Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Finding the sum of all three-digit multiples of 11 less than or equal to 500 that exclude the digit 1 involves first calculating the sum of every three-digit multiple of 11 in this range and subsequently subtracting those containing the digit 1. The multiples of 11 from 121 to 495 comprise 35 terms, calculated as $n = \frac{495-121}{11} + 1$, and their sum is determined by the formula $\frac{35}{2}(121+495) = 10780$.

Next, the subsets containing the digit 1 must be systematically excluded from this total. Multiples with 1 in the hundreds place range from 121 to 198, totaling 8 terms with a sum of $\frac{8}{2}(121+198) = 1276$. Multiples with 1 in the tens place, excluding those already accounted for in the hundreds place, include 319 and 418, which sum to 737. Finally, multiples with 1 only in the units place are 231, 341, and 451, with a sum of 1023. Subtracting these excluded values from the initial total of 10780 leaves 7744.

The partial fraction decomposition technique allows us to express a product in the denominator as a sum of simpler terms, where each numerator coefficient $A_k$ is found by ignoring the $(\alpha + k)$ term and evaluating the remaining expression at $\alpha = -k$. Applying this cover-up rule specifically to the given product of terms from $\alpha$ to $\alpha+20$, we find the coefficients $A_{14}$ and $A_{13}$ through direct substitution. By evaluating the ratio $\frac{A_{14}}{A_{13}}$, we determine the value to be $-\frac{1}{2}$, while the ratio $\frac{A_{15}}{A_{13}}$ evaluates to $\frac{1}{5}$.

Combining these ratios within the requested expression, we calculate $100\left(\frac{A_{14}}{A_{13}} + \frac{A_{15}}{A_{13}}\right)^2$ by substituting $-\frac{1}{2}$ and $\frac{1}{5}$ into the parentheses. This simplifies to $100\left(-\frac{1}{2} + \frac{1}{5}\right)^2$, which becomes $100\left(-\frac{3}{10}\right)^2$ and results in a final value of 9.

The sum of the first $k$ terms of an arithmetic progression follows the general formula $S_k = \frac{k}{2} [2a + (k-1)d]$. By examining the difference between the sum of the first $4n$ terms and the first $2n$ terms, we can identify a specific algebraic expression that relates directly to the sum of the first $6n$ terms.

Setting up the equation based on the provided difference gives:
$\frac{4n}{2}[2a + (4n-1)d] - \frac{2n}{2}[2a + (2n-1)d] = 1000$
Expanding this expression, we have:
$2n[2a + (4n-1)d] - n[2a + (2n-1)d] = 1000$
$4an + n(8n-2)d - 2an - n(2n-1)d = 1000$
Combining the like terms simplifies the equation to:
$2an + n(6n-1)d = 1000$
$n[2a + (6n-1)d] = 1000$

The sum of the first $6n$ terms is defined as $S_{6n} = \frac{6n}{2}[2a + (6n-1)d]$, which simplifies further to $3n[2a + (6n-1)d]$. By substituting the value of $1000$ into this derived expression, we find that the sum is $3 \times 1000$, resulting in 3000.

To find the minimum value of $f(x)=a^{a^{x}}+a^{1-a^{x}}$, we can utilize the arithmetic mean-geometric mean inequality, which states that the sum of two positive numbers is at least twice the square root of their product. When we treat the two terms $a^{a^x}$ and $a^{1-a^x}$ as our numbers, multiplying them together simplifies the expression to $a^{a^x} \cdot a^{1-a^x} = a^{a^x + 1 - a^x} = a$. Consequently, the minimum value of their sum is $2\sqrt{a}$.

The infinite series $\frac{3}{2} x^{2}+\frac{5}{3} x^{3}+\frac{7}{4} x^{4}+\dots$ consists of terms where the general coefficient $\frac{2n+1}{n+1}$ can be expressed as $2 - \frac{1}{n+1}$. By applying this decomposition, the series can be rewritten as $2(x^{2}+x^{3}+x^{4}+\dots) - (\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+\dots)$. The first grouping is a standard infinite geometric series starting with $x^2$ and a common ratio of $x$, which sums to $\frac{2x^2}{1-x}$.

