Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

The relationship between the arithmetic mean and the geometric mean allows for an efficient determination of these variables. Given the structure $x+y+z=12$ and the product $x^{3}y^{4}z^{5}=(0.1)(600)^{3}$, we apply the weighted AM-GM inequality by considering the terms $\frac{x}{3}, \frac{y}{4},$ and $\frac{z}{5}$ weighted by their respective exponents $3, 4,$ and $5$. The sum of these weights is $3+4+5=12$, which matches the total sum of the variables. Because the weighted arithmetic mean of $1$ aligns exactly with the geometric mean derived from the provided product, equality must hold, meaning $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$.

This proportionality implies $x=3k, y=4k,$ and $z=5k$. Substituting these expressions into the sum equation $3k+4k+5k=12$ results in $12k=12$, identifying $k=1$ and thus $x=3, y=4, z=5$. With the variables defined, the final expression $x^{3}+y^{3}+z^{3}$ is calculated as $3^{3}+4^{3}+5^{3}$, which simplifies to $27+64+125$, resulting in $216$.

The general term of the series is found by expressing the numerator as the sum of the first $r$ cubes and the denominator as the sum of the first $r$ odd integers. The sum of the first $r$ cubes simplifies to $\frac{r^2(r+1)^2}{4}$, while the sum of the first $r$ odd integers is $r^2$. Dividing these expressions simplifies the $r^{th}$ term of the series to $t_r = \frac{1}{4}(r+1)^2$.

Summing the first 9 terms corresponds to evaluating $S_9 = \sum_{r=1}^{9} \frac{1}{4}(r+1)^2$. This expansion results in $\frac{1}{4}(2^2 + 3^2 + \dots + 10^2)$, which is equivalent to $\frac{1}{4} \left( \sum_{r=1}^{10} r^2 - 1 \right)$. Since the sum of squares up to 10 is 385, subtracting 1 and dividing the difference by 4 yields 96.

When $l$ and $n$ have three geometric means $G_1, G_2, G_3$ inserted between them, the set $l, G_1, G_2, G_3, n$ forms a geometric progression. By representing the terms using a common ratio $r$, we define $G_1=lr$, $G_2=lr^2$, $G_3=lr^3$, and $n=lr^4$, which establishes that $r^4 = \frac{n}{l}$. Additionally, the definition of the arithmetic mean $m$ provides the relation $l+n = 2m$.

To evaluate the expression $G_1^4+2G_2^4+G_3^4$, we substitute these geometric terms to obtain $(lr)^4 + 2(lr^2)^4 + (lr^3)^4$. Factoring out $l^4r^4$ yields the expression $l^4r^4(1 + 2r^4 + r^8)$, which simplifies to the perfect square $l^4r^4(1+r^4)^2$. Replacing $r^4$ with $\frac{n}{l}$ transforms this into $l^4 \cdot \frac{n}{l} \cdot (1 + \frac{n}{l})^2$, which simplifies further to $l^3n \cdot \left(\frac{l+n}{l}\right)^2$. Since $l+n=2m$, we substitute this value into the equation to get $l^3n \cdot \left(\frac{2m}{l}\right)^2$. Calculating this final term results in $l^3n \cdot \frac{4m^2}{l^2}$, which simplifies to $4lm^2n$.

This expression represents an Arithmetico-Geometric series, which can be evaluated by leveraging the relationship between consecutive terms. By defining the series as $S = 10^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9$, one can multiply the entire sum by the common ratio $\frac{11}{10}$ to shift the terms. Subtracting the resulting expression from the original sum creates the equation $-\frac{1}{10}S = 10^9 + 11^1(10)^8 + 11^2(10)^7 + \dots + 11^9 - 11^{10}$. The sequence of terms before the final subtraction forms a geometric series with ten elements that simplifies to $11^{10} - 10^{10}$ when calculated using the geometric summation formula. Consequently, the equation simplifies to $-\frac{1}{10}S = 11^{10} - 10^{10} - 11^{10}$, which reduces to $-\frac{1}{10}S = -10^{10}$. This leads to $S = 10^{11}$, which is equivalent to $100(10)^9$, meaning $k = 100$.

