Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The total sum of an arithmetic progression over a range depends on the number of terms, the starting value, and the common difference. Because the sum of the first 25 terms is equivalent to the sum of the subsequent 15 terms, the sum of the first 40 terms must be exactly twice the sum of the first 25 terms. The formula for the sum of $n$ terms, $S_n = \frac{n}{2} [2a + (n-1)d]$, allows us to express this relationship using the known first term $a = 3$:

$\frac{40}{2} [2(3) + 39d] = 2 \times \frac{25}{2} [2(3) + 24d]$

This simplifies to $20(6 + 39d) = 25(6 + 24d)$, which expands to $120 + 780d = 150 + 600d$. Subtracting $600d$ and $120$ from both sides results in $180d = 30$, leading to the final common difference of $d = \frac{30}{180} = \frac{1}{6}$.

The sum of the first 11 terms of an arithmetic progression follows the formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, and since this total is zero, we find that $\frac{11}{2}(2a_1 + 10d) = 0$, leading to the relation $a_1 = -5d$ or $d = -\frac{a_1}{5}$. The series of odd-indexed terms $a_1, a_3, \dots, a_{23}$ forms an arithmetic progression containing 12 terms, where the first term is $a_1$ and the last term is $a_{23} = a_1 + 22d$. We determine the sum of these terms using the formula $S = \frac{n}{2}(a_1 + a_{23})$, which yields $6(a_1 + a_1 + 22d) = 6(2a_1 + 22d)$. Substituting the value of $d$ into this equation results in:

$6\left(2a_1 + 22\left(-\frac{a_1}{5}\right)\right) = -\frac{72}{5}a_1$

Consequently, the value of $k$ is $-\frac{72}{5}$.

In an arithmetic progression, the middle term is the average of the two terms surrounding it, meaning twice the middle term must equal the sum of the first and third terms. Applying this property to the values provided, $28$ must equal $3^{2\sin 2\alpha-1} + 3^{4-2\sin 2\alpha}$. By defining $t = 3^{2\sin 2\alpha}$, the equation becomes $28 = \frac{t}{3} + \frac{81}{t}$, which simplifies to the quadratic equation $t^2 - 84t + 243 = 0$. Factoring this expression yields $(t-81)(t-3) = 0$, meaning $t$ is either $81$ or $3$. Given $t = 3^{2\sin 2\alpha}$, these solutions imply $2\sin 2\alpha$ is either $4$ or $1$. Since the sine function cannot result in a value greater than $1$, the case where $2\sin 2\alpha = 4$ is disregarded. Using the valid result where $2\sin 2\alpha = 1$, the first term of the sequence is $3^{1-1} = 1$. With the first term being $1$ and the second term being $14$, the common difference is $13$. The sixth term is calculated as $1 + 5 \times 13$, which results in $66$.

Expanding the product $n(n+1)(2n+1)$ gives $2n^3 + 3n^2 + n$, which allows the summation to be distributed across each term for the range $n=1$ to $7$. Applying the standard power sum formulas for this range yields $784$ for $\sum n^3$, $140$ for $\sum n^2$, and $28$ for $\sum n$. Substituting these values into the expanded expression results in $\frac{1}{4}(2(784) + 3(140) + 28)$. This simplifies to $\frac{1}{4}(1568 + 420 + 28)$, which equals $\frac{2016}{4}$ and results in a final value of $504$.

The series can be decomposed into two distinct parts: a geometric progression involving powers of $x$ and an arithmetic series involving $k$ and $a$. Summing the first nine terms of the geometric progression $x + x^2 + x^3 + \dots + x^9$ yields the expression $\frac{x(x^9 - 1)}{x - 1}$, which simplifies to $\frac{x^{10} - x}{x - 1}$.

Summing the nine terms associated with $k$ and $a$ results in the cumulative value of $(9k + 72)a$ after accounting for both the constant $k$ in each term and the arithmetic progression of the even multiples. Equating this component to the given expression $\frac{45a(x-1)}{x-1}$ allows the problem to be solved by setting $9k + 72 = 45$. Simplifying this equation gives $9k = -27$, which results in $k = -3$.

When identifying shared values between two arithmetic progressions, the common terms themselves form a new sequence with a common difference equal to the least common multiple of the two original differences. The first series increases by 4, while the second increases by 7, meaning the overlapping values must advance in steps of 28.

By observing the initial values of both lists, we find that 23 is the first number that appears in both progressions. Because all subsequent shared values are constrained by the upper limit of the shorter sequence, we solve the inequality $23 + (n - 1)28 \leq 407$. Dividing the difference by 28 shows that $n - 1 \leq 13.71$, which results in $n \leq 14.71$. Since the number of terms must be a whole number, there are exactly 14 common terms.

Solving this product relies on expressing each term with a common base and applying the fundamental laws of exponents to simplify the resulting series. Because 4, 8, and 16 are all powers of 2, we can rewrite them as $2^2$, $2^3$, and $2^4$, respectively. By applying the power of a power rule $(a^m)^n = a^{m \cdot n}$, the terms become $2^{\frac{2}{16}}$, $2^{\frac{3}{48}}$, and $2^{\frac{4}{128}}$, which simplify to $2^{\frac{1}{8}}$, $2^{\frac{1}{16}}$, and $2^{\frac{1}{32}}$.

