Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The series can be analyzed by identifying its general term $u_n$, which corresponds to the expansion of a logarithmic power series. By substituting $t = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} = 1 - \sqrt{\frac{2}{3}}$, the series $S$ simplifies to $1 + \sum_{n=1}^\infty \frac{t^n}{n(n+1)}$. Using partial fractions, this expression can be rewritten as $1 + \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)t^n$, which splits into two separate summations.

The term $\sum_{n=1}^\infty \frac{t^n}{n}$ equates to $-\ln(1-t)$, while the term $\sum_{n=1}^\infty \frac{t^n}{n+1}$ can be expressed as $\frac{1}{t} \sum_{n=1}^\infty \frac{t^{n+1}}{n+1}$, which simplifies to $\frac{-\ln(1-t)-t}{t}$. Combining these into the original series yields $S = 2 + \left(\frac{1-t}{t}\right)\ln(1-t)$. Given $t = 1 - \sqrt{\frac{2}{3}}$, we find that $1-t = \sqrt{\frac{2}{3}}$ and the coefficient $\frac{1-t}{t}$ reduces to $\sqrt{6}+2$.

Substituting these values leads to $S = 2 + \left(\frac{\sqrt{6}+2}{2}\right)\ln\left(\frac{2}{3}\right)$, which simplifies to $2 + \left(\sqrt{\frac{3}{2}}+1\right)\ln\left(\frac{2}{3}\right)$. Comparing the resulting values of $a$ and $b$ to the required form identifies $a=2$ and $b=3$. The final calculation for the expression $11a+18b$ gives $11(2)+18(3)$, which is 76.

Telescoping series allow us to simplify these complex summations by causing intermediate terms to cancel out. For $m$, rationalizing the denominator of each term $\frac{1}{\sqrt{k}+\sqrt{k+1}}$ results in $\sqrt{k+1}-\sqrt{k}$, so the entire sum becomes $(\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + \cdot s + (\sqrt{100}-\sqrt{99})$, which simplifies to $\sqrt{100} - 1 = 9$. Similarly, the sum $n$ can be evaluated by decomposing each fraction $\frac{1}{k(k+1)}$ into $\frac{1}{k} - \frac{1}{k+1}$, resulting in the sequence $(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \cdot s + (\frac{1}{99} - \frac{1}{100}) = 1 - \frac{1}{100} = \frac{99}{100}$. Substituting these results into the linear equation $11x - 100y = 0$ yields $11(9) - 100(\frac{99}{100}) = 99 - 99 = 0$, demonstrating that the point $(m, n)$ lies on this line.

The arithmetic progression 3, $a, b, c$ can be described using a common difference $d$, where $a = 3+d$, $b = 3+2d$, and $c = 3+3d$. Substituting these expressions into the second sequence 3, $a-1$, $b+1$, $c+9$ transforms it into 3, $2+d$, $4+2d$, and $12+3d$, which must form a geometric progression. For the first three terms to satisfy the geometric property, the square of the middle term must equal the product of the flanking terms, leading to the relation $(2+d)^2 = 3(4+2d)$.

Solving the quadratic equation $d^2 + 4d + 4 = 12 + 6d$ results in $d^2 - 2d - 8 = 0$, which factors into $(d-4)(d+2) = 0$. Evaluating the case where $d=4$, the terms $a, b, c$ correspond to 7, 11, and 15, which creates the valid geometric progression 3, 6, 12, 24. Finding the arithmetic mean of $a, b,$ and $c$ gives $\frac{7+11+15}{3}$, resulting in 11.

The $n^{\text{th}}$ term of a geometric sequence is given by $ar^{n-1}$, where $a$ represents the first term and $r$ the common ratio. In the first progression, the third term $ar^2$ is equal to $b$, implying $r^2 = \frac{b}{a}$, so the eleventh term is $ar^{10} = a(r^2)^5 = a\left(\frac{b}{a}\right)^5$. For the second progression, the fifth term $aR^4$ is also $b$, so $R = \left(\frac{b}{a}\right)^{\frac{1}{4}}$, making the $p^{\text{th}}$ term $aR^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$. Equating these two values, $a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$, leads to the exponent equation $5 = \frac{p-1}{4}$, which solves to $p = 21$.

The fundamental property of a triangle is the triangle inequality, which dictates that the sum of the lengths of any two sides must be strictly greater than the third side. Representing the sides of the geometric progression as $a$, $ar$, and $ar^2$ for some positive $a$, the condition $a + ar > ar^2$ leads to the quadratic inequality $r^2 - r - 1 < 0$. Since we are given that $r > 1$, this constraint restricts the possible values of $r$ to the interval $(1, \frac{1+\sqrt{5}}{2})$.

