Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

For a quadratic equation $Ax^{2}+Bx+C=0$, the product of the roots is given by $C/A$ and the sum by $-B/A$. For the given equation, the product simplifies to $n^{2}-2n+2$, which is $(n-1)^{2}+1$, making the minimum value $\alpha = 1$. The sum of the roots is $\frac{3}{n^{2}-2n+2}$, which reaches a maximum value of $\beta = 3$ when the denominator is minimized.

With a geometric progression defined by first term $a=1$ and common ratio $r=\frac{1}{3}$, the sum of the first six terms is:
$S_6 = \frac{1(1-(\frac{1}{3})^{6})}{1-\frac{1}{3}} = \frac{3}{2} \left(1 - \frac{1}{729}\right) = \frac{364}{243}$

The recurrence relation becomes significantly easier to handle when expressed as a geometric progression. The term $\frac{n^2 - 2n - 1}{n^2 (n+1)^2}$ is equivalent to $\frac{2}{(n+1)^2} - \frac{1}{n^2}$. When this is substituted into the original equation $a_{n+1} = \frac{1}{2}a_n + \frac{2}{(n+1)^2} - \frac{1}{n^2}$, the terms can be rearranged as $a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2}(a_n - \frac{2}{n^2})$.

Setting $b_n = a_n - \frac{2}{n^2}$ yields $b_{n+1} = \frac{1}{2} b_n$, showing that $\{b_n\}$ is a geometric sequence with a common ratio of $r = \frac{1}{2}$. Given $a_1 = 1$, the first term is $b_1 = 1 - \frac{2}{1^2} = -1$. The sum of this infinite geometric progression is $\frac{-1}{1 - \frac{1}{2}}$, which evaluates to $-2$. The final required value is the absolute value of this sum, resulting in $2$.

The general term $T_r$ of the series is defined by the ratio of the sum of the first $r$ cubes to the sum of the first $r$ odd integers. Since the sum of the first $r$ cubes is expressed as $\left(\frac{r(r+1)}{2}\right)^{2}$ and the sum of the first $r$ odd integers is $r^{2}$, the general term reduces to $\frac{r^{2}(r+1)^{2}}{4r^{2}}$, which simplifies to $\frac{(r+1)^{2}}{4}$.

The summation $\sum_{r=1}^{8} \frac{(r+1)^{2}}{4}$ expands to $\frac{1}{4} \sum_{r=1}^{8} (r^{2}+2r+1)$. Applying standard summation identities for the first $n$ squares, integers, and constants with $n=8$ produces $\frac{1}{4} \left[ \frac{8 \times 9 \times 17}{6} + 2 \times \frac{8 \times 9}{2} + 8 \right]$. This expression simplifies to $\frac{1}{4} [204 + 72 + 8]$, yielding a final value of 71.

The expression in the denominator, $r^4+r^2+1$, can be factored as $(r^2-r+1)(r^2+r+1)$ by viewing it as the difference of squares $(r^2+1)^2 - r^2$. Using partial fraction decomposition, the general term $\frac{r}{r^4+r^2+1}$ can be rewritten as $\frac{1}{2}\left(\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right)$. If we define $f(r) = \frac{1}{r^2-r+1}$, then $f(r+1) = \frac{1}{(r+1)^2-(r+1)+1} = \frac{1}{r^2+r+1}$, transforming the summation into a telescoping series $\frac{1}{2}\sum_{r=1}^{25} (f(r) - f(r+1))$. This reduces to the simplified form $\frac{1}{2}(f(1) - f(26))$. Plugging in the specific values where $f(1) = 1$ and $f(26) = \frac{1}{26^2-26+1} = \frac{1}{651}$, the total sum becomes $\frac{1}{2}\left(1 - \frac{1}{651}\right)$, which simplifies to $\frac{325}{651}$. Since $325$ and $651$ are coprime, $p=325$ and $q=651$, resulting in a sum of $p+q = 976$.

