Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The general term of the series $\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots$ is $\frac{1}{n(n+1)}$, which decomposes into the partial fractions $\frac{1}{n} - \frac{1}{n+1}$. Expanding this alternating sum leads to $(1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) \dots$, which simplifies to $1 - 2(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} \dots)$.

Since the logarithmic series expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ evaluates to $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \dots$ at $x=1$, the infinite bracketed expression $(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots)$ is equivalent to $1 - \ln(2)$. Substituting this into the main expression results in $1 - 2(1 - \ln(2))$, which simplifies to $2\ln(2) - 1$. Converting this into a single logarithmic expression yields $\ln(4) - \ln(e)$, which is equal to $\log_e \left( \frac{4}{e} \right)$.

The determinant evaluates to zero because its first column consists of terms that are linearly dependent on the second and third columns. In a geometric progression where terms are defined by $l = A R^{p-1}$, $m = A R^{q-1}$, and $n = A R^{r-1}$, taking the logarithm of each term yields the linear equation $\log(\text{term}) = \log A + (\text{index}-1) \log R$.

Substituting these logarithmic expressions into the determinant, the first column entries $\log l$, $\log m$, and $\log n$ can be rewritten as $\log A + (p-1) \log R$, $\log A + (q-1) \log R$, and $\log A + (r-1) \log R$, respectively. Because every entry in the first column is now a linear combination of the entries in the second and third columns, the determinant property regarding linear dependence dictates that the value of the determinant is 0.

Calculating the sum of an alternating series of cubes is most efficient when considering the entire sequence as a sum of all cubes from $1^3$ through $9^3$ and then adjusting for the negative even terms. The standard formula for the sum of the first $n$ cubes is $\left[ \frac{n(n+1)}{2} \right]^2$, which for the range up to $9$ results in $\left[ \frac{9 \times 10}{2} \right]^2 = 2025$. Because the even terms $2^3, 4^3, 6^3,$ and $8^3$ were included as positive values in this initial sum, they must be removed twice to achieve the required alternating pattern. Factoring out $2^3$ from these terms yields $8 \times (1^3+2^3+3^3+4^3) = 800$, and subtracting twice this amount from the total leads to the final calculation:

$2025 - 1600 = 425$

For any three terms to form an arithmetic progression, the middle term must be the arithmetic mean of the surrounding terms, which is expressed by the property $2b = a + c$. Applying this principle to the sequence $1, \log_9 (3^{1-x}+2)$, and $\log_3 (4 \cdot 3^x-1)$ leads to the equation $2 \log_9 (3^{1-x}+2) = 1 + \log_3 (4 \cdot 3^x-1)$. By using the change of base rule where $\log_9 y = \frac{\log_3 y}{\log_3 9} = \frac{\log_3 y}{2}$, the expression simplifies to $\log_3 (3^{1-x}+2) = \log_3 3 + \log_3 (4 \cdot 3^x-1)$. Combining the logarithmic terms on the right side results in $\log_3 (3^{1-x}+2) = \log_3 (3(4 \cdot 3^x-1))$, which implies that the arguments must be equal: $3^{1-x} + 2 = 12 \cdot 3^x - 3$.

Substituting $t = 3^x$ converts the expression $3^{1-x}$ into $\frac{3}{t}$, resulting in the equation $\frac{3}{t} + 2 = 12t - 3$. Multiplying through by $t$ transforms this into the quadratic equation $12t^2 - 5t - 3 = 0$. Factoring this quadratic gives $(4t - 3)(3t + 1) = 0$, yielding solutions $t = 3/4$ and $t = -1/3$. Since $t = 3^x$ must be positive, we discard the negative root and solve $3^x = 3/4$. Taking the logarithm base 3 of both sides gives $x = \log_3 (3/4)$, which simplifies using the quotient rule to $\log_3 3 - \log_3 4$, ultimately resulting in $1 - \log_3 4$.

The sum of an infinite geometric series with first term $a$ and common ratio $r$ is defined by the formula $\frac{a}{1-r}$. When each term of the series is squared, the new sequence $a^2, a^2r^2, a^2r^4, \dots$ is also an infinite geometric progression, characterized by a first term of $a^2$ and a common ratio of $r^2$, resulting in a sum expressed as $\frac{a^2}{1-r^2}$. Based on the information provided, the first sum is $20$, which gives the equation $\frac{a}{1-r} = 20$, or equivalently $a = 20(1-r)$. The second sum is $100$, leading to the equation $\frac{a^2}{1-r^2} = 100$, which can be rearranged as $a^2 = 100(1-r^2)$.

