234$. Simplifying this yields:
$(n-1)12 \le 225$
Dividing by 12 results in $n-1 \le 18.75$, which confirms there are exactly 19 terms in the sequence. Finally, the sum is calculated using the formula $S_n = \frac{n}{2}[2a+(n-1)d]$ as:
$S_{19} = \frac{19}{2}[2(9) + (18)(12)] = 19(9 + 108) = 2223$

Recognizing the pattern within each term of the sequence allows us to express the $r$-th term as $a_r = 1 - (\frac{2}{3})^r$. When we sum these terms from $r=1$ to $100$, we can distribute the summation to separate the constant values from the geometric component, resulting in $100 - \sum_{r=1}^{100} (\frac{2}{3})^r$. This approach simplifies the total sum by isolating the integer portion from the fractional decay of the geometric series.

Calculating the geometric sum requires the standard formula for the first $n$ terms, $\frac{a(1-r^n)}{1-r}$, where the first term is $\frac{2}{3}$ and the ratio is $\frac{2}{3}$. This calculation yields:
$\frac{\frac{2}{3}(1-(\frac{2}{3})^{100})}{1-\frac{2}{3}} = 2\left(1-\left(\frac{2}{3}\right)^{100}\right) = 2 - 2\left(\frac{2}{3}\right)^{100}$
Substituting this value back into our original expression for the total sum, we obtain $S = 100 - (2 - 2(\frac{2}{3})^{100})$, which simplifies to $S = 98 + 2(\frac{2}{3})^{100}$. Since the term $2(\frac{2}{3})^{100}$ represents a very small positive value strictly between 0 and 1, the greatest integer less than or equal to $S$ is 98.

Geometric progressions are defined by their first term $a$ and a common ratio $r$, where the terms can be expressed as $a, ar, ar^2, ar^3, ar^4$. Substituting these terms into the relation $3a_2 + a_3 = 2a_4$ gives $3ar + ar^2 = 2ar^3$. Assuming $a \neq 0$, we divide by $ar$ to obtain the quadratic equation $3 + r = 2r^2$, which rearranges to $2r^2 - r - 3 = 0$. Factoring this equation as $(2r - 3)(r + 1) = 0$ yields potential ratios of $r = \frac{3}{2}$ and $r = -1$. Because the condition $a_1 > 0$ must be satisfied, $r = -1$ is excluded, leaving $r = \frac{3}{2}$ as the unique common ratio.

Using $r = \frac{3}{2}$ in the condition $a_2 + a_4 = 2a_3 + 1$ allows us to find the first term $a$ by rewriting the equation as $ar + ar^3 = 2ar^2 + 1$, or $a(r^3 - 2r^2 + r) = 1$. Substituting $\frac{3}{2}$ for $r$ leads to $a(\frac{27}{8} - 2(\frac{9}{4}) + \frac{3}{2}) = 1$, which simplifies to $a(\frac{27}{8} - \frac{9}{2} + \frac{3}{2}) = a(\frac{3}{8}) = 1$, so $a = \frac{8}{3}$. The final expression $a_2 + a_4 + 2a_5$ equates to $ar + ar^3 + 2ar^4 = a(r + r^3 + 2r^4)$. Plugging in the values results in $\frac{8}{3}(\frac{3}{2} + \frac{27}{8} + 2(\frac{81}{16})) = \frac{8}{3}(\frac{12}{8} + \frac{27}{8} + \frac{81}{8}) = \frac{8}{3}(\frac{120}{8}) = 40$.

The general term of an arithmetic progression is given by the formula $l = a + (n-1)d$, where $a$ is the first term, $l$ is the last term, $n$ is the number of terms, and $d$ is the common difference. Substituting the provided values of 100 and 199 results in the equation $199 = 100 + (n-1)d$, which simplifies to $(n-1)d = 99$. Since the common difference $d$ must be an integer, $(n-1)$ must necessarily be a divisor of 99.

The problem constraints specify that the number of terms $n$ satisfies $3 \le n \le 33$, which implies the value $(n-1)$ must fall within the range $2 \le (n-1) \le 32$. The positive divisors of 99 are 1, 3, 9, 11, 33, and 99. Comparing these divisors against the required range, the valid values for $(n-1)$ are 3, 9, and 11.