Given that $\sqrt{5}$ is approximately $2.236$, the value $\frac{1+\sqrt{5}}{2}$ is approximately $1.618$, meaning $r$ lies strictly between $1$ and $1.618$. Within this range, the greatest integer $[r]$ is clearly $1$. Similarly, because $r$ falls between $1$ and $1.618$, $-r$ must reside between $-1.618$ and $-1$, making the greatest integer $[-r]$ equal to $-2$. Substituting these specific integer values into the expression $3[r] + [-r]$ yields $3(1) + (-2)$, which simplifies to $1$.

The expression $S(x) = (1+x)+2(1+x)^2+3(1+x)^3+\ldots+60(1+x)^{60}$ is an arithmetico-geometric series, where each term consists of an arithmetic coefficient multiplied by a geometric power. Multiplying $S(x)$ by the common ratio $(1+x)$ and subtracting this from the original series creates a geometric progression, which significantly simplifies the calculation.

Subtracting $(1+x)S(x)$ from $S(x)$ results in the equation $-xS(x) = \sum_{k=1}^{60} (1+x)^k - 60(1+x)^{61}$. Using the standard formula for the sum of a geometric progression, $\sum_{k=1}^{n} r^k = \frac{r(r^n - 1)}{r-1}$, this expression becomes:
$-x S(x) = \frac{(1+x)((1+x)^{60} - 1)}{x} - 60(1+x)^{61}$
Substituting $x=60$ into this derived equation allows for the determination of $60^2 S(60)$. Through algebraic simplification, this yields $3600(61)^{61} - (61^{61} - 61)$, which is equivalent to $3599(61)^{61} + 61$. Comparing this result to the required form $a(b)^b + b$ identifies $a=3599$ and $b=61$, ultimately providing the value $(a+b) = 3660$.

The identity $\frac{{}^{n}C_{r}}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}$ allows us to transform the series $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10}$ by rewriting each individual term as $\frac{{}^{12}C_{r+1}}{12}$. Factoring out the constant $\frac{1}{12}$, the expression becomes $\frac{1}{12} \sum_{r=1}^{9} {}^{12}C_{r+1}$, which corresponds to the sum of binomial coefficients ${}^{12}C_2 + {}^{12}C_3 + \dots + {}^{12}C_{10}$.

The total sum of binomial coefficients for $n=12$ is $2^{12}$, which equals 4096. Because the series only includes terms from $k=2$ to $k=10$, we must subtract the terms for $k=0, 1, 11,$ and 12 from this total. The values for these excluded terms are ${}^{12}C_0 = 1$, ${}^{12}C_1 = 12$, ${}^{12}C_{11} = 12$, and ${}^{12}C_{12} = 1$, which sum to 26. Subtracting this from the total yields $2^{12} - 26 = 4096 - 26 = 4070$. Multiplying this value by the factor $\frac{1}{12}$ results in $\frac{4070}{12}$, which simplifies to $\frac{2035}{6}$. Given that 2035 and 6 are coprime, we have $n=2035$ and $m=6$, making the sum $n+m = 2041$.

The arrangement follows a pattern where the $n$-th row starts after all numbers in the preceding $n-1$ rows have been filled, with each row $k$ containing exactly $k$ integers. The total count of integers in the first $n-1$ rows is given by the sum $\frac{(n-1)n}{2}$, making the first number of the $n$-th row $T_n = 1 + \frac{n(n-1)}{2}$. Calculating for the specific row values, the $103^{\text{rd}}$ row starts at $T_{103} = 1 + \frac{103 \times 102}{2} = 5254$, whereas the $104^{\text{th}}$ row starts at $T_{104} = 1 + \frac{104 \times 103}{2} = 5357$. As 5310 falls between the starting values of the $103^{\text{rd}}$ and $104^{\text{th}}$ rows, it is located within the $103^{\text{rd}}$ row.

