Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The sum of coefficients in a polynomial expression is simply its value at $x=1$. Substituting this value into $(1-2x+2x^2)^{2023}(3-4x^2+2x^3)^{2024}$ yields $(1-2+2)^{2023}(3-4+2)^{2024}$, which simplifies to $(1)^{2023}(1)^{2024}$, resulting in $a=1$.

Evaluating the limit for $b$ presents an indeterminate form of $0/0$ as $x$ approaches $0$. Applying L'Hôpital's rule, the derivative of the numerator using the fundamental theorem of calculus is $\frac{\log(1+x)}{x^{2024}+1}$, and the derivative of the denominator is $2x$. The limit simplifies to $\lim_{x\to 0} \frac{\log(1+x)}{2x(x^{2024}+1)}$, which can be computed as $\frac{1}{2} \cdot \lim_{x\to 0} \frac{\log(1+x)}{x} \cdot \lim_{x\to 0} \frac{1}{x^{2024}+1}$. Given that the standard limit $\lim_{x\to 0} \frac{\log(1+x)}{x}$ equals $1$, the value of $b$ is $\frac{1}{2}$.

Substituting $a=1$ and $b=1/2$ into the quadratic equation $2bx^2+ax+4=0$ results in $x^2+x+4=0$. Since this quadratic has a negative discriminant, its roots are complex, and any equation $cx^2+dx+e=0$ sharing a root with it must have identical roots, implying the coefficients are proportional. The ratio of the coefficients satisfies $\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$, meaning $c=d$ and $e=4d$. Consequently, the ratio $d:c:e$ is $1:1:4$.

Exponents in the expression $x^{pq^2}=y^{qr}=z^{p^2 r}$ imply that the ratios of their logarithms define the progression terms. By setting the exponential expression equal to a common variable, we determine that $\log_y x = \frac{r}{pq}$, $\log_z y = \frac{p^2}{q}$, and $\log_x z = \frac{q^2}{pr}$. Applying these values to the sequence $3, 3 \log_y x, 3 \log_z y, 7 \log_x z$ creates an arithmetic progression where the common difference is $\frac{1}{2}$.

The difference between the first two terms dictates that $3 \frac{r}{pq} - 3 = \frac{1}{2}$, which simplifies to $\frac{3r}{pq} = \frac{7}{2}$, or $r = \frac{7}{6} pq$. Using the given identity $r = pq+1$, we substitute to obtain $pq+1 = \frac{7}{6} pq$, which confirms that $pq = 6$ and $r = 7$. Substituting these into the third term of the arithmetic progression, the equation $\frac{3p^2}{q} = 4$ combined with $pq=6$ allows us to isolate $p$ and $q$. Replacing $q$ with $\frac{6}{p}$ in the equation $\frac{3p^2}{q} = 4$ yields $3p^2 = 4 (\frac{6}{p})$, simplifying to $p^3 = 8$, which means $p = 2$ and $q = 3$. The final value is calculated by the equation $r-p-q = 7-2-3 = 2$.

The binomial expansion of $(2+x)^9$ follows the general form $\sum_{k=0}^9 \binom{9}{k} 2^{9-k} x^k$. By setting $x=1$, the expression becomes $(2+1)^9 = 3^9 = 19683$, which represents the sum of all coefficients in the expansion. Because the question specifically requires the mean of the coefficients for $x^1$ through $x^7$, the coefficients for $x^0$, $x^8$, and $x^9$ must be removed from this total to isolate the necessary sum.

The constant term, where $k=0$, is $\binom{9}{0} 2^9 = 512$, the coefficient of $x^8$, where $k=8$, is $\binom{9}{8} 2^1 = 18$, and the coefficient of $x^9$, where $k=9$, is $\binom{9}{9} 2^0 = 1$. Subtracting these values from the total sum results in $19683 - 512 - 18 - 1 = 19152$. Dividing this sum by the seven terms requested gives a mean of $19152 / 7 = 2736$.

