Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

An arithmetic progression starting with 1 has terms $a_n = 1 + (n-1)d$, while a geometric progression starting with 1 follows $g_n = r^{n-1}$. Using the first condition $a_2 + g_2 = 1$, we write $(1+d) + r = 1$, which reveals the relationship $d = -r$. Substituting this into the second equation $a_3 + g_3 = 4$ leads to $(1 + 2d) + r^2 = 4$, which simplifies to $1 - 2r + r^2 = 4$, resulting in the quadratic equation $r^2 - 2r - 3 = 0$. Factoring this quadratic gives $(r-3)(r+1) = 0$, yielding solutions of $r = 3$ or $r = -1$. Because the geometric progression is increasing, we must choose $r = 3$, which implies the common difference $d = -3$. With these parameters fixed, the tenth term of the arithmetic progression is $a_{10} = 1 + 9(-3) = -26$ and the fifth term of the geometric progression is $g_5 = 1 \cdot 3^4 = 81$, and their sum is 55.

Inserting $n$ terms between two values to form an arithmetic progression requires determining the common difference by dividing the total gap between the boundary values by the number of steps, which is $n+1$. Given the boundary values of 59 and 159 with 39 inserted terms, the common difference is defined by $d = \frac{159 - 59}{39 + 1} = \frac{100}{40} = 2.5$. Each arithmetic mean $A_k$ follows the pattern $A_k = a + kd$, where $a$ is the first term.

To find the average of the terms $A_{25}, A_{28}, A_{31}$, and $A_{36}$, calculate the mean of these expressions:
$\frac{A_{25} + A_{28} + A_{31} + A_{36}}{4} = \frac{(a + 25d) + (a + 28d) + (a + 31d) + (a + 36d)}{4}$
Simplifying the numerator results in $\frac{4a + 120d}{4}$, which reduces to $a + 30d$. Substituting the values $a = 59$ and $d = 2.5$ into this simplified form gives $59 + 30(2.5) = 59 + 75 = 134$.

The denominator of the general term $1+4r^4$ can be factored using the identity $(2r^2+1)^2 - (2r)^2$, which resolves into $(2r^2 - 2r + 1)(2r^2 + 2r + 1)$. By applying partial fraction decomposition, the general term $\frac{r}{1+4r^4}$ is equivalent to $\frac{1}{4} \left( \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} \right)$, which establishes that the series is telescopic.

Summing these terms from $r=1$ to $10$ causes the intermediate terms to cancel out, leaving only the first part of the expression evaluated at $r=1$ and the second part evaluated at $r=10$. This leaves $\frac{1}{4} \left( \frac{1}{2(1)^2 - 2(1) + 1} - \frac{1}{2(10)^2 + 2(10) + 1} \right)$, which simplifies to $\frac{1}{4} \left( 1 - \frac{1}{221} \right)$. The resulting fraction is $\frac{55}{221}$, where $m = 55$ and $n = 221$. Given these values are coprime, $m + n = 276$.

To determine the sum of the squares of the first 10 terms of an arithmetic progression, one must first identify the general expression for the $n$-th term. Given the sum of the first $n$ terms $S_n = 3n^2 + 5n$, the $n$-th term $a_n$ is found by calculating the difference $S_n - S_{n-1}$. Substituting the given formula, this results in $a_n = (3n^2 + 5n) - (3(n-1)^2 + 5(n-1))$, which simplifies to $a_n = 6n + 2$.

With the expression for the $n$-th term established, the task is to evaluate the summation $\sum_{n=1}^{10} (6n + 2)^2$. Factoring out a constant, this is equivalent to $4 \sum_{n=1}^{10} (3n + 1)^2$. Expanding the term inside the summation leads to $4 \sum_{n=1}^{10} (9n^2 + 6n + 1)$, which can be separated into three distinct parts: $36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$. Utilizing the standard summation formulas for the first $n$ squares and integers, where $\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$ and $\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$, the expression becomes $36(385) + 24(55) + 40$. Evaluating these values yields $13860 + 1320 + 40$, resulting in a final total of 15220.