To find the value of the common ratio, square the first expression to obtain $a^2 = 400(1-r)^2$. Equating this to the squared sum expression gives $400(1-r)^2 = 100(1-r^2)$. Factoring the right side using the difference of squares identity results in $400(1-r)^2 = 100(1-r)(1+r)$. Dividing both sides by $100(1-r)$ simplifies the equation to $4(1-r) = 1+r$. Expanding this leads to $4 - 4r = 1 + r$, and isolating the variable $r$ results in $5r = 3$, which confirms the common ratio is $\frac{3}{5}$.

In a geometric progression with an odd number of terms, the product of the sequence is equal to the middle term raised to the power of the total number of terms. For a sequence of 9 terms, the fifth term serves as the middle value, which means the product is equivalent to $(ar^4)^9$. Since the fifth term is 2, the expression simplifies to $2^9$, which equals 512.

The product relies on expressing each term with the same base, which allows the exponents to be added together. Rewriting the factors, $4^{\frac{1}{8}}$ becomes $(2^2)^{\frac{1}{8}} = 2^{\frac{2}{8}}$ and $8^{\frac{1}{16}}$ becomes $(2^3)^{\frac{1}{16}} = 2^{\frac{3}{16}}$, transforming the overall expression into $2^{\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots}$.

The exponent represents an infinite arithmetico-geometric series $S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$. Multiplying this series by $\frac{1}{2}$ produces $\frac{1}{2}S = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \dots$. Subtracting this equation from $S$ yields $\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$. The right side is now an infinite geometric series with a first term $a = \frac{1}{4}$ and a common ratio $r = \frac{1}{2}$, which sums to $\frac{a}{1-r} = \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{1}{2}$. Given that $\frac{1}{2}S = \frac{1}{2}$, the sum of the series $S = 1$, leading to a final result of $2^1 = 2$.

This problem relies on the properties of arithmetic progressions and the rationalization of radical denominators to resolve a telescoping sum. Given the relation $a_5 = 2a_7$, we substitute the standard forms $a_5 = a_1 + 4d$ and $a_7 = a_1 + 6d$ to get $a_1 + 4d = 2(a_1 + 6d)$, which simplifies to $a_1 + 8d = 0$. By solving this alongside the second condition $a_{11} = a_1 + 10d = 18$, we subtract the equations to find $2d = 18$, yielding the common difference $d = 9$ and the initial term $a_1 = -72$.

Each term within the summation follows the form $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$. By multiplying the numerator and denominator by the conjugate $\sqrt{a_{k+1}} - \sqrt{a_k}$, the expression simplifies to $\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k}$, which reduces to $\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$ since the difference between consecutive terms in an arithmetic progression is constant. The series thus becomes a telescoping sum: $\frac{1}{d} (\sqrt{a_{11}} - \sqrt{a_{10}} + \sqrt{a_{12}} - \sqrt{a_{11}} + \dots + \sqrt{a_{18}} - \sqrt{a_{17}})$, which ultimately simplifies to $\frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d}$.

We now compute the values $a_{10} = a_1 + 9d = -72 + 81 = 9$ and $a_{18} = a_1 + 17d = -72 + 153 = 81$. Substituting these values into the expression $12 \left( \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d} \right)$ yields $12 \left( \frac{\sqrt{81} - \sqrt{9}}{9} \right)$. This simplifies to $12 \left( \frac{9 - 3}{9} \right)$, which is equivalent to $12 \times \frac{6}{9}$ or $12 \times \frac{2}{3}$, resulting in 8.

The sum of coefficients in a binomial expansion is determined by setting all variables equal to 1, which for the expression $\left(\sqrt{x} - 6x^{-\frac{3}{2}}\right)^{n}$ yields $(1-6)^n = (-5)^n$. The constant term $\alpha$ represents the term in the expansion where the power of $x$ is zero; using the general term $T_{k+1} = {}^{n}C_{k} (\sqrt{x})^{n-k} (-6x^{-\frac{3}{2}})^k$, this condition simplifies to $n-4k=0$. Since the sum of the remaining coefficients is the total sum minus the constant term, we have the relation $(-5)^n - \alpha = 649$. Given $\alpha = {}^{n}C_{\frac{n}{4}}(-6)^{\frac{n}{4}}$, we test integer values for $n$ where $n \leq 15$ and find that $n=4$ satisfies the equation, as $(-5)^4 - {}^{4}C_1(-6)^1 = 625 + 24 = 649$.

With $n=4$, the constant term is calculated as $\alpha = {}^{4}C_1(-6)^1 = -24$. To find the coefficient of $x^{-n}$, we look for the term where the exponent of $x$ is $-4$. Setting the general exponent $\frac{4-4k}{2} = -4$ results in $4-4k = -8$, which gives $k=3$. The coefficient of this specific term is ${}^{4}C_3(-6)^3 = 4 \times (-216) = -864$. Setting this result equal to $\lambda \alpha$, we have $-864 = \lambda(-24)$, which confirms that $\lambda = 36$.