Simultaneously, the second grouping corresponds to the Maclaurin expansion for $-\log_{e}(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$, meaning the specific portion $\frac{x^2}{2}+\frac{x^3}{3}+\dots$ is equal to $-\log_{e}(1-x) - x$. Combining these components results in the expression $\frac{2x^2}{1-x} + x + \log_{e}(1-x)$. Bringing the term $x$ into the algebraic fraction yields $\frac{2x^2 + x(1-x)}{1-x} + \log_{e}(1-x)$, which simplifies further to $\frac{x^2+x}{1-x} + \log_{e}(1-x)$, ultimately resulting in $x\left(\frac{1+x}{1-x}\right)+\log_{e}(1-x)$.

The first terms of both sequences can be found by relating consecutive terms. Given $c_2 = a_2 + b_2 = 12$ and $c_3 = a_3 + b_3 = 13$, we can express the terms in the second equation using the properties $a_3 = a_2 - 3$ and $b_3 = 2b_2$. Substituting these relationships into the expression for $c_3$ gives $(a_2 - 3) + 2b_2 = 13$, which simplifies to $a_2 + 2b_2 = 16$. Subtracting the equation $a_2 + b_2 = 12$ from this result yields $b_2 = 4$, which implies $a_2 = 8$. From these values, the first term of the arithmetic progression is $a_1 = a_2 - (-3) = 11$ and the first term of the geometric progression is $b_1 = b_2 / 2 = 2$.

With the first terms identified, the total sum $\sum_{k=1}^{10} c_k$ is calculated as the sum of the arithmetic and geometric series. The sum of the first 10 terms of the AP is $\frac{10}{2} [2(11) + 9(-3)] = 5(22 - 27) = -25$. The sum of the first 10 terms of the GP is $2 \left( \frac{2^{10} - 1}{2 - 1} \right) = 2(1023) = 2046$. Adding these results together gives $2046 - 25 = 2021$.

The ratio of the two arithmetic series simplifies significantly because the numerator is $2(1+2+3+\dots+y)$ and the denominator is $3(1+2+3+\dots+y)$. Since the sum $1+2+3+\dots+y$ appears in both, it cancels out, leaving the equation $\frac{2}{3} = \frac{4}{\log_{10} x}$. Solving for the logarithmic term gives $\log_{10} x = 6$, which implies $x = 10^{6}$.

To find $y$, recognize that the sum of the logarithms equals the logarithm of the product of the terms, which simplifies to $y = \log_{10} x^{1 + 1/3 + 1/9 + \dots}$. The exponent is an infinite geometric series with a sum of $\frac{1}{1 - 1/3} = \frac{3}{2}$. Using the established value of $\log_{10} x = 6$, the expression for $y$ becomes $y = \frac{3}{2} \times 6$, confirming that $y = 9$.

The Principle of Inclusion-Exclusion provides the most efficient approach for determining the count of numbers that are neither multiples of 3 nor 7 by calculating the total set and subtracting the unwanted values. We first identify the total set of 4-digit integers, which ranges from 1000 to 9999, resulting in 9000 distinct numbers. To find how many are either multiples of 3 or 7, we add the individual counts of multiples for each, then subtract the multiples of their least common multiple-which is 21-because those have been counted twice.

The count of 4-digit numbers divisible by 3 is calculated as $\frac{9999-1002}{3}+1 = 3000$, and for those divisible by 7, the count is $\frac{9996-1001}{7}+1 = 1286$. We must identify the overlap by counting numbers divisible by 21, found by $\frac{9996-1008}{21}+1 = 429$. Applying the formula $9000 - (3000 + 1286 - 429)$ yields the total count of numbers that are not multiples of 3 or 7, which equals 5143.