When investigating the product of three terms in a geometric progression, it is highly efficient to denote the terms as $\frac{k}{r}$, $k$, and $kr$. This choice allows the common ratio $r$ to cancel during multiplication, leading directly to $k^3 = 27$, which confirms the middle term is 3. With the middle term identified, the sum of the sequence becomes

$S = 3\left(r + \frac{1}{r} + 1\right)$

The sum values are constrained by the properties of $r + \frac{1}{r}$. By the AM-GM inequality, when $r > 0$, the expression $r + \frac{1}{r}$ is at least 2, which implies $S \ge 9$. When $r < 0$, the expression $r + \frac{1}{r}$ is at most -2, which implies $S \le -3$. Consequently, the set of possible sums includes everything outside the interval $(-3, 9)$, meaning $a = -3$ and $b = 9$. Squaring and adding these values yields $a^2 + b^2 = 9 + 81 = 90$.

Arithmetic progressions are defined by the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$ and the nth term formula $a_n = a_1 + (n-1)d$. Substituting the given values $a_n = \frac{1}{4}a_1$ and $d = -\frac{3}{4}$ into these formulas generates a system of equations to determine $a_1$ and $n$. The sum formula $\frac{n}{2}(a_1 + \frac{1}{4}a_1) = \frac{525}{2}$ simplifies to the relation $a_1n = 420$, while the term formula $\frac{1}{4}a_1 = a_1 + (n-1)(-\frac{3}{4})$ leads to $a_1 = n-1$.

Substituting $n = a_1 + 1$ into the first relation creates the quadratic equation $a_1^2 + a_1 - 420 = 0$, which factors as $(a_1+21)(a_1-20) = 0$. Since the first term must be positive, we identify $a_1 = 20$ and consequently $n = 21$. Calculating the sum of the first 17 terms using $S_{17} = \frac{17}{2}(2a_1 + 16d)$ and substituting $a_1 = 20$ and $d = -\frac{3}{4}$ yields $S_{17} = \frac{17}{2}(40 - 12)$. This simplifies to $17 \times 14 = 238$.

This problem relies on the method of telescoping series, where the individual terms are decomposed such that most of them cancel each other out when summed. Focusing on the general term $\frac{1}{n(n+1)(n+2)}$, notice that the difference between the factors $(n+2)$ and $n$ is $2$, which allows us to rewrite the expression as $\frac{1}{2} \left( \frac{(n+2) - n}{n(n+1)(n+2)} \right)$. This simplifies the term into $\frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right)$.

When we sum this expression from $n=1$ to $10$, the sequence becomes a series of differences: $\frac{1}{2} \left( \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right) + \dots + \left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right) \right)$. In this structure, each negative part cancels with the positive part of the succeeding term. After all intermediate terms vanish, only the first fraction and the final fraction remain, leaving $\frac{1}{2} \left( \frac{1}{2} - \frac{1}{132} \right)$. Calculating the values within the parentheses yields $\frac{1}{2} \left( \frac{66 - 1}{132} \right) = \frac{65}{264}$. Multiplying this total sum by the constant $528$ gives $528 \times \frac{65}{264}$, which simplifies to $130$.

The relationship between the partial sum $S_n$ and the general term $a_n$ is defined by the subtraction $a_n = S_n - S_{n-1}$. Applying this to the given $S_n = 6n^3$ leads to $a_n = 6n^3 - 6(n-1)^3$. Expanding and simplifying this expression gives the quadratic formula $a_n = 18n^2 - 18n + 6$.

Calculating the difference $a_{k+1} - a_k$ allows for the simplification of the term inside the summation. With $a_{k+1} = 18(k+1)^2 - 18(k+1) + 6$, the expression expands to $18k^2 + 18k + 6$. Subtracting $a_k = 18k^2 - 18k + 6$ from this yields $36k$. Consequently, the fraction $\frac{a_{k+1} - a_k}{36}$ reduces simply to $k$.

Squaring this result provides $k^2$. Summing these values from $k=1$ to $k=6$ involves calculating $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$. Applying the standard sum of squares formula $\frac{n(n+1)(2n+1)}{6}$ with $n=6$ results in $\frac{6 \times 7 \times 13}{6}$, which equals 91.