For a sequence to form an arithmetic progression, the middle term must be the arithmetic mean of its neighbors, which is expressed as the middle term being half the sum of the flanking terms. When three numbers from a geometric progression are represented as $a, ar,$ and $ar^2$, doubling the middle term results in the sequence $a, 2ar,$ and $ar^2$. Applying the arithmetic progression condition to these terms gives the equation
$2(2ar) = a + ar^2$
Since $a$ cannot be zero in this geometric progression, we divide by $a$ to simplify the relationship to $4r = 1 + r^2$. Rearranging these terms results in the quadratic equation
$r^2 - 4r + 1 = 0$
Solving this quadratic equation yields the two possible values $r = 2 \pm \sqrt{3}$. Given the requirement that the original geometric progression is increasing, we identify the correct common ratio as $2 + \sqrt{3}$.

The sequence follows a pattern where each term is composed of repeating sevens, represented by the formula $t_r = \frac{7}{9}(1 - 10^{-r})$. Distributing the summation across these terms gives $S_{20} = \frac{7}{9} \sum_{r=1}^{20} 1 - \frac{7}{9} \sum_{r=1}^{20} 10^{-r}$, where the sum of $1$ over twenty terms is $20$ and the second part is a geometric series summing to $\frac{1}{9}(1 - 10^{-20})$. Substituting these values into the expression yields $\frac{7}{9} \left( 20 - \frac{1}{9}(1 - 10^{-20}) \right)$, which simplifies to $\frac{7}{9} \left( \frac{180 - 1 + 10^{-20}}{9} \right)$. This results in the final value of $\frac{7}{81}(179 + 10^{-20})$.

A telescoping series simplifies significantly because intermediate terms cancel out, leaving only the boundary values. In the expression $\sum_{k=1}^n (k^3 - (k-1)^3)$, expanding the summation produces $(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \dots + (n^3 - (n-1)^3)$. Every positive cube is neutralized by the corresponding negative cube in the subsequent term, reducing the entire expression to $n^3 - 0^3$, which confirms that the identity provided in the second statement is true for any natural number $n$.

The general term of the series, $T_k = (k-1)^2 + (k-1)k + k^2$, represents the expansion of a difference of cubes. Applying the algebraic identity $(k^3 - (k-1)^3) = (k - (k-1))(k^2 + k(k-1) + (k-1)^2)$, the general term simplifies exactly to $k^3 - (k-1)^3$. The series provided runs from $k=1$ to $k=20$, as the final term $361+380+400$ equates to $19^2 + 19(20) + 20^2$, which corresponds to $20^3 - 19^3$. Consequently, the entire sum is equivalent to the evaluation of the telescoping series at $n=20$, resulting in $20^3 = 8000$. Because the series sum is derived directly from the principle established in the second statement, that statement acts as the correct explanation.

This problem is essentially a summation of two distinct financial periods: a constant savings interval followed by an arithmetic progression. For the first three months, the saving amount remains a constant ₹ 200 per month, yielding a fixed total of $200 \times 3 = 600$. Beyond this initial period, the saving behavior shifts into an arithmetic progression where the amount increases by a common difference of 40 each month, starting with a value of 240 in the fourth month.

Subtracting the initial three months' savings from the total goal leaves a remainder of $11040 - 600 = 10440$, which must be covered by the arithmetic progression phase. Letting $m$ represent the number of months in this second phase, the sum of this arithmetic series is determined by the formula:

$\frac{m}{2}[2a + (m-1)d] = 10440$

Substituting the first term $a = 240$ and the common difference $d = 40$ into the equation yields $\frac{m}{2}[480 + (m-1)40] = 10440$, which simplifies to $m(240 + 20m - 20) = 10440$. This results in the quadratic equation $20m^2 + 220m - 10440 = 0$, or more simply, $m^2 + 11m - 522 = 0$. Solving this for the positive root gives $m = 18$ months. Adding the initial three months of service, the total time required is $18 + 3 = 21$.