The mathematical constant $e$ is defined by the infinite series $\sum_{k=0}^{\infty } \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots$, where the summation starting from $k=1$ yields $e-1$. Calculating the sum $S_n = \sum_{r=1}^{n-1} r(n-r)$ results in the expression $\frac{n(n^2-1)}{6}$, or $\frac{n(n-1)(n+1)}{6}$, which is necessary to condense the terms within the overall series. Substituting this into the term $\frac{2 S_n}{n!}$ simplifies it to $\frac{n+1}{3(n-2)!}$, making the expression inside the summation $\frac{n+1}{3(n-2)!} - \frac{1}{(n-2)!}$, which further reduces to $\frac{n-2}{3(n-2)!} = \frac{1}{3(n-3)!}$. Summing this result from $n=4$ to infinity creates the series $\frac{1}{3} \sum_{n=4}^{\infty } \frac{1}{(n-3)!}$, which corresponds to $\frac{1}{3} \left( \frac{1}{1!} + \frac{1}{2!} + \dots \right)$ and evaluates to $\frac{e-1}{3}$.

This problem is solved using the properties of a telescoping series, where the internal structure allows terms to cancel out sequentially, leaving only the boundary values. By examining the relationship between the numerator and the squares in the denominator, each term can be expressed as a difference of reciprocals.

The numerator of each term represents the difference between the two consecutive squares found in its denominator. For example, the first term $\frac{3}{1^{2} \times 2^{2}}$ is equivalent to $\frac{2^{2} - 1^{2}}{1^{2} \times 2^{2}}$, which simplifies to $\frac{1}{1^{2}} - \frac{1}{2^{2}}$. Extending this pattern to all ten terms, the series expands into the following expression:

$\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) + \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) + \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) + \dots + \left( \frac{1}{10^{2}} - \frac{1}{11^{2}} \right)$

Since the negative part of every fraction cancels the positive part of the next one, the sum collapses entirely. This leaves only the first value and the very last subtracted term, resulting in $1 - \frac{1}{121}$, which calculates to $\frac{120}{121}$.

Assigning a variable $S$ to represent the infinite summation $\sum_{n=1}^{\infty } \frac{a_n}{8^n}$ allows the recurrence relation to be viewed as an algebraic equation. Starting with $a_{n+2} = 2a_{n+1} + a_n$, we divide the entire relation by $8^n$ and manipulate the indices to align the terms with the series $S$. Multiplying by 64 to adjust the denominators appropriately, the relation transforms into the following form:
$64 \frac{a_{n+2}}{8^{n+2}} = 16 \frac{a_{n+1}}{8^{n+1}} + \frac{a_n}{8^n}$
Summing this expression from $n=1$ to infinity effectively produces the sum $S$, though we must account for the missing initial terms due to the index shifts. This process yields the equation:
$64\left(S - \frac{a_1}{8} - \frac{a_2}{64}\right) = 16\left(S - \frac{a_1}{8}\right) + S$
Substituting the given initial values $a_1=1$ and $a_2=1$ leads to $64(S - \frac{1}{8} - \frac{1}{64}) = 16(S - \frac{1}{8}) + S$. Expanding these terms results in $64S - 8 - 1 = 16S - 2 + S$, which simplifies to $47S = 7$.

The series behaves as an Arithmetico-Geometric Progression from the second term onward, with numerators following an arithmetic pattern of $2, 7, 12, \dots$ and denominators following a geometric pattern of $3, 3^2, 3^3, \dots$. Isolating the leading term allows the expression to be written as $S = 1 + S_1$, where $S_1 = \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \dots$. Multiplying $S_1$ by the common ratio $\frac{1}{3}$ yields $\frac{1}{3}S_1 = \frac{2}{3^2} + \frac{7}{3^3} + \frac{12}{3^4} + \dots$. Subtracting this shifted series from the original $S_1$ results in $\frac{2}{3}S_1 = \frac{2}{3} + (\frac{5}{3^2} + \frac{5}{3^3} + \frac{5}{3^4} + \dots)$.

