Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The sum of the first $n$ terms of an arithmetic progression is determined by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ represents the first term and $d$ is the common difference. Applying this to the provided information creates two equations: the first from $S_4 = 16$, which simplifies to $2a + 3d = 8$, and the second from $S_6 = -48$, which simplifies to $2a + 5d = -16$.

Subtracting the first equation from the second produces $2d = -24$, establishing that $d = -12$. Substituting this value into $2a + 3d = 8$ gives $2a - 36 = 8$, which means $2a = 44$.

With these values, the sum of the first ten terms is calculated using $S_{10} = \frac{10}{2}[2a + 9d]$. Substituting the known values yields the equation $S_{10} = 5[44 + 9(-12)]$, which simplifies to $S_{10} = 5[44 - 108] = -320$.

The series $1+2 \times 3+3 \times 5+4 \times 7+\dots$ follows a pattern where the $r$-th term is defined by the product of $r$ and $(2r-1)$, which simplifies to $2r^2 - r$. To find the sum of these 11 terms, we calculate $\sum_{r=1}^{11} (2r^2 - r)$, which expands into $2 \sum_{r=1}^{11} r^2 - \sum_{r=1}^{11} r$.

Applying the standard summation formulas for the sum of squares, $\frac{n(n+1)(2n+1)}{6}$, and the sum of integers, $\frac{n(n+1)}{2}$, with $n=11$, the expression becomes $2 \times \frac{11 \times 12 \times 23}{6} - \frac{11 \times 12}{2}$. This simplifies to $1012 - 66$, resulting in a final sum of 946.

To maximize the product of three specific terms in an Arithmetic Progression where $a_6 = 2$, we first express each term in terms of the first term $a_1$ and the common difference $d$. Given $a_6 = a_1 + 5d = 2$, we can define $a_1 = 2 - 5d$, $a_4 = 2 - 2d$, and $a_5 = 2 - d$. The product $P$ is thus a function of $d$ defined by $P(d) = (2 - 5d)(2 - 2d)(2 - d)$, which expands to the polynomial $P(d) = -10d^3 + 34d^2 - 32d + 8$.

To find the value of $d$ that maximizes this product, we determine the first derivative with respect to $d$, given by $P'(d) = -30d^2 + 68d - 32$. Setting $P'(d) = 0$ to identify the critical points, we arrive at the quadratic equation $15d^2 - 34d + 16 = 0$. We can factor this expression by splitting the middle term, resulting in $15d^2 - 10d - 24d + 16 = 0$, which simplifies to $(5d - 8)(3d - 2) = 0$. This gives us two critical values for the common difference: $d = \frac{8}{5}$ and $d = \frac{2}{3}$. Evaluating the second derivative $P''(d) = -60d + 68$ at these points shows that $d = \frac{8}{5}$ produces a negative value, confirming it as the local maximum.

The $n$-th term of an arithmetic progression is defined by the formula $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. Since the 19th term is zero, we have the equation $a + 18d = 0$, which establishes that $a = -18d$. To determine the ratio of the 49th term to the 29th term, we evaluate the expression $\frac{a + 48d}{a + 28d}$ by substituting $a = -18d$ to get $\frac{-18d + 48d}{-18d + 28d}$. Simplifying this fraction leads to $\frac{30d}{10d}$, which results in a final ratio of 3 : 1.

The problem requires calculating the sum of the floor function for 100 terms starting at $-\frac{1}{3}$ and decreasing by $\frac{1}{100}$ at each step, where the floor function maps a value to the greatest integer less than or equal to $x$. Since the terms range from approximately -0.33 down to -1.33, the result for each term must be either -1 or -2. To identify the transition point, we determine when $-\frac{1}{3} - \frac{k}{100}$ falls below -1, which occurs when $k > 66.67$. This means the 67 terms from $k=0$ to $k=66$ evaluate to -1, and the 33 terms from $k=67$ to $k=99$ evaluate to -2. Summing these results yields $67 \times (-1) + 33 \times (-2)$, which simplifies to $-67 - 66$ and results in -133.