The series $1^3 - 2^3 + 3^3 - \dots + 15^3$ is evaluated by grouping consecutive terms into pairs of the form $n^3 - (n+1)^3$ for $n = 1, 3, \dots, 13$, while leaving the last term $15^3$ isolated. Utilizing the expansion:
$n^3 - (n + 1)^3 = -3n^2 - 3n - 1$
the sum of these seven pairs results in -1519. Adding the final value of 3375 to this sum leads to the total of 1856.

The intersection of two arithmetic progressions is itself an arithmetic progression, and its common difference is the least common multiple of the individual common differences. To identify the common terms, we solve the system of linear congruences defined by the first sequence $x \equiv 1 \pm od 5$ and the second sequence $x \equiv 9 \equiv 2 \pm od 7$. This system identifies the common terms as those following the form $x \equiv 16 \pm od{35}$, which establishes a new sequence with a first term of 16 and a common difference of 35.

Since the first progression contains 101 terms ending at 501 and the second contains 71 terms ending at 499, the common terms must not exceed 499. By representing the intersection terms as $x = 35k + 16$ for non-negative integers $k$, we find the upper bound by solving the inequality $35k + 16 \le 499$. This results in $k \le 13.8$, meaning there are 14 terms in the intersection sequence, ranging from $k=0$ to $k=13$, which are $\{16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471\}$.

To determine which of these elements are divisible by 3, we require $35k + 16 \equiv 0 \pm od 3$. Simplifying this congruence, we observe that $2k + 1 \equiv 0 \pm od 3$, which implies $2k \equiv 2 \pm od 3$ or simply $k \equiv 1 \pm od 3$. Within our valid range for $k$ of 0 through 13, the values satisfying this condition are $k \in \{1, 4, 7, 10, 13\}$. Substituting these values back into the expression $35k + 16$ provides the set $\{51, 156, 261, 366, 471\}$, confirming that there are 5 such elements.

The sum 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. By applying this to the sum of the first four terms, $S_4 = \frac{4}{2}(2a + 3d) = 6$, we simplify the relationship to $2a + 3d = 3$. Similarly, for the sum of the first six terms, $S_6 = \frac{6}{2}(2a + 5d) = 4$, which simplifies to $6a + 15d = 4$.

Multiplying the first simplified equation by $3$ yields $6a + 9d = 9$. Subtracting this from the second equation, $6a + 15d = 4$, eliminates $a$ and results in $6d = -5$, giving $d = -\frac{5}{6}$. Substituting this value back into $2a + 3d = 3$ provides $2a - \frac{5}{2} = 3$, which resolves to $2a = \frac{11}{2}$ or $a = \frac{11}{4}$.

Finding the sum of the first twelve terms requires substituting $a = \frac{11}{4}$ and $d = -\frac{5}{6}$ into the expression $S_{12} = \frac{12}{2}[2a + 11d]$. This calculation becomes $6[2(\frac{11}{4}) + 11(-\frac{5}{6})]$, which simplifies further to $6[\frac{11}{2} - \frac{55}{6}]$. Converting the fractions to a common denominator results in $6[\frac{33}{6} - \frac{55}{6}]$, and completing the subtraction leads to $6(-\frac{22}{6})$, which equals $-22$.

The properties of an arithmetic progression link the sum of its terms directly to its first and last elements via the formula $S_n = \frac{n}{2}(a + l)$. With 30 terms and a first term $a = \frac{10}{3}$, the last term is expressed as $T_{30} = \frac{10}{3} + 29d$. The sum then becomes $S_{30} = \frac{30}{2} \left( \frac{10}{3} + T_{30} \right) = 15 \left( \frac{10}{3} + T_{30} \right)$. Since the problem requires this sum to equal the cube of the last term, the relationship is $15 \left( \frac{10}{3} + T_{30} \right) = (T_{30})^3$.

Substituting $x$ for the last term $T_{30}$ leads to the cubic equation $x^3 - 15x - 50 = 0$. Checking for potential roots shows that $x = 5$ satisfies the equality, as $125 - 75 - 50 = 0$. Equating the expression for the last term back to this result gives $\frac{10}{3} + 29d = 5$. Subtracting $\frac{10}{3}$ from both sides results in $29d = \frac{5}{3}$, which identifies the common difference as $d = \frac{5}{87}$.