The total time needed to count 4500 notes is found by splitting the task into the initial constant rate phase and the subsequent decreasing rate phase represented by an arithmetic progression. In the first 10 minutes, the person counts 150 notes each minute, accounting for 1500 notes in total. This leaves 3000 notes to be counted during the remaining time, where the number of notes counted each minute follows an arithmetic progression with a first term of 148 and a common difference of $-2$. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ for the remaining 3000 notes, we set $3000 = \frac{n}{2}[2(148) + (n-1)(-2)]$, which simplifies to $n^2 - 149n + 3000 = 0$. Solving this quadratic equation yields $n=24$ or $n=125$, and since counting for 125 minutes would imply a negative number of notes in later minutes, we determine that the progression lasts for 24 minutes. Adding these 24 minutes to the initial 10 minutes results in a total of 34 minutes.

This series is evaluated by shifting its terms to isolate a predictable geometric progression. Express the total sum as $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ and multiply the entire series by $\frac{1}{3}$ to obtain the shifted sequence: $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots$ Subtracting the second equation from the first isolates a repeating pattern, resulting \in $\frac{2}{3}S = 1 + \left( \frac{2}{3} - \frac{1}{3} \right) + \left( \frac{6}{3^2} - \frac{2}{3^2} \right) + \left( \frac{10}{3^3} - \frac{6}{3^3} \right) + \dots$ which simplifies to $\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$ Combining the first two constants yields $\frac{4}{3}$, and the remaining terms form an infinite geometric series where the first term $a = \frac{4}{3}$ and common ratio $r = \frac{1}{3}$. Applying the infinite \sum formula $\frac{a}{1-r}$, we find $\frac{2}{3}S = \frac{4/3}{1 - 1/3} = \frac{4/3}{2/3} = 2$ which confirms $S = 3$.

In any geometric progression, each term is created by multiplying the previous term by a common ratio, and the fact that terms alternate between positive and negative indicates that this ratio must be negative. The first two terms sum to $a + ar = 12$, while the sum of the third and fourth terms is $ar^2 + ar^3 = 48$, which can be factored as $ar^2(1 + r) = 48$. Dividing the second equation by the first allows the common factor $a(1 + r)$ to cancel out, leaving $r^2 = 4$, which implies $r = -2$ given the alternating sign condition. Substituting this ratio into the first equation $a(1 - 2) = 12$ yields the first term, $-12$.

A geometric progression relies on a fixed common ratio $r$ to generate subsequent terms from the preceding one. Expressing the first three terms as $a$, $ar$, and $ar^2$ leads to the equation $a = ar + ar^2$, which simplifies to $1 = r + r^2$ after dividing by the non-zero first term $a$. Rearranging this into the standard quadratic form $r^2 + r - 1 = 0$ and solving via the quadratic formula yields $r = \frac{-1 \pm \sqrt{5}}{2}$. Since all terms in the series must be positive, the common ratio $r$ must also be positive, identifying $r = \frac{1}{2}(\sqrt{5}-1)$ as the only acceptable value.

The foundation for solving this problem lies in the standard Maclaurin series expansion for the exponential function, which is defined as $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. By substituting $x = -1$ into this identity, we obtain the expression $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$. Because the first two terms cancel out, the remaining infinite series simplifies directly to $e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$, which identifies the sum as $e^{-1}$.

When terms $a_1, a_2, \ldots, a_n$ form a Harmonic Progression, their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \ldots, \frac{1}{a_n}$ form an Arithmetic Progression where the difference between consecutive terms is a constant value $d = \frac{1}{a_{k+1}} - \frac{1}{a_k}$. Each individual product in the sequence can be rewritten as $a_k a_{k+1} = \frac{a_k - a_{k+1}}{d}$, transforming the entire sum into a telescoping series that collapses into the form $\frac{a_1 - a_n}{d}$. Given that the $n$-th term of the associated Arithmetic Progression is defined as $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$, we can solve for $d$ to find that $(n-1)d = \frac{a_1 - a_n}{a_1 a_n}$. Substituting this identity back into the summation result yields the final expression $(n-1) a_1 a_n$.