This summation represents an arithmetico-geometric series where each term consists of an arithmetic factor increasing by one and a geometric factor increasing by a common ratio of $21/20$. Multiplying the entire series $S$ by $21/20$ and subtracting this new expression from the original $S$ causes the overlapping terms to collapse, reducing the equation to a standard geometric progression minus the final term. The subtraction simplifies to $-\frac{1}{20} S = (21^{20} - 20^{20}) - 21^{20}$, which further reduces to $-\frac{1}{20} S = -20^{20}$. Solving for $S$ yields $(20)^{21}$, and by comparing this result to the given form $k(20)^{19}$, the value of $k$ must be $20^2$, which equals $400$.

The numerator is a standard summation of cubes, which follows the formula $\left( \frac{n(n+1)}{2} \right)^2$. The denominator is the sum of terms where the $r$-th entry is $r(2r+1)$, which simplifies to $2r^2+r$. Summing this from 1 to $n$ involves the standard formulas for the sum of squares and the sum of integers, resulting in the expression $\frac{n(n+1)(4n+5)}{6}$.

Dividing the numerator by the denominator simplifies the expression to $\frac{3n(n+1)}{2(4n+5)}$. Setting this ratio equal to $\frac{9}{5}$ provides the relationship:

$\frac{3n(n+1)}{2(4n+5)} = \frac{9}{5}$

Simplifying this leads to $15n(n+1) = 18(4n+5)$, which reduces to the quadratic $5n^2 - 19n - 30 = 0$. Factoring this equation yields $n=5$ or $n=-\frac{6}{5}$; since the number of terms must be a positive integer, the value of $n$ is 5.

The alternating sum of the coefficients in the binomial expansion of $(1-x)^{100}$ equals zero, which is confirmed by substituting $x=1$ into the expression $(1-x)^{100} = 0$. This expansion contains 101 terms, and utilizing the symmetry property ${}^{100} C_{k} = {}^{100} C_{100-k}$, the sum of the terms from the first 50 indices is equal to the sum of the terms from the last 50 indices. Letting $S$ represent the sum of the first 50 coefficients, the total expansion can be expressed as $2S + {}^{100} C_{50} = 0$. Solving for $S$ provides the expression $- \frac{1}{2} {}^{100} C_{50}$. Expanding this into factorials as $-\frac{1}{2} \times \frac{100!}{50! 50!}$ allows the term to be simplified by rewriting $100!$ as $100 \times 99!$ and $50!$ as $50 \times 49!$, which reduces the expression to $- {}^{99} C_{49}$.

The evaluation of the series $\sum_{n=1}^{\infty } \frac{2 n^2+3 n+4}{(2n)!}$ relies on manipulating the numerator to create terms that allow the summation to be computed using the infinite series expansions of $e$ and $1/e$. By rewriting the numerator $2n^2+3n+4$ as $\frac{1}{2}(2n(2n-1) + 8n + 8)$, the original series is partitioned into three distinct parts: $\frac{1}{2} \sum_{n=1}^{\infty } \frac{1}{(2n-2)!} + 2 \sum_{n=1}^{\infty } \frac{1}{(2n-1)!} + 4 \sum_{n=1}^{\infty } \frac{1}{(2n)!}$.

Substituting the standard series values $\sum_{n=1}^{\infty } \frac{1}{(2n-2)!} = \frac{e+1/e}{2}$, $\sum_{n=1}^{\infty } \frac{1}{(2n-1)!} = \frac{e-1/e}{2}$, and $\sum_{n=1}^{\infty } \frac{1}{(2n)!} = \frac{e+1/e-2}{2}$, the expression becomes
$\frac{1}{2} \left(\frac{e+1/e}{2}\right) + 2 \left(\frac{e-1/e}{2}\right) + 4 \left(\frac{e+1/e-2}{2}\right)$
Expanding this results in $\frac{e+1/e}{4} + e - 1/e + 2e + 2/e - 4$. Combining the coefficients for $e$ and $1/e$ yields the final result of $\frac{13e}{4} + \frac{5}{4e} - 4$.