The terms inside the parentheses form an infinite geometric series with first term $a = \frac{5}{9}$ and common ratio $r = \frac{1}{3}$. Applying the summation formula $\frac{a}{1-r}$ produces $\frac{5/9}{1 - 1/3}$, which simplifies to $\frac{5/9}{2/3}$, or $\frac{5}{6}$. Substituting this value results in $\frac{2}{3}S_1 = \frac{2}{3} + \frac{5}{6}$, which equals $\frac{4+5}{6} = \frac{9}{6} = \frac{3}{2}$. Solving for $S_1$ provides $S_1 = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$. Adding the initial term $1$ gives a final result of $1 + \frac{9}{4} = \frac{13}{4}$.

The roots of the quadratic equation $4x^{2}-9x+5=0$ correspond to the arithmetic and geometric means of the sequence terms. Factoring this quadratic expression gives $(4x-5)(x-1)=0$, which identifies the two potential values as 1 and 5/4. Since the arithmetic mean is always greater than or equal to the geometric mean, we assign $\text{AM} = 5/4$ and $\text{GM} = 1$.

The given sequence is a geometric progression with the first term -16 and a common ratio of -1/2, defined by the general term formula $t_n = -16(-1/2)^{n-1}$. Applying the geometric mean condition $\sqrt{t_p t_q} = 1$ implies $t_p t_q = 1$. Substituting the formula for the $p$th and $q$th terms results in the equation $(-16 \cdot (-1/2)^{p-1})(-16 \cdot (-1/2)^{q-1}) = 1$, which simplifies to $256 \cdot (-1/2)^{p+q-2} = 1$. Rearranging this leads to the equality $(-2)^8 = (-2)^{p+q-2}$, and by equating the exponents, we find $p+q-2 = 8$, which means $p+q = 10$.

The expression for $S_n(x)$ relies on the logarithm property $\log_{a^{1/k}} x = k \log_a x$. By examining the sequence of denominators in the base, $2, 3, 6, 11, 18, 27, \dots$, we identify the $k$-th term as $2+(k-1)^2$. Summing these terms up to $n$ results in the general formula for $S_n(x)$:
$S_n(x) = \left( 2n + \frac{(n-1)n(2n-1)}{6} \right) \log_a x$
For $n=24$, the sum of the terms equals $4372$, leading to $4372 \log_a x = 1093$, which simplifies to $\log_a x = \frac{1}{4}$. For $n=12$, the sum of the terms is $24 + \frac{11 \cdot 12 \cdot 23}{6} = 530$, yielding $530 \log_a 2x = 265$, or $\log_a 2x = \frac{1}{2}$. Expanding $\log_a 2x$ as $\log_a 2 + \log_a x$ and substituting the previously determined value gives $\log_a 2 + \frac{1}{4} = \frac{1}{2}$. This implies $\log_a 2 = \frac{1}{4}$, which corresponds to $a^{1/4} = 2$, confirming that $a = 16$.

Determining the value of the given expression begins by calculating the sum of the arithmetico-geometric series in the base and the infinite geometric series in the exponent. Let the base be $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \dots$. By computing $S - \frac{S}{3}$, we obtain $1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{9} - \frac{2}{9}) + (\frac{10}{27} - \frac{6}{27}) + \dots$, which simplifies to $1 + \frac{1}{3} + \frac{4}{9} + \frac{4}{27} + \dots$. The terms following $1 + \frac{1}{3}$ form an infinite geometric progression with a first term of $\frac{4}{9}$ and a common ratio of $\frac{1}{3}$, which sums to $\frac{4/9}{1 - 1/3} = \frac{2}{3}$. Therefore, $\frac{2}{3} S = 1 + \frac{1}{3} + \frac{2}{3} = 2$, which confirms that the base of the expression is $S = 3$.