Representing the four terms of an arithmetic progression as $a-3d$, $a-d$, $a+d$, and $a+3d$ is an effective strategy because it simplifies the sum equation. Given that the sum of these terms equals 48, we can write $4a=48$, which directly determines $a=12$. Since the common difference $l$ is $2d$, the product of the terms combined with $l^4$ satisfies the equation $(12-3d)(12-d)(12+d)(12+3d) + (2d)^4 = 361$.

Expanding this expression by grouping the conjugate pairs leads to $(144-9d^2)(144-d^2) + 16d^4 = 361$, which simplifies further to $20736 - 144d^2 - 1296d^2 + 9d^4 + 16d^4 = 361$. Combining the like terms results in the equation $25d^4 - 1440d^2 + 20375 = 0$. Dividing this entire equation by 5 yields $5d^4 - 288d^2 + 4075 = 0$.

Using the quadratic formula to solve for $d^2$, we find $d^2 = \frac{288 \pm \sqrt{288^2 - 20(4075)}}{10}$, which simplifies to $d^2 = \frac{288 \pm 38}{10}$. Since the terms must be integers, we choose $d^2=25$, meaning $d=5$ and the common difference $l$ is 10. Substituting $a=12$ and $d=5$ back into our expressions for the four terms yields $-3$, 7, 17, and 27. The largest term in this sequence is 27.

Since the roots $\alpha, \beta, \gamma, \delta$ form a geometric progression, we denote them as $a, ar, ar^2, ar^3$. According to Vieta's formulas, the sum of the roots for the equation $x^2-x+p=0$ is given by $a(1+r)=1$, and for the equation $x^2-4x+q=0$, the sum is $ar^2(1+r)=4$. Dividing the second equation by the first yields $r^2=4$, which implies that the common ratio $r$ is either $2$ or $-2$.

Testing $r=2$ leads to $a=1/3$, which fails the condition that $p$ and $q$ must be integers. Choosing $r=-2$ results in $a=-1$, establishing the roots as -1, 2, -4, 8. The products $p = \alpha\beta = -2$ and $q = \gamma\delta = -32$ are consistent with the requirement for integer coefficients. The final value is calculated as
$\left| p+q \right| = \left| -2-32 \right| = 34$

Recognizing the structure of the Taylor series expansion for the exponential function, $e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!}$, is essential for evaluating this sum, specifically for the case where $x = -1$. The general term of the summation can be decomposed by simplifying the factorial expression $\frac{k(k+1)}{k!}$ into $\frac{k+1}{(k-1)!}$, which then expands to $\frac{k-1}{(k-1)!} + \frac{2}{(k-1)!}$, resulting in the simplified components $\frac{1}{(k-2)!} + \frac{2}{(k-1)!}$.

When distributing the summation over these terms, the first part $\sum_{k=2}^{\infty } \frac{(-1)^{k+1}}{(k-2)!}$ effectively shifts to the series expansion for $e^{-1}$ with a leading negative sign, yielding $-e^{-1}$ after accounting for the index change $j = k-2$. The second part $\sum_{k=1}^{\infty } \frac{2(-1)^{k+1}}{(k-1)!}$ similarly transforms into $2e^{-1}$ by setting $m = k-1$. Combining these two components, $-e^{-1} + 2e^{-1}$, leads to the final result of $e^{-1}$, which is equal to $\frac{1}{e}$.

Because the sequence follows a geometric progression, the product of any three consecutive terms is essentially the cube of the middle term. Expressing the terms $a_2$, $a_3$, and $a_4$ as $\frac{a_3}{r}$, $a_3$, and $a_3 r$ simplifies the product equation $a_2 \cdot a_3 \cdot a_4 = 64$ to $a_3^3 = 64$, which identifies the third term as $a_3 = 4$.

With $a_3$ determined, the sum $a_1 + a_3 + a_5 = \frac{813}{7}$ can be expressed in terms of the common ratio $r$. Given that $a_1 = \frac{a_3}{r^2}$ and $a_5 = a_3 r^2$, the equation becomes $\frac{4}{r^2} + 4 + 4 r^2 = \frac{813}{7}$. Substituting $t = r^2$ leads to the quadratic expression $4(\frac{1}{t} + 1 + t) = \frac{813}{7}$, which simplifies to $28t^2 - 785t + 28 = 0$. Factoring this quadratic yields $(28t - 1)(t - 28) = 0$. Since the sequence consists of increasing positive terms, the common ratio squared must be greater than one, establishing that $t = r^2 = 28$.