Each term in the infinite series represents a finite geometric progression of the form $\sum_{k=0}^n a^{n-k} b^k$, where $a = \frac{1}{3}$ and $b = \frac{4}{7}$. The sum of such a finite progression is $\frac{b^{n+1} - a^{n+1}}{b - a}$, and calculating the denominator gives $b - a = \frac{4}{7} - \frac{1}{3} = \frac{5}{21}$. Substituting this into the formula, each term simplifies to $T_n = \frac{21}{5} (b^{n+1} - a^{n+1})$. Summing this expression from $n=1$ to infinity requires evaluating two separate infinite geometric series, $\sum_{n=1}^\infty b^{n+1}$ and $\sum_{n=1}^\infty a^{n+1}$. The first series sums to $\frac{b^2}{1-b}$, which equals $\frac{16/49}{3/7} = \frac{16}{21}$, while the second series sums to $\frac{a^2}{1-a}$, which equals $\frac{1/9}{2/3} = \frac{1}{6}$. Combining these results, the total sum is $\frac{21}{5} \left( \frac{16}{21} - \frac{1}{6} \right)$, which simplifies to $\frac{21}{5} \left( \frac{32 - 7}{42} \right) = \frac{21}{5} \times \frac{25}{42} = \frac{5}{2}$.

The sequence $729, 81, 9, 1, \dots$ forms a geometric progression when viewed as powers of 9, specifically $9^3, 9^2, 9^1, 9^0$ and so on. The exponents follow an arithmetic progression with a starting term of 3 and a common difference of $-1$, meaning the $k$-th exponent is given by $3 - (k-1)$, or $4-k$. Consequently, the product of the first $n$ terms, denoted as $P_n$, is the base 9 raised to the sum of the first $n$ exponents, which is $\frac{n}{2}[2(3) + (n-1)(-1)] = \frac{n(7-n)}{2}$.

Expressing the product as a power of 3, we have $P_n = 9^{\frac{n(7-n)}{2}} = 3^{n(7-n)}$. Taking the $n$-th root yields $(P_n)^{\frac{1}{n}} = 3^{7-n}$. Summing these values from $n=1$ to $40$ results in the geometric series $3^6 + 3^5 + \dots + 3^{-33}$.

The sum of this geometric progression is found using the formula $S = a\frac{1-r^n}{1-r}$ with the first term $a = 3^6$, common ratio $r = \frac{1}{3}$, and 40 terms. This substitution provides $3^6 \cdot \frac{1-(1/3)^{40}}{1-1/3} = 3^6 \cdot \frac{3}{2}(1-3^{-40}) = \frac{3^7 - 3^{-33}}{2}$. Multiplying this result by 2 yields $\frac{3^7 - 3^{-33}}{1} = \frac{3^{40}-1}{3^{33}}$. Matching this expression to $\frac{3^\alpha-1}{3^\beta}$ reveals that $\alpha = 40$ and $\beta = 33$, and because their greatest common divisor is 1, the value of $\alpha+\beta$ is 73.

When three numbers form an arithmetic progression, they can be represented symmetrically around the middle term as $b-d$, $b$, and $b+d$. Summing these terms as $(b-d) + b + (b+d) = 3b$ leads to the total of $1$, identifying $b$ as $\frac{1}{3}$. Because $a^2$, $2b^2$, and $c^2$ form a geometric progression, the square of the middle term must equal the product of the outer terms, yielding $4b^4 = a^2 c^2$. This is equivalent to $4b^4 = ((b-d)(b+d))^2$, which simplifies to $4b^4 = (b^2 - d^2)^2$. Taking the square root, $2b^2 = |b^2 - d^2|$, and considering that $a < b < c$ implies a positive $d$, we determine that $d^2 = 3b^2$. Substituting the value of $b$ into this expression gives $d^2 = 3(\frac{1}{9})$, which is $\frac{1}{3}$. The expression $a^2 + b^2 + c^2$ expands from $(b-d)^2 + b^2 + (b+d)^2$ to $3b^2 + 2d^2$. Replacing the variables with the calculated values results in $3(\frac{1}{9}) + 2(\frac{1}{3})$, which equals $1$. Multiplying this sum by 9 results in 9.