Calculating the common difference for each of these valid factors yields $d = \frac{99}{3} = 33$, $d = \frac{99}{9} = 11$, and $d = \frac{99}{11} = 9$. Summing these distinct values of the common differences results in a total of $33 + 11 + 9 = 53$.

Inserting $n$ arithmetic means between two numbers $a$ and $100$ establishes an arithmetic progression with a total of $n+2$ terms. If $d$ represents the common difference, the first arithmetic mean $A_1$ is $a+d$, while the last arithmetic mean $A_n$ can be expressed as $100-d$. According to the given ratio $\frac{a+d}{100-d} = \frac{1}{7}$, cross-multiplication yields $7a+7d = 100-d$, which simplifies to $8d = 100-7a$ or $d = \frac{100-7a}{8}$.

Since $100$ is the $(n+2)$-th term of the sequence, the relationship $100 = a + (n+1)d$ must hold true. Using the provided condition $a+n=33$, we can substitute $n+1$ with $34-a$, resulting in the equation $100 = a + (34-a) \left(\frac{100-7a}{8}\right)$. Multiplying throughout by $8$ gives $800 = 8a + (34-a)(100-7a)$, which expands to $800 = 8a + 3400 - 238a - 100a + 7a^2$. This simplifies to the quadratic equation $7a^2 - 330a + 2600 = 0$. Solving for $a$ gives two potential values, $10$ and $\frac{260}{7}$. Since $n = 33-a$ and $n$ must be an integer, we accept $a=10$, which results in $n = 33 - 10 = 23$.

Rewriting each term in the sum $\sum_{n=1}^{20} \frac{1}{a_n a_{n+1}}$ as $\frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)$ transforms the entire expression into a telescoping sum $\frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{21}} \right)$. With the total value equal to $\frac{4}{9}$ and the difference $a_{21} - a_1 = 20d$, the expression simplifies to $\frac{20}{a_1 a_{21}} = \frac{4}{9}$, which means $a_1 a_{21} = 45$.

The sum of the 21 terms of the arithmetic progression is 189, so $\frac{21}{2} (a_1 + a_{21}) = 189$ results in $a_1 + a_{21} = 18$. Identifying $a_1$ and $a_{21}$ as the roots of the quadratic $x^2 - 18x + 45 = 0$ gives the values 3 and 15. The common difference $d$ is then derived from $a_{21} = a_1 + 20d$, resulting in $\pm \frac{3}{5}$. Calculating the product $a_6 a_{16}$ as $(a_1 + 5d)(a_1 + 15d)$ leads to $(3 + 3)(3 + 9) = 72$ or $(15 - 3)(15 - 9) = 72$.

To determine the sum of integers $n$ between 1 and 100 such that the $\text{H.C.F.}(n, 2040)$ is 1, we must first identify the prime factors of 2040. Since $2040 = 2^{3} \times 3 \times 5 \times 17$, the condition $\text{H.C.F.}(n, 2040) = 1$ implies that $n$ cannot be a multiple of 2, 3, 5, or 17. This principle requires us to systematically remove all multiples of these prime factors from the set of integers $\{1, 2, \ldots, 100\}$ before summing the remaining numbers.

Starting with the sum of all odd integers from 1 to 100, which is $1+3+5+\dots+99 = 2500$, we account for the elimination of all even numbers. Next, we subtract the sum of the odd multiples of 3 within this range, which forms the arithmetic progression $3, 9, \ldots, 99$; this sum is calculated as $\frac{17}{2}(3+99) = 867$. We then remove the sum of odd multiples of 5 that were not already excluded, specifically the set $\{5, 25, 35, 55, 65, 85, 95\}$, which totals 365. Finally, we account for 17, as other multiples of 17 such as 51 and 85 were already removed due to their divisibility by 3 and 5, respectively. Subtracting these values yields $2500 - 867 - 365 - 17 = 1251$.