In a geometric progression with terms $a, b, \frac{1}{18}$, the square of the middle term equals the product of the extremes, giving $b^2 = \frac{a}{18}$, or $a = 18b^2$. For the arithmetic progression with terms $\frac{1}{a}, 10, \frac{1}{b}$, the middle term is the arithmetic mean, which means $10 = \frac{1}{2}(\frac{1}{a} + \frac{1}{b})$, simplifying to $\frac{1}{a} + \frac{1}{b} = 20$.

Substituting $a = 18b^2$ into the expression $\frac{a+b}{ab} = 20$ leads to $a+b = 20ab$, and further replacement gives $18b^2 + b = 20(18b^2)b$. This simplifies to the quadratic equation $360b^2 - 18b - 1 = 0$. Solving for the positive value of $b$ using the quadratic formula yields $b = \frac{18 + \sqrt{324 + 1440}}{720} = \frac{60}{720} = \frac{1}{12}$. Consequently, calculating for $a$ gives $a = 18(\frac{1}{12})^2 = \frac{1}{8}$.

The value of the expression $16a + 12b$ is determined as $16(\frac{1}{8}) + 12(\frac{1}{12}) = 2 + 1 = 3$.

The core of this problem relies on the fundamental properties of sequences, where an arithmetic progression implies that the middle term is the arithmetic mean of its neighbors, and a geometric progression implies that the square of the middle term equals the product of the flanking terms. For the arithmetic sequence $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$, this condition provides the equation $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$, which simplifies to $\frac{2}{y} = \frac{x+z}{xz}$. For the geometric sequence $x, \sqrt{2} y, z$, the relationship is $(\sqrt{2} y)^2 = x z$, or simply $2 y^2 = x z$.

Substituting the geometric product into the arithmetic relationship yields $\frac{2}{y} = \frac{x+z}{2 y^2}$, which simplifies to $x+z = 4 y$. To solve for the variables, substitute these findings into the equation $x y + y z + z x = \frac{3}{\sqrt{2}} x y z$, rewriting the left side as $y(x+z) + x z$. Replacing $x+z$ with $4 y$ and $x z$ with $2 y^2$ leads to the equation
$y(4 y) + 2 y^2 = \frac{3}{\sqrt{2}} y (2 y^2)$
which simplifies to $6 y^2 = 3 \sqrt{2} y^3$. Solving for $y$ results in $y = \sqrt{2}$. The sum $x+y+z$ therefore equals $4 y + y = 5 y = 5 \sqrt{2}$. Calculating the final value $3(x+y+z)^2$ yields $3(5 \sqrt{2})^2 = 3(50) = 150$.

In any geometric progression, the product of two terms equidistant from a middle term is equal to the square of that middle term. Since the fourth and sixth terms are symmetrically positioned around the fifth term, their product $a_4 \cdot a_6$ is equivalent to $a_5^2$, which yields $a_5^2 = 9$. Given that the progression consists of increasing positive numbers, we identify $a_5 = 3$. Substituting this value into the sum $a_5 + a_7 = 24$ allows us to express the seventh term as $a_5 r^2$, leading to $3 + 3r^2 = 24$, which simplifies to $r^2 = 7$.

To evaluate the expression $a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7$, we determine each component using the properties of the progression. The product $a_1 a_9$ corresponds to $(ar^0)(ar^8) = a^2 r^8 = (ar^4)^2 = a_5^2$, which equals $3^2 = 9$. The product $a_2 a_4 a_9$ can be written as $(ar)(ar^3)(ar^8) = a^3 r^{12} = (ar^4)^3$, representing the cube of the fifth term, or $3^3 = 27$. Finally, calculating $a_5 + a_7 = 3 + (3 \times 7) = 3 + 21 = 24$ provides the remaining terms, resulting in an overall sum of $9 + 27 + 3 + 21 = 60$.