The exponent consists of the infinite geometric series $\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots$, which sums to $\frac{1/3}{1 - 1/3} = \frac{1}{2}$. Substituting these results into the original expression yields $l = 3^{\log_{0.25}(1/2)}$. Since $0.25$ is equivalent to $(1/2)^2$, the logarithmic term $\log_{0.25}(1/2)$ equals $\frac{1}{2}$ by applying the property $\log_{a^n}b = \frac{1}{n} \log_a b$. It follows that $l = 3^{1/2}$, and squaring both sides gives $l^2 = 3$.

Each of the variables is defined as the sum of an infinite geometric series with a common ratio less than $1$. Recalling that the sum of an infinite geometric series $a + ar + ar^2 + \dots$ equals $\frac{a}{1-r}$, we can represent $x$ and $y$ as $x = \frac{1}{1-\cos^2 \theta}$ and $y = \frac{1}{1-\sin^2 \phi}$. Rearranging these expressions to isolate the trigonometric terms gives $\cos^2 \theta = 1 - \frac{1}{x}$ and $\sin^2 \phi = 1 - \frac{1}{y}$.

Similarly, $z$ represents an infinite geometric series with the common ratio $\cos^2 \theta \sin^2 \phi$, which simplifies to $z = \frac{1}{1 - \cos^2 \theta \sin^2 \phi}$. By substituting the expressions for $\cos^2 \theta$ and $\sin^2 \phi$ into this formula, we obtain $z = \frac{1}{1 - (1 - \frac{1}{x})(1 - \frac{1}{y})}$. Expanding the product in the denominator yields $1 - (1 - \frac{1}{y} - \frac{1}{x} + \frac{1}{xy})$, which simplifies to $\frac{1}{x} + \frac{1}{y} - \frac{1}{xy}$. Consequently, $z = \frac{1}{\frac{x+y-1}{xy}}$, which further simplifies to $z = \frac{xy}{x+y-1}$. Rearranging this result by cross-multiplying leads directly to $z(x+y-1) = xy$, or $xy+z = z(x+y)$.

Finding the sum of terms in an arithmetic progression requires determining the first term $a$ and the common difference $d$ through the standard formula $S_n = \frac{n}{2}\{2a+(n-1)d\}$. Using the provided values, $S_{10}=530$ leads to the equation $2a+9d=106$, while $S_{5}=140$ simplifies to $2a+4d=56$. Subtracting the second equation from the first yields $5d=50$, identifying the common difference as $d=10$, which then determines the first term to be $a=8$. With these parameters established, $S_{20}-S_{6}$ is calculated by finding $S_{20} = \frac{20}{2}\{2(8)+19(10)\} = 2060$ and $S_{6} = \frac{6}{2}\{2(8)+5(10)\} = 198$, resulting in a final difference of 1862.

The sum of the first $n$ terms of an arithmetic progression is defined by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ represents the first term and $d$ is the common difference. Given the condition $S_{3n} = 3S_{2n}$, we equate the sum expressions to write $\frac{3n}{2}[2a+(3n-1)d] = 3 \cdot \frac{2n}{2}[2a+(2n-1)d]$. Canceling common terms on both sides leads to $2a+(3n-1)d = 2[2a+(2n-1)d]$, which simplifies to $2a+3nd-d = 4a+4nd-2d$ and ultimately reveals that $2a+(n-1)d = 0$.

To determine the value of the ratio $\frac{S_{4n}}{S_{2n}}$, we express the sums as $\frac{\frac{4n}{2}[2a+(4n-1)d]}{\frac{2n}{2}[2a+(2n-1)d]}$, which simplifies to $2 \cdot \frac{2a+(4n-1)d}{2a+(2n-1)d}$. By substituting the established identity $2a+(n-1)d = 0$ into the expression, we can rewrite the numerator as $2a+(n-1)d+3nd$ and the denominator as $2a+(n-1)d+nd$. Since the term $2a+(n-1)d$ evaluates to zero, the ratio reduces to $2 \cdot \frac{3nd}{nd}$, which results in 6.