The required sum $a_3 + a_5 + a_7$ can be rewritten as $a_3 + a_3 r^2 + a_3 r^4$, which is equivalent to $r^2(\frac{a_3}{r^2} + a_3 + a_3 r^2)$. This structure reveals that the sum is simply $r^2(a_1 + a_3 + a_5)$. Substituting the established values $r^2 = 28$ and $a_1 + a_3 + a_5 = \frac{813}{7}$ results in $28 \times \frac{813}{7} = 4 \times 813 = 3252$.

The characteristic behavior of arithmetic progressions relies on the fact that the difference between any two terms is proportional to the number of steps separating them, as defined by the relation $t_n - t_m = (n - m)d$. For the sequence with terms $b_n$, the gap between the 43rd and 31st positions spans 12 steps, meaning $-385 - (-277) = 12d_b$, which simplifies to $12d_b = -108$ and identifies the common difference as $-9$.

With the first progression utilizing a common difference 13 units higher, calculating $-9 + 13$ yields a common difference of 4. The general term formula for the 78th term indicates that $327 = a_1 + 77 \times 4$, which simplifies to $327 = a_1 + 308$. Subtracting 308 from 327 leaves 19 as the first term.

The sequence $1, 3, 11, 25, 45, 71, \dots$ displays a constant second difference of 6, establishing that the general term is defined by the quadratic expression $T_r = 3r^2 - 7r + 5$. To determine the sum up to $n=20$ terms, one calculates the series $\sum_{r=1}^{20} (3r^2 - 7r + 5)$.

This summation expands to $3 \sum_{r=1}^{20} r^2 - 7 \sum_{r=1}^{20} r + \sum_{r=1}^{20} 5$. Substituting the values into the standard summation formulas $\frac{n(n+1)(2n+1)}{6}$ and $\frac{n(n+1)}{2}$ yields $3 \left( \frac{20 \cdot 21 \cdot 41}{6} \right) - 7 \left( \frac{20 \cdot 21}{2} \right) + 5(20)$. Evaluating this expression gives $8610 - 1470 + 100$, which totals 7240.

The sum 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 $d$ represents the common difference. To solve this problem, we first equate the sum of the initial four terms, calculated as $\frac{4}{2}(2a + 3d)$, to one-fifth of the sum of the subsequent four terms, which is expressed as $4a + 22d$.

Establishing this relationship yields the equation $4a + 6d = \frac{1}{5}(4a + 22d)$. Multiplying both sides by 5 simplifies the expression to $20a + 30d = 4a + 22d$, which reduces to $16a = -8d$. Given that the first term $a$ is 3, we substitute this value to find that $48 = -8d$, resulting in a common difference of $d = -6$. With both $a$ and $d$ determined, calculating the sum of the first 20 terms involves substituting these values into the sum formula: $S_{20} = \frac{20}{2}[2(3) + 19(-6)]$. This simplifies to $10(6 - 114)$, which equals $10(-108)$, yielding a final result of $-1080$.

The sum of an arithmetic progression depends on the first term $a$ and the common difference $d$. The sum of the first 12 odd-indexed terms forms a new progression with a first term of $a$ and a common difference of $2d$. Applying the sum formula $S = \frac{k}{2}[2a + (k-1)D]$ with $k = 12$ and $D = 2d$ results in $12a + 132d$. Setting this equal to $-\frac{72}{5}a$ allows us to determine the relation between $a$ and $d$, which simplifies to $a = -5d$. When the total sum of the first $n$ terms, expressed as $\frac{n}{2}[2a + (n-1)d]$, is zero, the term within the brackets must be zero, implying $2a + (n-1)d = 0$. Substituting $a = -5d$ yields $-10d + (n-1)d = 0$, and since $d$ is non-zero, the expression simplifies to $-10 + n - 1 = 0$, which results in $n = 11$.