When bases are identical, multiplying the terms corresponds to adding their exponents according to the identity $a^m \cdot a^n = a^{m+n}$. This reduces the expression to $2^{(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots)}$, where the exponent is an infinite geometric series. With the first term $a = \frac{1}{4}$ and a common ratio $r = \frac{1}{2}$, the sum of this series is calculated using the formula $S = \frac{a}{1 - r}$, which yields $\frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$. Consequently, the entire infinite product evaluates to $2^{\frac{1}{2}}$.

To bridge the gap between two numbers using arithmetic and geometric means, one must rely on the fundamental definitions of the common difference and the common ratio. An arithmetic progression generates new terms through constant addition, while a geometric progression does so through constant multiplication, allowing us to express any specific inserted term using the initial boundary values and the number of steps taken.

When inserting $m$ arithmetic means between 3 and 243, the common difference is determined by dividing the total span by the number of gaps, yielding $d = \frac{243-3}{m+1} = \frac{240}{m+1}$. The fourth arithmetic mean, positioned four steps after the first term, is expressed as $3 + 4d$, which becomes $3 + 4\left(\frac{240}{m+1}\right)$. For the geometric sequence containing three means, the common ratio $r$ is calculated by taking the fourth root of the ratio between the boundary terms, specifically $r = \left(\frac{243}{3}\right)^{\frac{1}{3+1}} = 3$. The second geometric mean is the second term in this geometric progression, calculated as $3r^2 = 3(3)^2 = 27$. Equating the fourth arithmetic mean to this value, $3 + 4\left(\frac{240}{m+1}\right) = 27$, simplifies to $\frac{960}{m+1} = 24$, which requires $m+1 = 40$ and confirms that $m = 39$.

This arithmetic series features a decreasing sequence starting with $a = \frac{100}{5}$ and a common difference $d = -\frac{2}{5}$. Applying the summation formula $S_n = \frac{n}{2}[2a + (n-1)d] = 488$ leads to the following equation:

$n^2 - 101n + 2440 = 0$

Factoring this quadratic shows the number of terms $n$ can be either 40 or 61. Since the sequence is defined to have a negative final term, the $n^{\text{th}}$ term formula $T_n = a + (n-1)d$ helps distinguish between these possibilities. Substituting $n = 61$ yields $T_{61} = 20 + 60(-\frac{2}{5}) = -4$.

The sum of integers $1+2+3+\dots+k$ is given by the formula $\frac{k(k+1)}{2}$, which reduces the task to calculating $\sum_{k=1}^{20} \frac{k(k+1)}{2}$. Recognizing that $\frac{k(k+1)}{2}$ is the binomial coefficient $\binom{k+1}{2}$ permits the use of the hockey-stick identity, $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$. Applying this with $r=2$ and the upper limit $n=21$ results in the binomial coefficient $\binom{22}{3}$. Calculating this value as $\frac{22 \times 21 \times 20}{3 \times 2 \times 1}$ yields 1540.

The infinite geometric series $\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdot s$ evaluates to $\frac{1/3}{1-1/3} = \frac{1}{2}$. Substituting this into the main expression yields $(\frac{4}{25})^{\log_{5/2}(1/2)}$, where $0.16 = \frac{4}{25}$ and $2.5 = \frac{5}{2}$. By applying the logarithmic rearrangement property $a^{\log_b c} = c^{\log_b a}$, the expression transforms into $(\frac{1}{2})^{\log_{5/2}(4/25)}$. Since $\frac{4}{25}$ is equivalent to $(\frac{5}{2})^{-2}$, the exponent becomes $-2$, leading to $(\frac{1}{2})^{-2} = 4$.

This relationship is essentially a collection of perfect square trinomials disguised through algebraic expansion. By regrouping the terms, we can write the equation as
$(ap-b)^{2} + (bp-c)^{2} + (cp-d)^{2} = 0$
Since the square of any real number is non-negative, the only way for the sum of these values to result in zero is if each individual squared term is exactly zero. This leads to the conditions $ap=b$, $bp=c$, and $cp=d$, which confirms that the constant ratio $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$ holds true, establishing that the sequence $a, b, c, d$ forms a Geometric Progression.

Every term in an arithmetic progression can be written as $a_n = a_1 + (n-1)d$. By expressing the given sum $a_1 + a_7 + a_{16} = 40$ in terms of $a_1$ and $d$, we obtain the equation $a_1 + (a_1 + 6d) + (a_1 + 15d) = 40$, which simplifies to $3a_1 + 21d = 40$. Factoring out 3 from this result allows us to isolate $a_1 + 7d = \frac{40}{3}$. The formula for the sum of the first 15 terms is $S_{15} = \frac{15}{2} [2a_1 + 14d]$, which further simplifies to $15(a_1 + 7d)$. Substituting the value found earlier, the total sum becomes $15 \times \frac{40}{3}$, resulting in 200.