Evaluating this series requires identifying it as an arithmetico-geometric progression, where each term is the product of an increasing integer and a power of $1/2$. We define $S$ as the total sum, then subtract $\frac{S}{2}$ from $S$ to simplify the expression into a standard geometric series. This subtraction produces the terms $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{20}}$ alongside a final term of $-\frac{20}{2^{21}}$. Using the geometric sum formula, the series portion equates to $1 - \frac{1}{2^{20}}$, which simplifies the equation to $\frac{S}{2} = 1 - \frac{11}{2^{20}}$. Doubling this result isolates $S$ and yields the final value of $2 - \frac{11}{2^{19}}$.

An arithmetic progression is fundamentally defined by its first term $a_1$ and a constant common difference $d$, such that any term $a_n$ is expressed as $a_1 + (n-1)d$. When evaluating sums of such sequences, the total sum $S$ of thirty terms is given by $\frac{30}{2}(a_1 + a_{30})$, which simplifies to $15(2a_1 + 29d)$, while the sum $T$ of the fifteen odd-indexed terms $a_1, a_3, \dots, a_{29}$ can be calculated as $15a_1 + 210d$.

Substituting these expressions into the relation $S - 2T = 75$ gives $15(2a_1 + 29d) - 2(15a_1 + 210d) = 75$, which simplifies directly to $15d = 75$, resulting in $d = 5$. With the common difference determined, the value of the first term is found using $a_5 = a_1 + 4d = 27$, meaning $a_1 + 20 = 27$, so $a_1 = 7$. Applying these values to the tenth term, $a_{10} = a_1 + 9d$ becomes $7 + 9(5)$, which equals 52.

The sum of the first 15 terms of the given series follows the algebraic structure $\frac{27}{64} \left[ \frac{15(15+1)}{2} \right]^2$. Evaluating the term within the square brackets leads to $120$, which produces $14400$ when squared. Multiplying this by the fraction $\frac{27}{64}$ results in the value of $27 \times 225$. By equating this total sum to the expression $225 k$, we can isolate the variable by dividing both sides by $225$, which reveals that $k$ equals $27$.

Connecting arithmetic and geometric progressions relies on the formula for the $n$-th term, $T_n = A + (n-1)d$, and the property that three terms $a, b, c$ in a geometric progression satisfy $b^2 = ac$. Using the given indices, the terms are $a = A + 6d$, $b = A + 10d$, and $c = A + 12d$. The relationship for these terms follows the equality:

$(A + 10d)^2 = (A + 6d)(A + 12d)$

Expanding both sides results in $A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2$, which simplifies to $2Ad = -28d^2$. Since the progression is non-constant, $d \neq 0$, allowing us to determine that $\frac{A}{d} = -14$. Calculating the ratio $\frac{a}{c}$ as $\frac{A + 6d}{A + 12d}$ leads to $\frac{\frac{A}{d} + 6}{\frac{A}{d} + 12}$, which becomes $\frac{-14 + 6}{-14 + 12} = \frac{-8}{-2} = 4$.

The fundamental property of a geometric progression states that any term $a_n$ is expressed as $a_1 r^{n-1}$, where $r$ is the common ratio. By applying this to the given relationship $\frac{a_3}{a_1} = 25$, the expression simplifies to $\frac{a_1 r^2}{a_1} = r^2$, establishing that $r^2 = 25$. The target ratio $\frac{a_9}{a_5}$ is similarly expanded to $\frac{a_1 r^8}{a_1 r^4}$, which reduces to $r^4$. Since $r^4$ is equivalent to $(r^2)^2$, substituting our known value yields $25^2$, which corresponds to $5^4$.

When dealing with three consecutive terms of a geometric progression, representing them as $\frac{a}{r}$, $a$, and $ar$ simplifies the algebra because their product eliminates the common ratio. Given that this product equals 512, we can quickly determine the middle term by solving $a^3 = 512$, which yields $a = 8$.

Modifying the sequence by adding 4 to the first two terms produces the values $\frac{8}{r} + 4$, 12, and $8r$. Because these now form an arithmetic progression, the arithmetic mean property applies, requiring that twice the central term equals the sum of the surrounding terms. This creates the relationship $24 = \frac{8}{r} + 4 + 8r$ which simplifies to the quadratic $2r^2 - 5r + 2 = 0$. Solving for $r$ gives values of 2 or 0.5, both of which generate the same set of numbers, 4, 8, and 16, resulting in a total sum of 28.