The sequence $\frac{1}{16}, a, b$ forms a geometric progression, which implies that the square of the middle term equals the product of the outer terms, resulting in the relationship $a^2 = \frac{b}{16}$, or $b = 16a^2$. The arithmetic progression $\frac{1}{a}, \frac{1}{b}, 6$ indicates that the middle term is the arithmetic mean of the surrounding terms, expressed by the equation $\frac{2}{b} = \frac{1}{a} + 6$. Substituting the expression $b = 16a^2$ into this arithmetic condition yields $\frac{2}{16a^2} = \frac{1+6a}{a}$, which simplifies to $\frac{1}{8a^2} = \frac{1+6a}{a}$.

Multiplying both sides by $8a^2$ leads to the equation $1 = 8a(1+6a)$, which expands to $48a^2 + 8a - 1 = 0$. Factoring this quadratic equation gives $(12a-1)(4a+1) = 0$, providing two possible roots for $a$, specifically $\frac{1}{12}$ and $-\frac{1}{4}$. Given the constraint that $a > 0$, the only valid solution is $a = \frac{1}{12}$. Using this value, $b$ is determined to be $16(\frac{1}{12})^2$, which simplifies to $16 \cdot \frac{1}{144} = \frac{1}{9}$. Consequently, evaluating the expression $72(a+b)$ results in $72\left(\frac{1}{12} + \frac{1}{9}\right)$, which equals $6 + 8$, producing a final value of 14.

The series $y=\frac{1}{2}x^{2}+\frac{2}{3}x^{3}+\frac{3}{4}x^{4}+\dots$ is most effectively handled by rewriting the coefficient of each term, $\frac{n-1}{n}$, as $1-\frac{1}{n}$. By splitting the summation into two distinct parts, $\sum_{n=2}^{\infty } x^n - \sum_{n=2}^{\infty } \frac{x^n}{n}$, we can identify one part as a standard geometric series and the other as the Taylor expansion for the natural logarithm. The geometric component simplifies to $\frac{x^2}{1-x}$, and the logarithmic part equates to $-\ln(1-x)-x$, which combines to provide the closed-form expression for $y$:
$y = \frac{x}{1-x} + \ln(1-x)$
Plugging in $x=\frac{1}{2}$ produces $y = 1 + \ln(1/2)$, which simplifies to $1 - \ln 2$. Applying this result to the final expression $e^{1+y}$ yields $e^{1+1-\ln 2} = e^{2-\ln 2}$, which further resolves to $\frac{1}{2}e^{2}$.

The condition $\text{gcd}(x, 18) = 3$ means that the four-digit number $x$ must be a multiple of 3, but it cannot be divisible by 2 or by the square of 3, which is 9. Effectively, this requires $x$ to be an odd multiple of 3 that is not a multiple of 9.

There are 1499 such four-digit numbers that are odd multiples of 3. Among these, we must exclude any that are also multiples of 9, as they would have a greatest common divisor with 18 that is at least 9. There are 499 such odd multiples of 9. Subtracting these from the total gives $1499 - 499 = 1000$.

The relationship between the sum of the first $n$ terms of an arithmetic progression, denoted by $S_n$, and its individual terms is governed by the fundamental identity $a_n = S_n - S_{n-1}$. Rearranging the provided equality $\frac{S_{10}}{S_p} = \frac{100}{p^2}$ allows for any sum $S_p$ to be expressed as a function of $S_{10}$ through the equation $S_p = S_{10} \cdot \frac{p^2}{100}$.

Determining the ratio $\frac{a_{11}}{a_{10}}$ involves expressing both terms through this sum formula, where $a_{11} = S_{11} - S_{10}$ and $a_{10} = S_{10} - S_9$. By substituting the appropriate values for $p$-specifically $p=11$ for $S_{11}$ and $p=9$ for $S_9$-the numerator transforms into $S_{10}(\frac{121}{100} - 1) = S_{10} \cdot \frac{21}{100}$. Similarly, the denominator becomes $S_{10}(1 - \frac{81}{100}) = S_{10} \cdot \frac{19}{100}$. Dividing these values effectively cancels $S_{10}$, resulting in $\frac{21}{19}$.