Decomposing the summand $\frac{n^3(2n)! + (2n-1)n!}{n!(2n)!}$ into $\frac{n^3}{n!} + \frac{2n-1}{(2n)!}$ reveals two independent series that correspond to known exponential expansions. The numerator $n^3$ can be expressed as $n(n-1)(n-2) + 3n(n-1) + n$, which transforms the sum $\sum_{n=0}^{\infty } \frac{n^3}{n!}$ into $\sum \frac{1}{(n-3)!} + \sum \frac{3}{(n-2)!} + \sum \frac{1}{(n-1)!}$. Since each of these terms effectively sums to $e$, this part of the series yields $5e$.

The second term $\sum_{n=0}^{\infty } \frac{2n-1}{(2n)!}$ splits into $\sum \frac{1}{(2n-1)!} - \sum \frac{1}{(2n)!}$. Using the identities $\sum_{n=0}^{\infty } \frac{1}{(2n-1)!} = \frac{e - e^{-1}}{2}$ and $\sum_{n=0}^{\infty } \frac{1}{(2n)!} = \frac{e + e^{-1}}{2}$, the difference becomes $\frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2}$, which simplifies to $-\frac{1}{e}$. Combining the two results gives $5e - \frac{1}{e}$, identifying the integer coefficients as $a=5$, $b=-1$, and $c=0$. Evaluating $a^2 - b + c$ yields $5^2 - (-1) + 0$, which equals 26.

An arithmetic progression is defined by its first term $a$ and a common difference $d$, with the sum of the first $n$ terms given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$. By substituting the expressions for the sum of the first five and first nine terms into the given ratio $\frac{S_5}{S_9} = \frac{5}{17}$, the relationship between the variables can be determined. The expression $\frac{\frac{5}{2}(2a + 4d)}{\frac{9}{2}(2a + 8d)} = \frac{5}{17}$ simplifies to $\frac{2a + 4d}{9(2a + 8d)} = \frac{1}{17}$, which rearranges to $17(2a + 4d) = 18a + 72d$. Expanding this leads to $34a + 68d = 18a + 72d$, confirming the relationship $16a = 4d$, or more simply, $d = 4a$.

Considering the condition $110 < a_{15} < 120$ where the fifteenth term is defined as $a_{15} = a + 14d$, we substitute $d = 4a$ to get $110 < a + 14(4a) < 120$. This simplifies to the inequality $110 < 57a < 120$. Since the sequence consists of natural numbers, $a$ must be an integer, and the only integer value satisfying this range is $a = 2$. With $a$ determined, the common difference is calculated as $d = 4(2) = 8$.

To find the sum of the first ten terms, we use the formula $S_{10} = \frac{10}{2}[2a + 9d]$. Substituting the known values of $a = 2$ and $d = 8$ results in $S_{10} = 5[2(2) + 9(8)]$, which simplifies to $5[4 + 72]$. This yields a total of 380.

The key to evaluating complex algebraic expressions involving terms of a recurrence relation is to factor the expression into components that match the structure of the recurrence itself. By rearranging the given relation $a_{n+2} = 3 a_{n+1} - 2 a_n + 1$ into the form $a_{n+2} - 2 a_{n+1} = a_{n+1} - 2 a_n + 1$, we can define a new sequence $b_n = a_{n+1} - 2 a_n$. This transformation simplifies the recurrence into $b_{n+1} = b_n + 1$, identifying it as a simple arithmetic progression.

Given $a_0 = 0$ and $a_1 = 0$, we find the starting value of this sequence as $b_0 = a_1 - 2 a_0 = 0$. Since $b_{n+1} = b_n + 1$, the general term is simply $b_n = n$. The target expression $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ can be factored by grouping terms, yielding $(a_{25} - 2 a_{24})(a_{23} - 2 a_{22})$. Substituting our formula for $b_n$, this expression is equivalent to $b_{24} \cdot b_{22}$, which is $24 \times 22$, resulting in 528.