When working with the sum of terms in an arithmetic progression, selecting terms symmetrically around the middle value significantly simplifies the algebra. By representing the first three terms as $a-d$, $a$, and $a+d$, their sum equals $3a$, which immediately reveals the middle term to be $33 / 3 = 11$. Substituting this middle term into the product equation yields $(11-d)(11)(11+d) = 1155$, which reduces to $121 - d^2 = 105$ after dividing both sides by 11.

The resulting equation, $d^2 = 16$, indicates that the common difference $d$ can be either 4 or -4. Selecting the case where $d = -4$, the first term of the sequence becomes $11 - (-4) = 15$. Using the general formula for the $n^{\text{th}}$ term, $t_n = a + (n-1)d$, the eleventh term is found by calculating $15 + 10(-4)$, which results in -25.

Infinite geometric series depend on the first term $a$ and common ratio $r$, where the sum is defined by the formula $\frac{a}{1-r}$. When each term of the series is cubed, the new series $a^3, a^3r^3, a^3r^6, \dots$ also forms an infinite geometric series with a first term of $a^3$ and a common ratio of $r^3$.

For the original series, $\frac{a}{1-r} = 3$, which implies that $a = 3(1-r)$. Substituting this value for $a$ into the formula for the sum of the cubed series, $\frac{a^3}{1-r^3} = \frac{27}{19}$, we obtain the equation $\frac{27(1-r)^3}{1-r^3} = \frac{27}{19}$. By applying the algebraic identity $1-r^3 = (1-r)(1+r+r^2)$ to the denominator, we can divide both sides by 27 and cancel the $(1-r)$ factor, leading to the simplified expression $\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$.

Cross-multiplying this equation yields $19(1-2r+r^2) = 1+r+r^2$, which expands to $19 - 38r + 19r^2 = 1+r+r^2$. Rearranging all terms to one side results in the quadratic equation $18r^2 - 39r + 18 = 0$, which reduces further by dividing by 3 to $6r^2 - 13r + 6 = 0$. Factoring this quadratic gives $(3r-2)(2r-3) = 0$, suggesting potential ratios of $r = \frac{2}{3}$ or $r = \frac{3}{2}$. Because an infinite geometric series converges only when the absolute value of the common ratio is less than one, we discard the value greater than one, leaving the common ratio as $\frac{2}{3}$.

The equation $\sin^4 \alpha + 4 \cos^4 \beta + 1 + 1 = 4\sqrt{2} \sin \alpha \cos \beta$ is an application of the Arithmetic Mean-Geometric Mean inequality, which states that for positive values, the arithmetic mean is greater than or equal to the geometric mean, with equality occurring only when all terms are equivalent. By setting each component on the left-hand side equal, we find that $\sin^4 \alpha = 1$ and $4 \cos^4 \beta = 1$. This implies $\sin \alpha = 1$ and $\cos \beta = \pm \frac{1}{\sqrt{2}}$, and because $\beta$ is within the interval $[0, \pi]$, $\sin \beta$ must be the positive value $\frac{1}{\sqrt{2}}$. Substituting these values into the trigonometric identity $\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2 \sin \alpha \sin \beta$ results in $-2(1)(\frac{1}{\sqrt{2}})$, which simplifies to $-\sqrt{2}$.

Each term in the sequence can be simplified by recognizing that the sum of the first $n$ cubes is equal to the square of the sum of the first $n$ integers. Expressing the $n$-th term as $\frac{\sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} k}$, we substitute the known formulas $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. This results in the $n$-th term reducing to $\frac{n(n+1)}{2}$, which is equivalent to $\frac{n^2 + n}{2}$.

Summing these terms from $n=1$ to $15$ gives us $\sum_{n=1}^{15} \frac{n^2 + n}{2}$. The problem also requires subtracting $\frac{1}{2} \sum_{n=1}^{15} n$ from this total. Combining these expressions, the linear terms $\frac{n}{2}$ cancel out, leaving us with $\frac{1}{2} \sum_{n=1}^{15} n^2$. Using the standard formula for the sum of squares, $\frac{n(n+1)(2n+1)}{6}$, we substitute $n=15$ to calculate $\frac{1}{2} \times \frac{15 \times 16 \times 31}{6}$. Performing this arithmetic, $\frac{1240}{2}$ equals $620$.