Summing infinite series with factorial denominators often requires expressing the numerator in terms of the factors present in the denominator, allowing for cancellations that reveal simpler, standard series. By manipulating the quadratic term $n^2 + 6n + 10$ to align with $(2n+1)!$, we can decompose the series into a combination of known power series expansions for exponential functions.

We begin by scaling the numerator by 4, transforming $n^2 + 6n + 10$ into $4n^2 + 24n + 40$. This allows us to rewrite the numerator using the term $(2n+1)$ as $(2n+1)^2 + 10(2n+1) + 29$. Dividing this by $(2n+1)!$ gives us the summation $\frac{1}{4} \sum_{n=1}^{\infty } \left( \frac{(2n+1)^2}{(2n+1)!} + \frac{10(2n+1)}{(2n+1)!} + \frac{29}{(2n+1)!} \right)$, which simplifies to $\frac{1}{4} \sum_{n=1}^{\infty } \left( \frac{2n+1}{(2n)!} + \frac{10}{(2n)!} + \frac{29}{(2n+1)!} \right)$. By splitting $\frac{2n+1}{(2n)!}$ into $\frac{2n}{(2n)!} + \frac{1}{(2n)!}$, which equals $\frac{1}{(2n-1)!} + \frac{1}{(2n)!}$, the series becomes $\frac{1}{4} \sum_{n=1}^{\infty } \left( \frac{1}{(2n-1)!} + \frac{11}{(2n)!} + \frac{29}{(2n+1)!} \right)$.

We evaluate these individual series using the Taylor series expansions of $e^x$ and $e^{-x}$ evaluated at $x=1$. These standard sums are $\sum_{n=1}^{\infty } \frac{1}{(2n-1)!} = \frac{e - e^{-1}}{2}$, $\sum_{n=1}^{\infty } \frac{1}{(2n)!} = \frac{e + e^{-1} - 2}{2}$, and $\sum_{n=1}^{\infty } \frac{1}{(2n+1)!} = \frac{e - e^{-1} - 2}{2}$. Substituting these values back into our expression results in $\frac{1}{4} \left( \frac{e - e^{-1}}{2} + 11 \left( \frac{e + e^{-1} - 2}{2} \right) + 29 \left( \frac{e - e^{-1} - 2}{2} \right) \right)$. Aggregating the coefficients for $e$ and $e^{-1}$ along with the constants, we obtain $\frac{1}{8} \left( (1 + 11 + 29)e + (-1 + 11 - 29)e^{-1} - (22 + 58) \right)$, which yields $\frac{41e - 19e^{-1} - 80}{8}$, or $\frac{41}{8}e - \frac{19}{8}e^{-1} - 10$.

The intersection of an arithmetic and a geometric sequence is found by equating their respective general terms to identify values present in both sets. From the provided elements $\{11, 8, 21, 16, 26, 32, 4\}$, the arithmetic sequence follows the formula $a_m = 5m + 6$, while the geometric sequence follows the formula $g_n = 2^{n+1}$. Equating these expressions gives $5m + 6 = 2^{n+1}$, which necessitates that the value $2^{n+1} - 6$ be a multiple of $5$.

This divisibility requirement dictates that the unit digit of $2^{n+1}$ must be $6$, as only values ending in $6$ produce a result divisible by $5$ when $6$ is subtracted. Since the unit digit of powers of $2$ follows the repeating pattern $\{2, 4, 8, 6\}$, this condition is met only when the exponent $n+1$ is a multiple of $4$. Calculating the powers $2^4$, $2^8$, and $2^{12}$ yields the terms $16$, $256$, and $4096$, while the subsequent power, $2^{16}$, results in a number that exceeds the four-digit limit. Thus, there are $3$ common terms shared between these series.