The geometric mean of a collection of numbers is calculated as the $N$-th root of their product, where $N$ is the total count of terms. For the two geometric progressions provided, the first sequence is $2^1, 2^2, \dots, 2^{60}$ and the second sequence, consisting of $n$ terms with a base of 4, is $4^1, 4^2, \dots, 4^n$. We can express the second sequence in base 2 as $2^2, 2^4, \dots, 2^{2n}$. The product of the first sequence is $2^{1+2+\dots+60} = 2^{\frac{60 \times 61}{2}} = 2^{1830}$, and the product of the second sequence is $2^{2+4+\dots+2n} = 2^{2(1+2+\dots+n)} = 2^{n(n+1)}$. Multiplying these gives a total product of $2^{n^2+n+1830}$ for all $60+n$ terms, so the geometric mean is $2^{\frac{n^2+n+1830}{60+n}}$.

Equating this expression to the given value of $2^{\frac{225}{8}}$ results in the equality:
$\frac{n^2+n+1830}{60+n} = \frac{225}{8}$
Cross-multiplying leads to $8n^2 + 8n + 14640 = 13500 + 225n$, which simplifies to the quadratic equation $8n^2 - 217n + 1140 = 0$. Solving this for $n$ yields the roots $n=20$ and $n=\frac{57}{8}$. Since $n$ must be an integer, we have $n=20$. We then calculate the sum $\sum_{k=1}^{20} k(20-k)$, which is $20\sum_{k=1}^{20} k - \sum_{k=1}^{20} k^2$. Substituting the standard summation formulas, this is $20 \times \frac{20 \times 21}{2} - \frac{20 \times 21 \times 41}{6}$, which simplifies to $4200 - 2870$, giving a final value of 1330.

Recognizing the pattern in a complex series often requires expressing the $n$-th term, $T_n$, in a simplified form before attempting a summation. In this sequence, the numerator of the $n$-th term consists of the sum $\sum_{k=1}^n [(2k)^3 - (2k-1)^3]$, while the denominator is $n(4n+3)$. By reducing this general term, the complex structure collapses into a simple linear expression, making the total sum straightforward to evaluate across the 15 terms provided.

The numerator relies on the algebraic expansion $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, where $a=2k$ and $b=2k-1$. Because $a-b=1$, the numerator simplifies to $\sum_{k=1}^n [(2k)^2 + 2k(2k-1) + (2k-1)^2]$, which equates to $\sum_{k=1}^n (12k^2 - 6k + 1)$. Utilizing standard power sum formulas, this expression evaluates to $2n(2n^2+3n+1) - 3n(n+1) + n$, which reduces to $n^2(4n+3)$. Dividing this result by the denominator $n(4n+3)$ confirms that the $n$-th term is simply $T_n = n$.

Computing the total sum requires adding these terms from $n=1$ to $15$. Using the standard formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$, for $n=15$ yields $\frac{15 \times 16}{2}$, resulting in a final value of 120.

This infinite series can be evaluated using the method of subtraction for arithmetico-geometric series. Multiplying the entire series $S$ by $\frac{1}{6}$ and subtracting it from the original series yields $\frac{5}{6}S = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \dots$. Defining $T = \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \dots$, the expression becomes $\frac{5}{6}S = 1 + T$.

Multiplying $T$ by $\frac{1}{6}$ results in $\frac{1}{6}T = \frac{4}{6^2} + \frac{7}{6^3} + \frac{10}{6^4} + \dots$, and subtracting this from $T$ produces $\frac{5}{6}T = \frac{4}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \frac{3}{6^4} + \dots$. The latter part is an infinite geometric series with first term $\frac{3}{36}$ and common ratio $\frac{1}{6}$, which sums to $\frac{3/36}{1-1/6} = \frac{1}{10}$.

Consequently, $\frac{5}{6}T = \frac{2}{3} + \frac{1}{10} = \frac{23}{30}$, which simplifies to $T = \frac{23}{30} \times \frac{6}{5} = \frac{23}{25}$. Substituting this value into the equation $\frac{5}{6}S = 1 + T$ gives $\frac{5}{6}S = 1 + \frac{23}{25} = \frac{48}{25}$. Solving for $S$ results in $S = \frac{48}{25} \times \frac{6}{5} = \frac{288}{125}$.