For any three distinct real numbers in a geometric progression, we can represent them as $\frac{b}{r}$, $b$, and $br$, where $r$ is the common ratio. Since the terms must be distinct, $r$ cannot equal 1, and substituting these into the equation $a + b + c = xb$ leads to the relationship $x - 1 = r + \frac{1}{r}$. Because $r$ is a real number not equal to 1 or -1, the expression $r + \frac{1}{r}$ must either be greater than 2 or less than -2. This implies that $x - 1$ must be greater than 2 or less than -2, which means $x > 3$ or $x < -1$, effectively excluding 2 as a possible value for $x$.

Numbers $n$ satisfying $100 < n < 200$ with $H.C.F. (91, n) > 1$ must be divisible by at least one of the prime factors of 91, which are 7 and 13. According to the inclusion-exclusion principle, the total sum is found by adding the sum of all multiples of 7 to the sum of all multiples of 13, and then subtracting the sum of multiples of 91 to correct for double-counting those common terms.

The multiples of 7 in this range are 105, 112, ..., 196, which represent 14 terms. Their sum is calculated as $\frac{14}{2} \times (105 + 196) = 2107$. The multiples of 13 are 104, 117, ..., 195, totaling 8 terms. Their sum is $\frac{8}{2} \times (104 + 195) = 1196$. Within this interval, the only multiple of 91 is 182. Subtracting this overlap from the combined total yields the final result: $2107 + 1196 - 182 = 3121$.

The $n$-th term of any arithmetic sequence relates to the sum of its terms through the property $a_n = S_n - S_{n-1}$, allowing us to derive the general term formula from the provided expression for $S_n$. Subtracting $S_{n-1}$ from $S_n = 50n + \frac{n(n-7)}{2}A$ leads to $a_n = (n-4)A + 50$, which reveals that the common difference $d$ corresponds to the coefficient of $n$, equal to $A$. Calculating the 50th term by substituting 50 for $n$ yields $a_{50} = (50-4)A + 50$, which simplifies to $50 + 46A$, establishing the final ordered pair as $(d, a_{50}) = (A, 50 + 46A)$.

Identifying the general term $T_n$ by observing how the numerator and denominator components evolve with the index $n$ allows for a dramatic simplification of the series structure. For this sequence, the $n$-th term is defined as
$T_n = \frac{3n \cdot \frac{n(n+1)(2n+1)}{6}}{2n+1} = \frac{n^2(n+1)}{2}$
This reduction converts the complex series into the summation of simpler cubic and quadratic terms.

Calculating the sum for 15 terms requires applying standard summation identities within the equation
$S_{15} = \frac{1}{2} \left( \sum_{n=1}^{15} n^3 + \sum_{n=1}^{15} n^2 \right)$
Substituting the specific totals where the sum of cubes is 14400 and the sum of squares is 1240, the expression becomes $\frac{1}{2} (14400 + 1240)$. Completing the final arithmetic yields 7820.

The sum of the first $k$ integers is given by $\frac{k(k+1)}{2}$, which simplifies the expression for $S_k$ to $\frac{k+1}{2}$. Calculating the sum of squares $S_1^2 + S_2^2 + \dots + S_{10}^2$ thus becomes equivalent to computing $\sum_{k=1}^{10} \left(\frac{k+1}{2}\right)^2$, which can be factored as $\frac{1}{4} (2^2 + 3^2 + \dots + 11^2)$.

To evaluate the sequence of squares from $2^2$ to $11^2$, it is efficient to find the sum of squares from $1^2$ to $11^2$ and subtract the initial term $1^2$. Using the standard formula $\frac{n(n+1)(2n+1)}{6}$ where $n=11$, the sum is $\frac{11 \times 12 \times 23}{6} = 506$. Subtracting $1$ yields $505$, making the entire sum of the original series $\frac{505}{4}$. Setting this equal to $\frac{5}{12} A$ allows for the final calculation of $A = \frac{505 \times 12}{4 \times 5} = 101 \times 3 = 303$.