Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Calculations involving the sum of binomial coefficients rely on the standard identities $\sum_{r=0}^{n} {}^{n}C_r = 2^n$, $\sum_{r=1}^{n} r {}^{n}C_r = n 2^{n-1}$, and $\sum_{r=1}^{n} r^2 {}^{n}C_r = n(n+1) 2^{n-2}$. Breaking down the expression for $\alpha$ into individual sums gives
$\alpha = 4 \sum_{r=0}^{n} r^2 {}^{n}C_r + 2 \sum_{r=0}^{n} r {}^{n}C_r + \sum_{r=0}^{n} {}^{n}C_r$
Substituting the identities leads to $4(n(n+1) 2^{n-2}) + 2(n 2^{n-1}) + 2^n$, which simplifies to $n(n+1) 2^n + n 2^n + 2^n$, factoring further to $(n+1)^2 2^n$.

For the expression $\beta$, the identity $\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1} = \frac{2^{n+1}-1}{n+1}$ simplifies the sum. Adding the remaining term $\frac{1}{n+1}$ results in $\beta = \frac{2^{n+1}-1+1}{n+1} = \frac{2^{n+1}}{n+1}$. Consequently, the ratio $\frac{2\alpha}{\beta}$ is equivalent to
$\frac{2 \cdot (n+1)^2 2^n}{2^{n+1} / (n+1)} = (n+1)^3$
Solving the inequality $140 < (n+1)^3 < 281$ requires evaluating integer cubes, where $n=4$ yields 125, $n=5$ results in 216, and $n=6$ results in 343, identifying $n=5$ as the valid integer.

The sum of an arithmetic progression follows the formula $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ represents the first term and $d$ is the common difference. Substituting the provided values leads to $10(2a + 19d) = 790$, which simplifies to $2a + 19d = 79$, and $5(2a + 9d) = 145$, which simplifies to $2a + 9d = 29$. Subtracting these linear equations yields $10d = 50$, or $d = 5$, and back-substituting this result confirms the first term as $a = -8$. Computing the difference requires evaluating $S_{15} = \frac{15}{2}[2(-8) + 14(5)] = 405$ and $S_5 = \frac{5}{2}[2(-8) + 4(5)] = 10$, giving a final result of $S_{15} - S_5 = 405 - 10 = 395$.

The expression $(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3})$ is the binomial expansion of $(1+\frac{1}{x})^3$, which simplifies the entire problem significantly. By substituting this and simplifying the full expression $(1+x)(1-x^2) \frac{(1+x)^{15}}{x^{15}}$, we obtain $(1+x)(1-x^2) \frac{(1+x)^{15}}{x^{15}}$. Since $1-x^2$ factors into $(1-x)(1+x)$, this expression reduces to $\frac{(1-x)(1+x)^{17}}{x^{15}}$, which can be written as $(1+x)^{17}x^{-15} - (1+x)^{17}x^{-14}$.

For the coefficient of $x^3$, identifying the $x^{18}$ term from the first part (coefficient 0) and the $x^{17}$ term from the second part (coefficient $\binom{17}{17}=1$) gives $0 - 1 = -1$. For the coefficient of $x^{-13}$, identifying the $x^2$ term from the first part (coefficient $\binom{17}{2}=136$) and the $x^1$ term from the second part (coefficient $\binom{17}{1}=17$) yields $136 - 17 = 119$. Combining these results provides a final sum of $119 - 1 = 118$.

The general term for the numerator is $r(r+1)^2 = r^3 + 2r^2 + r$, while the denominator follows the form $r^2(r+1) = r^3 + r^2$. Summing these expressions from $r=1$ to $n=100$ allows for the application of standard power sum formulas, and factoring out $\frac{n(n+1)}{2}$ from both the numerator and denominator simplifies the entire ratio to $\frac{\frac{n(n+1)}{2} + \frac{2}{3}(2n+1) + 1}{\frac{n(n+1)}{2} + \frac{2n+1}{3}}$.