The structure of this series reveals that the $r$-th term is composed of the sum of the first $r$ odd numbers in the numerator, which simplifies to $r^2$, divided by $r!$ in the denominator. This gives us the general term $T_r = \frac{r^2}{r!}$. By simplifying this expression, we reduce it to $\frac{r}{(r-1)!}$, which can be further decomposed by rewriting the numerator as $(r-1) + 1$, yielding the split form $\frac{1}{(r-2)!} + \frac{1}{(r-1)!}$. Summing these components from $r=1$ to infinity equates to the sum of two separate infinite series each converging to $e$, resulting in $2e$.

Vieta's formulas establish that for a quadratic equation $3x^2 - px + q = 0$, the sum of its roots equals $\frac{p}{3}$ and their product equals $\frac{q}{3}$. By calculating the $10^{\text{th}}$ and $11^{\text{th}}$ terms of the arithmetic progression, which serve as these roots, we can directly determine the values of $p$ and $q$. Given the sum of the first 11 terms is 88 with a common difference $d = \frac{3}{2}$, we use the formula
$S_{11} = \frac{11}{2}(2a + 10d) = 88$
to solve for the first term $a$. Simplifying this leads to $a + 5d = 8$, and substituting $d = \frac{3}{2}$ yields $a + \frac{15}{2} = 8$, so $a = \frac{1}{2}$.

With $a$ and $d$ determined, the $10^{\text{th}}$ term $T_{10} = a + 9d = \frac{1}{2} + \frac{27}{2} = 14$ and the $11^{\text{th}}$ term $T_{11} = a + 10d = \frac{1}{2} + \frac{30}{2} = \frac{31}{2}$ represent the roots of the quadratic equation. Using Vieta's relations, the sum of the roots is
$\frac{p}{3} = 14 + \frac{31}{2} = \frac{59}{2}$
which results in $p = \frac{177}{2}$. Similarly, the product of the roots is
$\frac{q}{3} = 14 \cdot \frac{31}{2} = 217$
resulting in $q = 651$. Evaluating the expression $q - 2p$ yields $651 - 2\left(\frac{177}{2}\right) = 651 - 177$, which equals 474.

The sum of the first $n$ terms of an arithmetic progression follows the formula $S_n = \frac{n}{2}[2a+(n-1)d]$, where $a$ denotes the first term and $d$ denotes the common difference. Applying this to the provided sums, $S_{40}=1030$ becomes $20(2a+39d)=1030$, which simplifies to $2a+39d=51.5$. Similarly, $S_{12}=57$ transforms into $6(2a+11d)=57$, simplifying to $2a+11d=9.5$.

Subtracting the second equation from the first isolates the variable $d$ as $28d=42$, establishing that the common difference is $d=1.5$. Inserting this value back into the equation $2a+11(1.5)=9.5$ results in $2a+16.5=9.5$, or $2a=-7$. Calculating the difference $S_{30}-S_{10}$ involves evaluating $15(2a+29d)-5(2a+9d)$. Substituting $2a=-7$ and $d=1.5$ into this expression yields $15(-7+29 \times 1.5)-5(-7+9 \times 1.5)$, which expands to $15(-7+43.5)-5(-7+13.5)$. This equals $15(36.5)-5(6.5)$, providing the final result of $547.5-32.5=515$.

For any arithmetic progression of three terms summing to 36, the terms are symmetric around the average value of 12, taking the form $12-d, 12, 12+d$. Consequently, the products $p$ and $q$ for sets $A$ and $B$ are defined as $p = 12(144-d^2)$ and $q = 12(144-D^2)$. The ratio $\frac{p+q}{p-q} = \frac{19}{5}$ simplifies to $5(p+q) = 19(p-q)$, which yields the relationship $12q = 7p$

By substituting the definitions of $p$ and $q$ and utilizing the constraint $D=d+3$, we obtain the equation $12(144-(d+3)^2) = 7(144-d^2)$ Substituting the expansion $D^2 = d^2+6d+9$ into this expression simplifies the left side to $12(135-d^2-6d) = 7(144-d^2)$, leading to the quadratic equation $5d^2+72d-612 = 0$ Solving this quadratic results \in $d=6$, which implies $D=9$. Finally, using these common differences, the products are $p = 12(144-36) = 1296$ and $q = 12(144-81) = 756$, resulting \in the difference $p-q = 540$.