The algebraic identity $a^2 - b^2 = (a-b)(a+b)$ provides an efficient way to simplify the alternating series by grouping terms in pairs. Applying this to the sequence $(1^2-2^2) + (3^2-4^2) + \dots + (2021^2-2022^2) + 2023^2$ transforms each bracketed pair into $-(a+b)$, which results in $-(1+2) - (3+4) - \dots - (2021+2022) + 2023^2$. This simplifies to the sum of the first 2022 integers negated, represented as $-\frac{2022 \times 2023}{2} + 2023^2$, which further reduces to $-1011 \times 2023 + 2023^2$.

Factoring out $2023$ from this expression yields $2023 \times (2023 - 1011)$, or $2023 \times 1012$. Comparing this product to the given expression $1012 m^2 n$, we establish that $m^2 n = 2023$. Since the prime factorization of $2023$ is $17^2 \times 7$ and the condition $\gcd(m, n) = 1$ must hold, we identify $m = 17$ and $n = 7$. Consequently, the value of $m^2 - n^2$ is calculated as $17^2 - 7^2$, which results in $289 - 49 = 240$.

Geometric progressions rely on a constant ratio $r$ between consecutive terms, whereas arithmetic progressions require a constant difference $d$. When transforming a geometric progression into an arithmetic one by modifying a term, the fundamental property of arithmetic progressions-that the middle term is the average of the two adjacent terms-provides the necessary tool to solve for the unknowns.

Representing the three numbers as $\frac{a}{r}, a, ar$, we double the middle term to create the new sequence $\frac{a}{r}, 2a, ar$. The arithmetic property requires the following equality:
$2(2a) = \frac{a}{r} + ar$
Dividing this equation by $a$ yields $4 = \frac{1}{r} + r$, which rearranges into the quadratic equation $r^{2} - 4r + 1 = 0$. Solving for $r$ using the quadratic formula results in $r = 2 \pm \sqrt{3}$. Since the progression is increasing, the only valid value is $r = 2 + \sqrt{3}$.

The problem states that the fourth term of the original geometric progression is $3r^{2}$. Given that the fourth term is $ar^{3}$, we establish that $ar^{3} = 3r^{2}$, which confirms that $a = 3$. The common difference $d$ of the resulting arithmetic progression is calculated by subtracting the first term from the middle term, so $d = 2a - \frac{a}{r}$. Substituting the known values leads to $d = 2(3) - \frac{3}{2 + \sqrt{3}}$, which simplifies through rationalization to $6 - 3(2 - \sqrt{3})$, resulting in $d = 3\sqrt{3}$. Finally, calculating $r^{2} - d$ involves substituting the values into $(2 + \sqrt{3})^{2} - 3\sqrt{3}$, which expands to $7 + 4\sqrt{3} - 3\sqrt{3}$ to equal $7 + \sqrt{3}$.

In a geometric progression defined by the first term $a$ and common ratio $r$, any term $T_n$ can be expressed as $ar^{n-1}$. The product of the third and fifth terms is given by $T_3 \cdot T_5 = (ar^2)(ar^4) = a^2r^6 = 25$, which implies $ar^3 = 5$ for an increasing series where terms are positive. The sum of the second and sixth terms is $T_2 + T_6 = ar + ar^5 = ar(1 + r^4) = \frac{25}{2}$. By substituting $a = \frac{5}{r^3}$ into this sum, we get $\frac{5}{r^3} \cdot r(1 + r^4) = \frac{5(1 + r^4)}{r^2} = \frac{25}{2}$, which simplifies to the quadratic form $2(1 + r^4) = 5r^2$ or $2r^4 - 5r^2 + 2 = 0$. Solving this for $r^2$ yields $r^2 = 2$ as the required value for an increasing progression. The sum of the fourth, sixth, and eighth terms is $T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4)$, which calculates to $5(1 + 2 + 4) = 35$.

The sum of an infinite geometric series with first term $a$ and common ratio $r$ is determined by the formula $S = \frac{a}{1-r}$. Given that the sum of the original series $a, ar, ar^2, \ldots$ is 15, we can write $\frac{a}{1-r} = 15$. When we consider the series formed by the squares of these terms, $a^2, a^2r^2, a^2r^4, \ldots$, it becomes a new geometric series with first term $a^2$ and common ratio $r^2$, which sums to $\frac{a^2}{1-r^2} = 150$. By factoring the denominator as $(1-r)(1+r)$, this expression becomes $\frac{a}{1-r} \cdot \frac{a}{1+r} = 150$. Since the first sum is 15, substituting this value allows us to find that $\frac{a}{1+r} = 10$.