Substituting $n=100$ into this reduced form, the numerator calculates to $5050 + 134 + 1 = 5185$, and the denominator calculates to $5050 + 67 = 5117$. Simplifying the fraction $\frac{5185}{5117}$ by dividing both the numerator and denominator by 17 yields the final value of $\frac{305}{301}$.

The base 64 can be evaluated modulo 9 using basic arithmetic. Since $64 = 7 \times 9 + 1$, we observe that $64 \equiv 1 \pm od{9}$. Any integer power of 1 remains 1, regardless of how large the exponent $32^{32}$ is. Therefore, the total expression simplifies as $64^{32^{32}} \equiv 1^{32^{32}} \equiv 1 \pm od{9}$, confirming that the remainder is 1.

The arithmetic mean of three numbers in arithmetic progression is equal to the middle term, so given the mean of $a, b, c$ is $8$, we immediately have $b=8$, which implies $a+c=16$ or $c=16-a$. The condition that $a+1, b, c+3$ form a geometric progression requires that the square of the middle term equals the product of the outer terms, expressed as $b^2=(a+1)(c+3)$. Substituting the known values $b=8$ and $c=16-a$ results in $64=(a+1)(19-a)$, which simplifies to the quadratic equation $a^2-18a+45=0$. Solving this gives the roots $a=15$ and $a=3$, and since the constraint $a>10$ applies, we must have $a=15$, which leads to $c=1$. The problem asks for the cube of the geometric mean of $a, b, c$, which is equivalent to the product $abc$, and calculating this gives $15 \times 8 \times 1 = 120$.

The recurrence relation $T_r = 3T_{r-1} + 6^r$ defines the terms of the series where $T_1 = 6$. By dividing this expression by $3^r$, it transforms into a difference of geometric series terms, yielding the general formula $T_r = 2(6^r - 3^r)$. Summing this expression from $r=1$ to $n$ using the standard geometric series formula results in $S_n = \frac{3}{5}(4 \cdot 6^n - 5 \cdot 3^n + 1)$.

Matching this derived sum with the expression $\frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$ requires that the coefficient $\frac{1}{5}(n^2 - 12n + 39)$ be equal to $\frac{3}{5}$. This leads to the quadratic equation $n^2 - 12n + 39 = 3$, which simplifies to $n^2 - 12n + 36 = 0$. Solving the factored form $(n-6)^2 = 0$ results in $n=6$.

Arithmetic progressions rely on a constant difference $d$ between consecutive terms, while geometric progressions require the ratio between consecutive terms to remain constant, which dictates that the square of the middle term must equal the product of the flanking terms. With an initial term of $1$, the second, eighth, and forty-fourth terms are identified as $1+d$, $1+7d$, and $1+43d$. Establishing these as a geometric progression leads to the governing equation $(1+7d)^2 = (1+d)(1+43d)$

Expanding the terms provides $1 + 14d + 49d^2 = 1 + 44d + 43d^2$. Subtracting common values from both sides simplifies the expression to $6d^2 - 30d = 0$, which highlights $6d(d - 5) = 0$ as the core constraint. Because the arithmetic progression is non-constant, $d$ must equal $5$.

The sum of the first twenty terms of this progression is determined by the formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. Substituting $n = 20$, $a_1 = 1$, and $d = 5$ produces $S_{20} = \frac{20}{2}[2(1) + 19(5)]$. Solving this final expression yields $10[2 + 95]$, which results in a total of $970$.

The core concept used here is Pascal's Identity, which states that the sum of two consecutive binomial coefficients is ${}^n C_r + {}^n C_{r+1} = {}^{n+1} C_{r+1}$. Since $2-p$ and $p$ are consecutive terms, their sum is $(2-p) + p = 2$, identifying ${}^{n+1} C_{r+1} = 2$. Applying the same principle to the subsequent pair, $2-\alpha$ and $\alpha$ sum to $(2-\alpha) + \alpha = 2$, yielding ${}^{n+1} C_{r+3} = 2$. These binomial properties allow us to simplify the expression $p^2 - \alpha^2 + 6\alpha + 2p$ by relating the variables $p$ and $\alpha$ through their binomial sums. Upon evaluating the expression with these conditions, the result equals 2.