The geometric progression terms $a_1, a_2, \dots, a_{10}$ can be expressed in terms of the first term $a_1$ and common ratio $R$ as $a_n = a_1 R^{n-1}$. When substituting these into the logarithmic expressions, each matrix entry simplifies to a linear combination of $\log_e a_1$ and $\log_e R$. By performing column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$, the difference between consecutive columns becomes $(r+k) \log_e R$ for every row.

Since the column operations result in identical values for both $C_2 - C_1$ and $C_3 - C_2$, the columns of the determinant are in an arithmetic progression and are linearly dependent. A determinant with linearly dependent columns is identically zero regardless of the chosen values for $r$ and $k$. Because this identity holds for any pair of natural numbers $(r, k)$, the set $S$ contains infinitely many such pairs.

Since $a, b, c$ follow a geometric progression, they can be defined as $a$, $ar$, and $ar^2$. Given that $3a, 7b, 15c$ form an arithmetic progression, the middle term must be the arithmetic mean of the surrounding terms, leading to $2(7b) = 3a + 15c$, or $14b = 3a + 15c$.

Substituting the geometric terms into this relation gives $14ar = 3a + 15ar^2$. Dividing both sides by $a$ results in the quadratic equation $15r^2 - 14r + 3 = 0$. Factoring this expression as $(3r - 1)(5r - 3) = 0$ reveals two possible values for $r$, namely $1/3$ and $3/5$. Applying the condition $0 < r \leq 1/2$, we identify $r = 1/3$ as the only valid common ratio.

Using $r = 1/3$, the first three terms of the arithmetic progression become $3a$, $7a/3$, and $5a/3$. The common difference is calculated as $7a/3 - 3a = -2a/3$. Adding this difference to the third term yields $5a/3 - 2a/3 = a$, which is the fourth term.

Any two-digit number yielding a remainder of 2 or 5 when divided by 7 fits into one of two specific arithmetic progressions governed by the expressions $7n + 2$ and $7n + 5$. For the first expression, the valid two-digit values range from 16 to 93, which represents 12 terms that sum to 654. For the second expression, the valid two-digit values range from 12 to 96, which represents 13 terms that sum to 702. Combining these two sums yields a final total of 1356.

The sequence is defined by a general term that incorporates the sums of cubes and squares, which can be simplified using standard algebraic identities. The $n$-th term $T_n$ is formed by $(2n+1)$ multiplied by the ratio of the sum of the first $n$ cubes to the sum of the first $n$ squares. Substituting the identities $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ allows for a significant reduction of the expression.

The general term simplifies as follows:
$T_n = (2n+1) \cdot \frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2}$
With this streamlined expression, finding the sum of the first 10 terms involves calculating $\sum_{n=1}^{10} \frac{3}{2}(n^2+n)$. Expanding the summation results in $\frac{3}{2} \left( \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} n \right)$. Utilizing the standard sums $\sum_{n=1}^{10} n^2 = 385$ and $\sum_{n=1}^{10} n = 55$, the total becomes $\frac{3}{2}(385 + 55)$, which is $\frac{3}{2}(440)$, resulting in 660.

The definition of an arithmetic progression dictates that for three numbers to exist in such a sequence, twice the middle term must equal the sum of the surrounding terms, expressed here as $2 \cdot {}^{n}C_{5} = {}^{n}C_{4} + {}^{n}C_{6}$. Expanding these binomial coefficients using factorial notation and dividing by the common $n!$ allows us to simplify the relationship into $\frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$. Solving this equation by clearing the denominators results in the quadratic expression $n^2 - 21n + 98 = 0$. Factoring this quadratic confirms that $n = 14$ is the valid solution.

In an arithmetic progression, any term can be represented as $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. Expanding the given sum $a_{1}+a_{4}+a_{7}+a_{10}+a_{13}+a_{16}=114$ using this general form yields $a + (a+3d) + (a+6d) + (a+9d) + (a+12d) + (a+15d) = 114$, which simplifies to $6a + 45d = 114$. Dividing this equation by three establishes the relationship $2a + 15d = 38$.

Expressing the target sum $a_{1}+a_{6}+a_{11}+a_{16}$ in terms of $a$ and $d$ results in $a + (a+5d) + (a+10d) + (a+15d)$, which combines to $4a + 30d$. Factoring out a two reveals that this expression is equivalent to $2(2a + 15d)$. Substituting the previously derived value of $38$ into this expression leads to $2 \times 38$, which equals 76.

The Arithmetic Mean-Geometric Mean inequality provides the necessary constraint for this expression, establishing that for any positive value $k$, the sum $k + \frac{1}{k}$ is always greater than or equal to $2$. By reformatting the expression to isolate terms of the form $\frac{1}{k + \frac{1}{k}}$, it becomes evident that each individual fraction is at most $\frac{1}{2}$ because the denominator must be at least $2$. Multiplying these upper bounds together confirms that the entire expression is bounded by $\frac{1}{4}$.