The identity $\frac{{}^nC_k}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$ allows us to transform the summations by converting them into standard forms related to Vandermonde's Identity. For the expression $\alpha$, rewriting the terms as $\frac{1}{n+1} {}^{n+1}C_{k+1} \cdot {}^nC_{n-k}$ leads to the simplified result:
$\alpha = \frac{1}{n+1} {}^{2n+1}C_{n+1}$
Similarly, substituting $\frac{{}^nC_{k+1}}{k+2}$ with $\frac{{}^{n+1}C_{k+2}}{n+1}$ in the expression for $\beta$ transforms the sum into:
$\beta = \frac{1}{n+1} {}^{2n+1}C_{n+2}$
Calculating the ratio $\frac{\beta}{\alpha}$ involves evaluating the quotient $\frac{{}^{2n+1}C_{n+2}}{{}^{2n+1}C_{n+1}}$, which simplifies to $\frac{n}{n+2}$. Setting this equal to the condition derived from $5\alpha = 6\beta$ produces:
$\frac{n}{n+2} = \frac{5}{6}$
Cross-multiplying results in $6n = 5n + 10$, which identifies $n=10$.

The alternating sum $A_k = \sum_{i=1}^k (a_{2i-1}^2 - a_{2i}^2)$ can be simplified by applying the difference of squares identity, $x^2 - y^2 = (x-y)(x+y)$. For an arithmetic progression with common difference $d$, each pair $(a_{2i-1}^2 - a_{2i}^2)$ factors into $(a_{2i-1} - a_{2i})(a_{2i-1} + a_{2i})$, which equals $(-d)(2a + (4i-3)d)$. Summing these $k$ terms leads to the consolidated expression $A_k = -kd(2a + (2k-1)d)$.

By setting this formula equal to the provided values, we form the equations $153 = 3d(2a + 5d)$ and $435 = 5d(2a + 9d)$, which simplify to $51 = d(2a + 5d)$ and $87 = d(2a + 9d)$. Subtracting the first resulting equation from the second eliminates the $2ad$ term, yielding $36 = d(9d - 5d)$, or $4d^2 = 36$. Given that the terms are positive, we have $d=3$. Substituting $d=3$ into $51 = d(2a + 5d)$ gives $17 = 2a + 15$, which confirms $a=1$. These values satisfy the additional condition $a_1^2 + a_2^2 + a_3^2 = 1^2 + 4^2 + 7^2 = 66$.

To reach the final answer, we calculate $a_{17} = a + 16d = 1 + 16(3) = 49$. Using our derived formula for $k=7$, we find $A_7 = -7(3)(2(1) + (13)(3)) = -21(2 + 39) = -21(41) = -861$. Performing the final subtraction $a_{17} - A_7$ results in $49 - (-861)$, which equals 910.

An infinite geometric progression with first term $a$ and common ratio $r$ converges to $\frac{a}{1-r}$ provided the magnitude of $r$ is less than 1. For the first given series, this relationship implies $\frac{a}{1-r} = 57$. The second series, consisting of terms $a^3, a^3r^3, a^3r^6, \dots$, forms its own geometric progression with a first term of $a^3$ and a common ratio of $r^3$, resulting in the sum $\frac{a^3}{1-r^3} = 9747$.

By cubing the first equation, we obtain $\frac{a^3}{(1-r)^3} = 57^3$. Dividing this resulting expression by the second equation allows us to isolate the variable $r$ as $\frac{1-r^3}{(1-r)^3} = \frac{57^3}{9747} = 19$. Expanding the numerator as $(1-r)(1+r+r^2)$, we simplify the equation to $\frac{1+r+r^2}{(1-r)^2} = 19$, which rearranges into the quadratic form $18r^2 - 39r + 18 = 0$. Dividing by 3 simplifies this to $6r^2 - 13r + 6 = 0$, which factors into $(2r-3)(3r-2) = 0$. Since an infinite geometric series requires $|r| < 1$ for convergence, we reject $r = \frac{3}{2}$ and accept $r = \frac{2}{3}$. Substituting this ratio into our first expression, $a = 57(1 - \frac{2}{3}) = 19$. Consequently, the value of $a + 18r$ is $19 + 18(\frac{2}{3})$, which results in 31.

The expression $(1-x)^{2008} (1+x+x^2)^{2007}$ is evaluated by recognizing that $(1-x)(1+x+x^2)$ is the difference of cubes identity, $1-x^3$, which allows the expression to be rewritten as $(1-x)(1-x^3)^{2007}$. By applying the binomial theorem to $(1-x^3)^{2007}$, the general term is expressed as $(-1)^r \binom{2007}{r} x^{3r}$, and multiplying this by the leading factor of $(1-x)$ produces two specific series containing terms of the form $x^{3r}$ and $x^{3r+1}$.