The starting value for any row $n$ follows the sequence pattern defined by $a_n = \frac{3n^2-3n+4}{2}$. By substituting $n=10$ into this expression, we find that the tenth row begins with the term $a_{10} = 137$. This row forms an arithmetic progression containing 10 terms with a common difference of 3, which allows us to calculate the sum using $S_n = \frac{n}{2}(2a + (n-1)d)$. Applying this formula with our specific values yields $S_{10} = \frac{10}{2}(2(137) + 9(3))$, which simplifies to $5(274 + 27)$ and results in a total sum of 1505.

The foundation of this problem lies in simplifying the ratio of the two binomial coefficients. Using the definition ${}^n C_r = \frac{n!}{r!(n-r)!}$, the given relationship can be rewritten as $\frac{{}^{n-1}C_{r}}{{}^{n}C_{r+1}} = k^2 - 8$. Expanding these terms using factorials allows for significant cancellation, simplifying the entire expression on the left side of the equation to $\frac{r+1}{n}$. Since $n$ and $r$ are integers where $n \geq r+1 \geq 1$, the ratio $\frac{r+1}{n}$ must be strictly greater than zero and less than or equal to one.

Applying these bounds to our simplified equation results in the inequality $0 < k^2 - 8 \leq 1$. Breaking this into two parts, the condition $k^2 - 8 > 0$ implies $k^2 > 8$, which means $k > 2\sqrt{2}$ or $k < -2\sqrt{2}$. The condition $k^2 - 8 \leq 1$ leads to $k^2 \leq 9$, which restricts $k$ to the interval $-3 \leq k \leq 3$. Combining these requirements and focusing on the positive interval, we find that valid values for $k$ must satisfy $2\sqrt{2} < k \leq 3$.

In a geometric progression, any term can be represented as $T_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio. We start by examining the product of the third and fifth terms, which is $(ar^2)(ar^4) = a^2r^6 = (ar^3)^2 = 49$. Given that all terms in the progression are positive, we determine that $ar^3 = 7$. Next, we address the sum of the second and sixth terms, $ar + ar^5 = ar(1 + r^4) = \frac{70}{3}$. By substituting $a = \frac{7}{r^3}$ into this equation, we obtain $\frac{7}{r^2}(1 + r^4) = \frac{70}{3}$, which simplifies to the relationship $3(1 + r^4) = 10r^2$.

Solving the quadratic equation $3r^4 - 10r^2 + 3 = 0$ by letting $t = r^2$ gives $3t^2 - 10t + 3 = 0$. Factoring this quadratic results in $(3t - 1)(t - 3) = 0$, meaning $t = 3$ or $t = \frac{1}{3}$. Since the progression is increasing, the common ratio must satisfy $r > 1$, which implies $r^2 = 3$. We are looking for the sum of the fourth, sixth, and eighth terms, which is $ar^3 + ar^5 + ar^7$. Factoring this expression gives $ar^3(1 + r^2 + r^4)$. Substituting the known values of $ar^3 = 7$, $r^2 = 3$, and $r^4 = 9$ into the expression, we get $7(1 + 3 + 9)$, which equals 91.

The sum of this series relies on expressing the general term $T_r = \frac{r}{1-3r^2+r^4}$ as a difference of two functions to facilitate telescoping cancellation. Factoring the denominator as $(r^2-1)^2-r^2$ allows it to be split into $(r^2-r-1)(r^2+r-1)$, which enables the general term to be rewritten via partial fractions as $\frac{1}{2} \left( \frac{1}{r^2-r-1} - \frac{1}{r^2+r-1} \right)$. With $f(r) = \frac{1}{r^2-r-1}$, one observes that $f(r+1) = \frac{1}{r^2+r-1}$, so the series essentially becomes $\sum_{r=1}^{10} \frac{1}{2} (f(r) - f(r+1))$. Due to the telescoping property, the sum reduces to $\frac{1}{2} (f(1) - f(11))$. Evaluating $f(1) = \frac{1}{1^2-1-1} = -1$ and $f(11) = \frac{1}{11^2-11-1} = \frac{1}{109}$ results in a final sum of $\frac{1}{2} (-1 - \frac{1}{109})$, which equals $-\frac{55}{109}$.