The polynomial $f(x)=a_0 x^2+a_1 x+a_2$ has a derivative given by $f'(x)=2 a_0 x+a_1$. By applying the given conditions $f'(0)=1$ and $f'(1)=0$, we identify that $a_1=1$ and $2 a_0+1=0$, which yields the coefficient $a_0=-\frac{1}{2}$.

The coefficients $a_0, a_1, a_2$ form an arithmetico-geometric progression, which combines an arithmetic sequence with a common difference of $1$ and a geometric sequence with a common ratio of $2$. Letting the starting terms of these sequences be $a$ and $y$ respectively, the coefficients can be expressed as $a_0=ay$, $a_1=(a+1)(2y)$, and $a_2=(a+2)(4y)$. Substituting $a_0=-\frac{1}{2}$ and $a_1=1$ into these expressions, we solve for $a$ and $y$ by noting that $ay=-\frac{1}{2}$ and $2y(a+1)=1$. Expanding the second equation gives $2ay+2y=1$, and substituting $ay=-\frac{1}{2}$ results in $-1+2y=1$, which simplifies to $y=1$ and consequently $a=-\frac{1}{2}$.

With these variables determined, the third coefficient is calculated as $a_2=(-\frac{1}{2}+2)(4)(1)$, which simplifies to $1.5 \times 4$, resulting in $6$. Substituting the values $a_0=-\frac{1}{2}$, $a_1=1$, and $a_2=6$ back into the original quadratic expression yields $f(x)=-\frac{1}{2} x^2+x+6$. Evaluating this function at $x=4$ gives $f(4)=-\frac{1}{2}(16)+4+6$, leading to $-8+10$, which results in $2$.

The sum of any arithmetic progression with $n$ terms is given by the formula $\frac{n}{2}(a_1 + a_n)$. Since the total sum of the entire sequence is $192$, we can write $\frac{n}{2}(a_1 + a_n) = 192$, which simplifies to $a_1 + a_n = \frac{384}{n}$.

We are also given the sum of the even-indexed terms, which are $a_2, a_4, \dots, a_n$. This sub-series contains exactly $\frac{n}{2}$ terms and acts as its own arithmetic progression where the first term is $a_2$ and the last term is $a_n$. The sum of this series is calculated as $\frac{n/2}{2}(a_2 + a_n) = 120$. Because the common difference of the original progression is $1$, we can substitute $a_2 = a_1 + 1$ into this expression, which leads to $\frac{n}{4}(a_1 + 1 + a_n) = 120$, or $a_1 + a_n + 1 = \frac{480}{n}$.

Subtracting the first relation, $a_1 + a_n = \frac{384}{n}$, from the second, $a_1 + a_n + 1 = \frac{480}{n}$, isolates the variable $n$. This arithmetic operation results in $1 = \frac{480}{n} - \frac{384}{n}$, which simplifies to $1 = \frac{96}{n}$, establishing that $n = 96$.

The sum of an infinite geometric series with a common ratio $r$, where $|r|<1$, is given by the formula $\frac{1}{1-r}$. Applying this to the given variables, we have $x = \frac{1}{1-a}$, $y = \frac{1}{1-b}$, and $z = \frac{1}{1-c}$. Since $a$, $b$, and $c$ form an arithmetic progression, the transformed sequence $1-a$, $1-b$, and $1-c$ also maintains the properties of an arithmetic progression.

By the fundamental definition of a harmonic progression, if a set of numbers forms an arithmetic progression, their reciprocals form a harmonic progression. Consequently, the terms $x$, $y$, and $z$ form a harmonic progression because their constituent parts-the denominators $1-a$, $1-b$, and $1-c$-are in an arithmetic progression. Because a sequence is harmonic if and only if the reciprocals of its terms form an arithmetic progression, the values $\frac{1}{x}$, $\frac{1}{y}$, and $\frac{1}{z}$ must form an arithmetic progression.