Determining the coefficient of $x^{2012}$ requires checking whether the exponents $3r$ or $3r+1$ can equal 2012 for some integer $r$. Since 2012 is not divisible by 3, and 2011 is also not divisible by 3, there is no integer $r$ that satisfies either condition. Because neither exponent can yield 2012, the coefficient of $x^{2012}$ is 0.

The relationship between these sequences is anchored in the fundamental properties of geometric ratios and arithmetic differences. For the geometric progression $2, p, q$, the definition of a common ratio dictates that $p^2 = 2q$. When these values represent the $7^{\text{th}}, 8^{\text{th}}$, and $13^{\text{th}}$ terms of an arithmetic progression with first term $a$ and common difference $d$, they can be expressed as $2 = a + 6d$, $p = a + 7d$, and $q = a + 12d$.

Subtracting the arithmetic equations reveals that $d = p - 2$ and $q - p = 5d$, which together imply $q = p + 5(p - 2) = 6p - 10$. Substituting this expression back into the geometric condition $p^2 = 2q$ yields the quadratic equation $p^2 - 12p + 20 = 0$, resulting in roots $p = 10$ or $p = 2$. Because the problem specifies $q \neq 2$, we discard $p = 2$ and conclude that $p = 10$, which implies $d = 8$.

With these parameters defined, the first term is found by $a = 2 - 6(8) = -46$. Extending the geometric progression $2, 10, 50, 250, 1250$ identifies the fifth term as $1250$. Setting this value equal to the general arithmetic formula $-46 + (n - 1)8 = 1250$ leads to $8(n - 1) = 1296$, which simplifies to $n - 1 = 162$ and results in $n = 163$.

Shared terms between two arithmetic sequences form a distinct progression where the common difference is the least common multiple of the original differences. Because the first sequence increases by 4 and the second by 3, the new progression shares a common difference of $\operatorname{LCM}(4,3)=12$. Starting with 11 as the first common term, the progression continues as 11, 23, 35, and so on, up to the point where the value remains at or below 403. Solving $11 + (n-1)12 \leq 403$ results in $n-1 \leq 32.66$, confirming there are 33 terms in this sub-sequence. Utilizing the arithmetic sum formula $\frac{n}{2}(2a+(n-1)d)$ with $n=33$, $a=11$, and $d=12$ yields $\frac{33}{2}(22+32 \times 12)$, which results in a total of 6699.

The problem can be solved by simplifying the second group of terms using partial fraction decomposition. Each term of the form $\frac{1}{2k(2k-1)}$ can be rewritten as $\frac{1}{2k-1} - \frac{1}{2k}$, allowing the entire summation to be expressed as an alternating harmonic series:
$\left( 1 - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdot s + \left( \frac{1}{2023} - \frac{1}{2024} \right) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdot s - \frac{1}{2024}$

Using the property of alternating harmonic series, we know that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdot s - \frac{1}{2n}$ is equivalent to the sum of reciprocals from $n+1$ to $2n$. By substituting $n=1012$ into this identity, the series simplifies to the sum $\frac{1}{1013} + \frac{1}{1014} + \cdot s + \frac{1}{2024}$.

Substituting this result back into the original equation, we have $\sum_{k=1}^{1012} \frac{1}{\alpha+k} - \sum_{k=1013}^{2024} \frac{1}{k} = \frac{1}{2024}$. Moving the summation to the right-hand side gives $\sum_{k=1}^{1012} \frac{1}{\alpha+k} = \frac{1}{2024} + \sum_{k=1013}^{2024} \frac{1}{k}$, which simplifies to $\sum_{k=1012}^{2023} \frac{1}{k}$. Comparing both sides of the equation, we can see that $\alpha+1$ must correspond to $1012$, leading to $\alpha = 1011$.

The middle term in an arithmetic progression is the arithmetic mean of the surrounding terms, meaning $2(\frac{K}{2}) = K$ must equal the sum of the first and third terms, $(4^{1+x} + 4^{1-x}) + (16^x + 16^{-x})$. Simplifying these expressions by letting $t = 4^x + 4^{-x}$, we observe that $4^{1+x} + 4^{1-x} = 4t$ and $16^x + 16^{-x} = t^2 - 2$, leading to the function $K = t^2 + 4t - 2$. Because $x \ge 0$, $4^x \ge 1$ and thus $t = 4^x + 4^{-x} \ge 2$, meaning the expression $t^2 + 4t - 2$ reaches its minimum when $t = 2$. Substituting this minimum value yields $K = 2^2 + 4(2) - 2$, which confirms the least value of 10.