To determine the value of the given expression, first calculate the sum of the arithmetic series $\alpha = 3+4+8+9+13+14+\ldots$ which consists of 40 terms. This series is composed of two distinct arithmetic progressions interleaved, each containing 20 terms. The first sequence begins with 3 and has a common difference of 5, summing to $\frac{20}{2}[2(3)+(20-1)5] = 10(6+95) = 1010$. The second sequence begins with 4 and has a common difference of 5, summing to $\frac{20}{2}[2(4)+(20-1)5] = 10(8+95) = 1030$. Combining these yields a total value of $\alpha = 1010 + 1030 = 2040$.

The term $(\tan \beta)^{\frac{\alpha}{1020}}$ simplifies using $\alpha = 2040$ to $(\tan \beta)^{\frac{2040}{1020}} = (\tan \beta)^2$. Solving the quadratic equation $x^2 + x - 2 = 0$ involves factoring the expression into $(x+2)(x-1) = 0$, which identifies the roots as $x = -2$ and $x = 1$. Because $\beta$ is restricted to the interval $\left(0, \frac{\pi}{2}\right)$, $\tan \beta$ must be positive, which implies that $(\tan \beta)^2$ must also be positive. Consequently, the only valid root is 1, so $(\tan \beta)^2 = 1$, which leads to $\tan \beta = 1$ and thus $\beta = \frac{\pi}{4}$.

Substituting $\beta = \frac{\pi}{4}$ into the final expression $\sin^2 \beta + 3 \cos^2 \beta$ yields $\sin^2\left(\frac{\pi}{4}\right) + 3\cos^2\left(\frac{\pi}{4}\right)$. Using the trigonometric values, this becomes $\left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2$, which simplifies to $\frac{1}{2} + 3\left(\frac{1}{2}\right) = \frac{4}{2} = 2$.

The minimum value of $f(\theta) = \alpha \tan^{2}\theta + \beta \cot^{2}\theta$ is established via the Arithmetic Mean-Geometric Mean inequality, which results in $2\sqrt{\alpha\beta}$ because the product of the two terms is constant. Rewriting $g(\theta) = \alpha \sin^{2}\theta + \beta \cos^{2}\theta$ as $(\alpha-\beta)\sin^{2}\theta + \beta$ reveals that, given $\alpha > \beta$, the function attains a maximum value of $\alpha$ at $\sin^2\theta = 1$. Equating the minimum of $f(\theta)$ with the maximum of $g(\theta)$ produces $2\sqrt{\alpha\beta} = \alpha$, which simplifies to $4\alpha\beta = \alpha^2$ and establishes the relationship $\alpha = 4\beta$.

Applying this relationship to the geometric progression defines the first term as $a = \frac{\alpha}{2\beta} = \frac{4\beta}{2\beta} = 2$ and the common ratio as $r = \frac{2\beta}{\alpha} = \frac{2\beta}{4\beta} = \frac{1}{2}$. The sum of the first 10 terms is found using the formula $S_{10} = \frac{a(1-r^{10})}{1-r}$, which substitutes to $S_{10} = 2 \frac{1-(1/2)^{10}}{1-1/2}$. Evaluating this expression results in $4 \left( 1 - \frac{1}{1024} \right)$, simplified further to $4 \left( \frac{1023}{1024} \right) = \frac{1023}{256}$. Given the fraction is $\frac{m}{n}$ with $\gcd(m,n)=1$, the values $m=1023$ and $n=256$ combine to yield $m+n = 1279$.

We can approach this by separating the initial fraction from the rest of the summation to reveal a clear geometric pattern. First, simplify the starting term $\frac{6}{3^{26}}$ by noting that $6 = 2 \cdot 3$, which allows us to rewrite it as $\frac{2 \cdot 3}{3^{26}}$, or simply $\frac{2}{3^{25}}$. Now, examine the remaining series, which consists of terms starting from $\frac{10}{3^{25}}$ up to $\frac{10 \cdot 2^{24}}{3}$. By factoring out a constant $\frac{10}{3^{25}}$ from this sequence, we are left with the sum $1 + 2 \cdot 3 + 2^2 \cdot 3^2 + \dots + 2^{24} \cdot 3^{24}$.