The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. When we have a ratio of the sum of $p$ terms to the sum of $q$ terms, we can write the expression as:

$\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^2}{q^2}$

By canceling the common factor of $\frac{1}{2}$ and dividing both sides by the ratio $\frac{p}{q}$, the equation simplifies to:

$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$

Dividing the numerator and denominator on the left by $2$ produces the standard form for individual terms $a_n = a_1 + (n-1)d$:

$\frac{a_1 + \frac{p-1}{2}d}{a_1 + \frac{q-1}{2}d} = \frac{p}{q}$

To determine the ratio of the sixth term to the twenty-first term, we set the term indices equal to the coefficients of $d$ within the formula. Specifically, setting $\frac{p-1}{2} = 6-1$ gives $p=11$, and setting $\frac{q-1}{2} = 21-1$ gives $q=41$. Substituting these values into the ratio yields $\frac{a_6}{a_{21}} = \frac{11}{41}$.

The series $1 + \frac{1}{4 \cdot 2!} + \frac{1}{16 \cdot 4!} + \frac{1}{64 \cdot 6!} + \dots$ represents the Maclaurin expansion of the hyperbolic cosine function, which follows the general form $\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots$. By inspecting the given terms, it becomes evident that the series matches this expansion when $x$ is set to $1/2$, as the powers of $1/2$ produce the denominators $1/4, 1/16,$ and $1/64$. Substituting this value into the identity results in the expression $\frac{e^{1/2} + e^{-1/2}}{2}$, which is equivalent to $\frac{\sqrt{e} + 1/\sqrt{e}}{2}$. Multiplying the numerator and denominator by $\sqrt{e}$ simplifies the fraction to $\frac{e+1}{2\sqrt{e}}$.

The infinite sum of a geometric series with a common ratio $|r|<1$ is expressed as $\frac{1}{1-r}$, which allows us to identify $x = \frac{1}{1-a}$, $y = \frac{1}{1-b}$, and $z = \frac{1}{1-c}$. Rearranging these equations to express the common ratios in terms of the sums gives $a = 1 - \frac{1}{x}$, $b = 1 - \frac{1}{y}$, and $c = 1 - \frac{1}{z}$. Since $a, b, c$ form an arithmetic progression, they must satisfy the condition $2b = a + c$, which becomes $2(1 - \frac{1}{y}) = (1 - \frac{1}{x}) + (1 - \frac{1}{z})$ upon substitution. Simplifying this result leads to
$\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$
This final relation is the standard definition for $x, y, z$ to form a harmonic progression.

The exponential function $e^x$ is defined by its Maclaurin series expansion, which provides the identity $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdot s$ alongside $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdot s$. Adding these two series cancels out the odd-denominator terms, resulting in the equality:
$e + e^{-1} = 2\left(1 + \frac{1}{2!} + \frac{1}{4!} + \cdot s\right)$
Isolating the requested series requires dividing by 2 and subtracting 1 from the result, yielding $\frac{e + e^{-1}}{2} - 1 = \frac{e^2 + 1 - 2e}{2e}$, which simplifies to $\frac{(e-1)^2}{2e}$.

The sum of an odd number of terms is efficiently determined by splitting the sequence into an even component consisting of the first $n-1$ terms and an independent final $n$-th term, which is $n^2$. Since the provided formula for an even number of terms, $k$, is $\frac{k(k+1)^2}{2}$, we can directly substitute $k = n-1$ to find the sum of all terms preceding the last one.

Substituting $n-1$ into the given expression yields $\frac{(n-1)n^2}{2}$. Adding the value of the final $n$-th term, $n^2$, completes the calculation:
$\frac{(n-1)n^2}{2} + n^2$
By factoring out $\frac{n^2}{2}$, we combine the remaining values to reach $\frac{n^2(n-1+2)}{2}$, which simplifies directly to $\frac{n^2(n+1)}{2}$.

Any term in an arithmetic progression follows the linear structure $T_r = a + (r-1)d$, where $a$ is the starting value and $d$ is the constant difference. By establishing the equations $a + (m-1)d = \frac{1}{n}$ and $a + (n-1)d = \frac{1}{m}$, we subtract the second from the first to obtain $(m-n)d = \frac{1}{n} - \frac{1}{m}$, which simplifies to $(m-n)d = \frac{m-n}{mn}$ and confirms $d = \frac{1}{mn}$. Substituting this value back into the first equation yields $a + \frac{m-1}{mn} = \frac{1}{n}$, which simplifies to $a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n}$ and reveals that $a = \frac{1}{mn}$. Consequently, the difference $a - d$ results in $\frac{1}{mn} - \frac{1}{mn} = 0$.