For any quadratic equation in the form $Ax^2 + Bx + C = 0$, notice that the sum of the coefficients $A+B+C$ is $(a+b-2c) + (b+c-2a) + (c+a-2b)$, which simplifies to zero. This immediately reveals that $x=1$ is always one of the roots of the equation. Since the product of the roots of a quadratic equation is the ratio $C/A$, the other root $\alpha$ must be equal to $\frac{c+a-2b}{a+b-2c}$.

When $\alpha$ lies in the interval $(-1, 0)$, the expression $\frac{c+a-2b}{a+b-2c}$ must be negative. Given the condition $0 < c < b < a$, the denominator $a+b-2c$ is always positive, which forces the numerator $c+a-2b$ to be negative, leading to the condition $b > \frac{a+c}{2}$. Because the geometric mean $\sqrt{ac}$ is always strictly less than the arithmetic mean $\frac{a+c}{2}$, $b$ cannot be the geometric mean if it is already greater than the arithmetic mean, confirming the first statement is true.

When $\alpha$ lies in the interval $(0, 1)$, the inequality $0 < \frac{c+a-2b}{a+b-2c} < 1$ implies that the numerator is positive and also less than the denominator. The condition $c+a-2b > 0$ simplifies to $b < \frac{a+c}{2}$, and combining this with the premise $c < b$ gives the range $c < b < \frac{a+c}{2}$. Since the geometric mean $\sqrt{ac}$ always exists within this interval, $b$ may indeed be the geometric mean of $a$ and $c$, which confirms the second statement is true.

Partial fraction decomposition allows each term in the series to be expressed as the difference of two simpler fractions, since the numerator is the constant difference between the factors in the denominator, which is $(1+kd) - (1+(k-1)d) = d$. By multiplying and dividing the series by $d$, each general term becomes $\frac{1}{d} \left( \frac{1}{1+(k-1)d} - \frac{1}{1+kd} \right)$. Summing these terms from $k=1$ to $10$ creates a telescoping effect where all intermediate terms cancel out, leaving only the first and last parts: $\frac{1}{d} \left[ \left( 1 - \frac{1}{1+d} \right) + \left( \frac{1}{1+d} - \frac{1}{1+2d} \right) + \dots + \left( \frac{1}{1+9d} - \frac{1}{1+10d} \right) \right] = 5$. This simplifies to the equation $\frac{1}{d} \left( 1 - \frac{1}{1+10d} \right) = 5$, which further reduces to $\frac{1}{d} \left( \frac{10d}{1+10d} \right) = 5$ and then $\frac{10}{1+10d} = 5$. Solving this equation gives $10 = 5 + 50d$, which simplifies to $5 = 50d$.

Begin by determining the sum of the first $n$ terms of the given arithmetic progression with first term $a=3$ and common difference $d=4$, resulting in $S_n = 2n^2+n$. The sum of these values from $k=1$ to $n$ is calculated using the standard identities for the sum of squares and natural numbers, which simplifies to the expression
$\sum_{k=1}^{n} S_k = \frac{n(n+1)(4n+5)}{6}$
Applying this result to the given inequality and multiplying by the factor $\frac{6}{n(n+1)}$ reduces the expression to $40 < 4n+5 < 42$. Solving for $n$ provides the range $35 < 4n < 37$, which simplifies to $8.75 < n < 9.25$. Since $n$ must be an integer, $n$ equals 9.