Comparing the two equations $\frac{a}{1-r} = 15$ and $\frac{a}{1+r} = 10$ by division shows that $\frac{1-r}{1+r} = \frac{10}{15} = \frac{2}{3}$. This ratio simplifies to $3 - 3r = 2 + 2r$, yielding $r = \frac{1}{5}$. Substituting this common ratio back into the equation for the first sum gives $\frac{a}{1 - 1/5} = 15$, which results in $a = 12$. The target series $ar^2, ar^4, ar^6, \ldots$ is an infinite geometric series with a first term of $ar^2$ and a common ratio of $r^2$. The sum of this series is calculated as $\frac{ar^2}{1-r^2}$, which, upon substituting our values, becomes $\frac{12 \cdot (1/25)}{24/25} = \frac{1}{2}$.

The ratio of the sum of four terms in a geometric progression, $a(1+r+r^2+r^3) = \frac{65}{12}$, to the sum of their reciprocals, $\frac{1+r+r^2+r^3}{ar^3} = \frac{65}{18}$, provides an immediate way to isolate the variables. Dividing these two equations cancels the common factor $(1+r+r^2+r^3)$, revealing that $a^2r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$.

Given that the product of the first three terms, $a \cdot ar \cdot ar^2 = a^3r^3$, equals 1, we can substitute $r^3 = \frac{3}{2a^2}$ into this product expression. This substitution yields $a^3 \left( \frac{3}{2a^2} \right) = 1$, which simplifies to $\frac{3a}{2} = 1$, confirming the first term is $a = \frac{2}{3}$. Consequently, the common ratio satisfies $r^3 = \frac{3}{2(2/3)^2} = \frac{3}{2(4/9)} = \frac{27}{8}$, meaning $r = \frac{3}{2}$. The third term $\alpha$ is defined as $ar^2 = \frac{2}{3} \cdot \left( \frac{3}{2} \right)^2 = \frac{3}{2}$, and therefore $2\alpha$ equals 3.

When three terms form an arithmetic progression, the middle term is the arithmetic mean of the surrounding terms, which means that twice the middle term equals the sum of the first and last terms. Applying this property to the given sequence yields $2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - \frac{7}{2})$. Utilizing logarithmic laws to combine the terms on both sides, the equation simplifies to $(2^x - 5)^2 = 2(2^x - \frac{7}{2})$. By substituting $t = 2^x$, the relationship becomes $(t - 5)^2 = 2t - 7$, which rearranges into the quadratic equation $t^2 - 12t + 32 = 0$. Factoring this expression gives $(t - 4)(t - 8) = 0$, resulting in potential values of $t = 4$ or $t = 8$. When $t = 4$, the equation $2^x = 4$ implies $x = 2$, but this value makes the argument of the middle logarithmic term negative, which is undefined. Therefore, the only viable solution is $t = 8$, which corresponds to $2^x = 8$ and yields $x = 3$.

The series $(100)(100)+(99)(101)+(98)(102)+\dots+(1)(199)$ can be expressed as the sum of $(100-x)(100+x)$ for $x$ ranging from $0$ to $99$. Expanding this expression yields $100^2 - x^2$, which allows the total sum to be calculated by evaluating the two components separately:
$\sum_{x=0}^{99} (100^2 - x^2) = \sum_{x=0}^{99} 100^2 - \sum_{x=0}^{99} x^2$
The first part consists of $100$ terms of $100^2$, which equals $100^3$. The second part utilizes the sum of squares formula $\frac{n(n+1)(2n+1)}{6}$ for $n=99$, resulting in $\frac{99(100)(199)}{6}$, which simplifies to $1650(199)$. Substituting these values into the given equation $100^{\alpha}-199 \beta = 100^{3}-199(1650)$ allows for a direct comparison to identify that $\alpha = 3$ and $\beta = 1650$. The slope of the line passing through the origin and $(\alpha, \beta)$ is defined by the ratio $\frac{\beta}{\alpha}$, and calculating $\frac{1650}{3}$ yields $550$.