The total work required to complete the assignment remains constant at $17m$ units. When the number of operational computer systems decreases by 4 every day starting from the second, the project completion time extends by 8 days, resulting in a total of 25 days of work. This creates a series where the daily work performed is $m$ on the first day, $(m - 4)$ on the second, $(m - 8)$ on the third, and finally $(m - 4 \times 24)$ on the twenty-fifth day.

Setting the original work requirement equal to the sum of the work done over 25 days produces the following equation:

$17m = 25m - 4(1 + 2 + \dots + 24)$

Applying the arithmetic series sum formula to the numbers 1 through 24, the equation becomes:

$17m = 25m - 4 \left( \frac{24 \times 25}{2} \right)$

Simplifying this yields $17m = 25m - 1200$, which rearranges to $8m = 1200$. Dividing by 8 results in $m = 150$.

When connecting the midpoints of an equilateral triangle with side length $a$, each subsequent triangle has side lengths exactly half those of the previous one. The perimeters follow an infinite geometric progression starting at $3a$, where each term is half the preceding one, resulting in a total sum $P = 3a + \frac{3a}{2} + \frac{3a}{4} + \dots = \frac{3a}{1 - 1/2} = 6a$.

Areas scale with the square of the side lengths, meaning each successive triangle possesses $1/4$ the area of the previous one. The total area sum becomes $Q = \frac{\sqrt{3}}{4}a^2 + \frac{\sqrt{3}}{16}a^2 + \frac{\sqrt{3}}{64}a^2 + \dots = \frac{\sqrt{3}a^2}{4} \left( \frac{1}{1 - 1/4} \right) = \frac{\sqrt{3}a^2}{4} \cdot \frac{4}{3} = \frac{a^2}{\sqrt{3}}$. Substituting $a = P/6$ derived from the perimeter equation into this area formula yields $Q = \frac{(P/6)^2}{\sqrt{3}} = \frac{P^2}{36\sqrt{3}}$. Rearranging this expression confirms that $P^2 = 36 \sqrt{3} Q$.

The core of the problem lies in the properties of the given progressions: the arithmetic property $a^3 + c^3 = 2b^3$ and the geometric property $(\log_c a)^2 = (\log_a b)(\log_b c)$, which together force $a=b=c$. Applying the change of base formula, the product $(\log_a b)(\log_b c)$ simplifies to $\frac{\log c}{\log a} = \log_a c = \frac{1}{\log_c a}$, and equating this to $(\log_c a)^2$ gives $(\log_c a)^3 = 1$, or $a = c$. Substituting $a=c$ into the arithmetic condition $a^3 + c^3 = 2b^3$ yields $2a^3 = 2b^3$, which confirms $a = b = c$ since $a, b, c > 1$.

With all variables equal, the first term $a_1$ of the arithmetic series is $\frac{a + 4a + a}{3} = 2a$, and the common difference $d$ is $\frac{a - 8a + a}{10} = -\frac{3}{5}a$. The sum of the first 20 terms is given by $S_{20} = 10(2(2a) + 19(-\frac{3}{5}a)) = -74a$. Setting $S_{20} = -444$ gives $a = 6$, so $abc = 6 \times 6 \times 6 = 216$.

To determine the coefficient of a specific power of $x$ in a binomial expansion, we focus on isolating the variable component using the general term formula. By breaking down the expression into its variable and numerical parts, we can determine the index $r$ that balances the powers of $x$ to match the required exponent. This process involves grouping the exponents effectively so that the variable simplifies into a single power, allowing us to solve a simple linear equation for the necessary value of $r$.

The general term of the expansion is represented as $T_{r+1} = {}^5 C_r (2 x^3)^{5-r} \left(-\frac{1}{3 x^2}\right)^r$. Extracting the constants and combining the powers of $x$, this simplifies to $T_{r+1} = {}^5 C_r (2)^{5-r} \left(-\frac{1}{3}\right)^r x^{15-3r-2r}$, which further reduces to $T_{r+1} = {}^5 C_r (2)^{5-r} \left(-\frac{1}{3}\right)^r x^{15-5r}$. We set the exponent $15-5r$ equal to $5$, which yields $5r = 10$, identifying $r = 2$ as the required index. Substituting $r = 2$ into the coefficient expression gives ${}^5 C_2 \cdot 2^3 \cdot \left(-\frac{1}{3}\right)^2$, which calculates to $10 \cdot 8 \cdot \frac{1}{9}$, resulting in $\frac{80}{9}$.