The expression inside the parentheses represents the powers of $6$ ranging from $6^0$ to $6^{24}$. Using the geometric series formula $\frac{a(r^n - 1)}{r - 1}$ with $a=1$, $r=6$, and $n=25$, this sum becomes $\frac{6^{25} - 1}{6 - 1}$, which is $\frac{6^{25} - 1}{5}$. Multiplying this by our factored constant $\frac{10}{3^{25}}$ gives $\frac{10}{3^{25}} \cdot \frac{6^{25} - 1}{5}$, simplifying to $\frac{2(6^{25} - 1)}{3^{25}}$. Distributing the $2$ in the numerator results in $\frac{2 \cdot 6^{25}}{3^{25}} - \frac{2}{3^{25}}$. Since $6^{25}$ is equivalent to $2^{25} \cdot 3^{25}$, the first part of this expression becomes $2 \cdot 2^{25}$, which equals $2^{26}$. Adding back the first term of the original series, $\frac{2}{3^{25}}$, cancels the negative term and leaves $2^{26}$ as the final result.

The arithmetic progression is defined by a first term $a$ and a common difference $d$, while the geometric progression has a first term $b$ and a common ratio $r$. The sum of the first ten terms of the arithmetic sequence follows the formula:

$\frac{10}{2}(2a + 9d) = 160$

Simplifying this gives the relation $2a + 9d = 32$, which allows us to express the first term as $a = \frac{32 - 9d}{2}$.

Given that the first term of the arithmetic progression is equal to the common ratio of the geometric progression ($a = r$) and the first term of the geometric progression is equal to the common difference of the arithmetic progression ($b = d$), the sum of the first two terms of the geometric progression, $b + br = 8$, becomes:

$d + da = 8$

Substituting the expression for $a$ into this equation leads to:

$d + d\left(\frac{32 - 9d}{2}\right) = 8$

Multiplying both sides by $2$ results in $2d + 32d - 9d^2 = 16$, which rearranges into the quadratic equation:

$9d^2 - 34d + 16 = 0$

Since the first term of the geometric progression is $d$, the sum of its all possible values is equivalent to the sum of the roots of this quadratic equation, which is $\frac{34}{9}$.

Any sequence constructed by scaling terms of a geometric progression by powers of a constant, such as $T_n = \frac{a_n}{2^{n-1}}$, preserves the geometric progression property with a modified common ratio. Since the given sequence $T_n$ has a common ratio of $\frac{1}{\sqrt{2}}$, the ratio between consecutive terms satisfies $\frac{a_{n+1}/2^n}{a_n/2^{n-1}} = \frac{1}{\sqrt{2}}$. Multiplying this expression by $2$ demonstrates that the underlying sequence $a_n$ has a common ratio of $\frac{2}{\sqrt{2}} = \sqrt{2}$.

With the common ratio established, the sum of the first 10 terms is determined using the geometric series formula $S_n = \frac{a_1(r^n - 1)}{r - 1}$. Substituting the values $S_{10} = 62$ and $r = \sqrt{2}$ yields the equation:
$62 = \frac{a_1 ((\sqrt{2})^{10} - 1)}{\sqrt{2} - 1}$
Because $(\sqrt{2})^{10} = 32$, the equation simplifies to $62 = \frac{31 a_1}{\sqrt{2} - 1}$. Dividing both sides by 31 and multiplying by the denominator shows that $a_1 = 2(\sqrt{2} - 1)$.