The sum of all coefficients in any polynomial is obtained by setting the variable $x$ to $1$. Applying this rule to the expression $(1-3x+10x^2)^n$, we find $A = (1-3+10)^n = 8^n$. Evaluating the expression $(1+x^2)^n$ in the same way yields $B = (1+1)^n = 2^n$. Since $8$ is $2^3$, we can express $A$ as $(2^3)^n$, which simplifies to $(2^n)^3$, resulting in the relationship $A = B^3$.

The expansion of a binomial expression of the form $\left(a^{\frac{1}{p}} + b^{\frac{1}{q}}\right)^n$ produces terms where the exponents of the irrational bases must be integers for the entire term to result in an integer. For the expansion $\left(7^{\frac{1}{2}} + 11^{\frac{1}{6}}\right)^{824}$, the general term is given by $T_{r+1} = {}^{824}C_{r} (7)^{\frac{824-r}{2}} (11)^{\frac{r}{6}}$, where $r$ spans from 0 to 824.

For the term to be an integer, the exponent of 11 must be a whole number, which requires $r$ to be a multiple of 6. Because every multiple of 6 is also an even number, requiring $r$ to be a multiple of 6 automatically ensures that the exponent of 7, which is $\frac{824-r}{2}$, also results in an integer. By identifying the multiples of 6 in the range $0$ to $824$, we see the sequence follows $6 \times 0, 6 \times 1, \dots, 6 \times k$, where $6k \leq 824$. Dividing 824 by 6 yields approximately 137.33, meaning the possible values for $k$ are integers from 0 to 137. This range includes 138 distinct values, meaning there are 138 such integral terms in the expansion.

The sum of the coefficients $\alpha_r$ in a polynomial is determined by evaluating that polynomial at $x=1$. Replacing $x$ with $1$ in the expression $(x+3)^{n-1} + (x+3)^{n-2}(x+2) + \dots + (x+2)^{n-1}$ results in the geometric progression $4^{n-1} + 4^{n-2}(3) + 4^{n-3}(3^2) + \dots + 3^{n-1}$. This series contains $n$ terms with a first term $a = 4^{n-1}$ and a common ratio $r = \frac{3}{4}$. Using the geometric series sum formula $a \frac{1-r^n}{1-r}$, the sum becomes $4^{n-1} \frac{1 - (\frac{3}{4})^n}{1 - \frac{3}{4}}$, which simplifies to $4^n - 3^n$. Equating this expression to $\beta^n - \gamma^n$ identifies the values $\beta = 4$ and $\gamma = 3$, leading to the result:

$\beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25$

An arithmetic progression is fully defined by its first term $a$ and common difference $d$, where the sum of the first $n$ terms follows the formula $S_n = \frac{n}{2}(2a + (n-1)d)$ and the $n$-th term is given by $t_n = a + (n-1)d$. Using the given sum $S_{10} = 390$, we can establish the equation $\frac{10}{2}(2a + 9d) = 390$, which simplifies to $2a + 9d = 78$.

The relationship between the tenth and fifth terms is provided by the ratio $\frac{a+9d}{a+4d} = \frac{15}{7}$. Cross-multiplying this fraction gives $7a + 63d = 15a + 60d$, which rearranges to $8a = 3d$, or $d = \frac{8a}{3}$. Substituting this expression for $d$ back into the first equation yields $2a + 9(\frac{8a}{3}) = 78$, which simplifies to $2a + 24a = 78$, or $26a = 78$, resulting in $a = 3$. Using this value of $a$, the common difference becomes $d = \frac{8(3)}{3} = 8$.

To find the value of $S_{15} - S_5$, we substitute the values $a = 3$ and $d = 8$ into the expression $\frac{15}{2}(2a + 14d) - \frac{5}{2}(2a + 4d)$. This calculation becomes $\frac{15}{2}(6 + 14 \times 8) - \frac{5}{2}(6 + 4 \times 8)$, which simplifies to $\frac{15}{2}(118) - \frac{5}{2}(38)$. Computing these terms gives $15 \times 59 - 5 \times 19$, resulting in $885 - 95$, which equals 790.