Arithmetic progressions are defined by a constant difference between consecutive terms, while the arithmetic mean provides a linear average that links the reciprocal values of the variables involved. The arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ being $\frac{5}{16}$ implies that the sum of these reciprocals is $\frac{5}{8}$, which simplifies to the relation $8(a+b) = 5ab$. Given the sequence $a, 4, \alpha, b$ is in arithmetic progression, the terms can be represented as $a$, $a+d$, $a+2d$, and $a+3d$ where $d = 4-a$. Expressing $\alpha$ as $8-a$ and $b$ as $12-2a$ allows substitution into the initial relation, resulting in $10a^2 - 68a + 96 = 0$. Solving this quadratic yields $a = 4.8$ or $a = 2$, and since the condition $a > 2$ is specified, $a = 4.8$ is the only valid solution.

With $a = 4.8$, the remaining variables are determined as $\alpha = 3.2$ and $b = 2.4$. Inserting these into the quadratic equation $\alpha x^2 - a x + 2(\alpha - 2 b) = 0$ produces $3.2 x^2 - 4.8 x + 2(3.2 - 4.8) = 0$, which reduces to $3.2 x^2 - 4.8 x - 3.2 = 0$. Dividing throughout by $1.6$ simplifies the equation to $2 x^2 - 3 x - 2 = 0$. Factoring this expression as $(2 x+1)( x-2) = 0$ reveals the roots to be $x = 2$ and $x = -0.5$. These roots fall precisely within the intervals $(1, 4)$ and $(-2, 0)$ respectively.

The $n$-th term of this series is defined by $T_n = \frac{1}{n} \sum_{k=1}^{n} k^2$. Using the standard formula for the sum of the first $n$ squares, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, we can simplify this term to $T_n = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$, which further expands to the expression $\frac{2n^2 + 3n + 1}{6}$.

To determine the total sum up to 10 terms, we calculate the summation of $T_n$ as $n$ ranges from 1 to 10. This requires evaluating the expression $\frac{1}{6} \left( 2 \sum_{n=1}^{10} n^2 + 3 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 1 \right)$. Utilizing standard summation identities, the sum of the first 10 squares is 385, the sum of the first 10 integers is 55, and summing the constant 1 ten times yields 10. Substituting these values into the expression leads to $\frac{1}{6} [2(385) + 3(55) + 10]$, which calculates to $\frac{1}{6} [770 + 165 + 10]$, resulting in a total value of $\frac{945}{6}$, which simplifies to $\frac{315}{2}$.

The core principle involves transforming a series of product terms into a single quadratic equation where the value of the parameter $n$ determines the roots. Expanding the summation $\sum_{k=1}^n (x+2k-2)(x+2k) = \frac{8n}{3}$ leads to $nx^2 + 2n^2x + \frac{4n(n^2-1)}{3} = \frac{8n}{3}$, which simplifies to the general quadratic $3x^2 + 6nx + 4n^2 - 12 = 0$. Testing $n = 3$ results in $3x^2 + 18x + 24 = 0$, or $x^2 + 6x + 8 = 0$. Since this factors as $(x+2)(x+4) = 0$, the roots are -2 and -4, which are consecutive even integers.

When the sum of the first $n$ terms of a sequence is represented by a quadratic expression without a constant term, the sequence necessarily forms an arithmetic progression. To find the expression for the $n$-th term, we use the relationship $a_n = S_n - S_{n-1}$, which, after substituting the quadratic form $\alpha n^2 + \beta n$, simplifies to the general formula $a_n = 2\alpha n + \beta - \alpha$.

Applying this relationship to the provided conditions, we use $a_{10} = 59$ to establish the equation $2\alpha(10) + \beta - \alpha = 19\alpha + \beta = 59$. The condition $a_6 = 7a_1$ translates to $2\alpha(6) + \beta - \alpha = 7(2\alpha(1) + \beta - \alpha)$, which simplifies to $11\alpha + \beta = 7(\alpha + \beta)$, or $4\alpha = 6\beta$. Since $2\alpha = 3\beta$, we can substitute $\alpha = 1.5\beta$ into the first equation, resulting in $19(1.5\beta) + \beta = 59$. This leads to $29.5\beta = 59$, yielding $\beta = 2$ and consequently $\alpha = 3$, which makes the sum $\alpha + \beta = 5$.