Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Because the diagonal of a square is $\sqrt{2}$ times its side length, the condition that the side of $A_n$ equals the diagonal of $A_{n+1}$ implies that each subsequent square has a side length equal to the previous side length divided by $\sqrt{2}$. This creates a geometric progression where the side length of the $n$th square is $s_n = \frac{12}{(\sqrt{2})^{n-1}}$.

Squaring this side length gives the area of $A_n$ as $\frac{144}{2^{n-1}}$, and we seek the smallest integer $n$ such that $\frac{144}{2^{n-1}} < 1$. Multiplying both sides by $2^{n-1}$ results in the inequality $144 < 2^{n-1}$. Since $2^7 = 128$ and $2^8 = 256$, the term $n-1$ must be at least 8 to exceed 144, which implies that $n = 9$.

This series is a classic example of a telescoping sum where each term is engineered to simplify the next. Adding $\frac{1}{1-x}$ to the total sum $S$ allows us to consolidate the first two terms:
$\frac{1}{1-x} + \frac{1}{1+x} = \frac{2}{1-x^2}$
Because $\frac{2}{1-x^2} + \frac{2}{x^2+1}$ further simplifies to $\frac{4}{1-x^4}$, a clear pattern emerges where the sum of any two adjacent components consolidates into the next power of two. By extending this logic across the entire sequence up to the $100^{th}$ term, the expression condenses into:
$S + \frac{1}{1-x} = \frac{2^{101}}{1-x^{2^{101}}}$
Rearranging to solve for $S$ gives $S = \frac{1}{x-1} + \frac{2^{101}}{1-x^{2^{101}}}$. Substituting $x=2$ leads to $1 + \frac{2^{101}}{1-2^{2^{101}}}$. Since $2^{2^{101}}$ is equivalent to $(2^2)^{101}$, or $4^{101}$, the final value is $1 - \frac{2^{101}}{4^{101}-1}$.

This problem relies on the method for solving an arithmetico-geometric series, where successive terms are generated by the product of terms from an arithmetic progression and a geometric progression. To evaluate $S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \cdot s$, begin by shifting the sequence by dividing the entire series by $5$ to obtain $\frac{S}{5} = \frac{7}{5^2} + \frac{9}{5^3} + \frac{13}{5^4} + \cdot s$. Subtracting this shifted equation from the original yields $\frac{4S}{5} = \frac{7}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \cdot s$. To simplify further, define the latter portion as $K = \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \cdot s$ and repeat the shifting process by dividing $K$ by $5$ to get $\frac{K}{5} = \frac{2}{5^3} + \frac{4}{5^4} + \cdot s$. Subtracting these two expressions for $K$ produces $\frac{4K}{5} = \frac{2}{5^2} + \frac{2}{5^3} + \frac{2}{5^4} + \cdot s$, which is a standard infinite geometric series. Calculating the sum of this geometric series results in $\frac{2/25}{1 - 1/5} = \frac{1}{10}$, implying that $K = \frac{1}{8}$. Substituting this back into the expression $\frac{4S}{5} = \frac{7}{5} + K$ gives $\frac{4S}{5} = \frac{7}{5} + \frac{1}{8} = \frac{61}{40}$, which simplifies to $S = \frac{61}{32}$. Finally, the value of $160 S$ is calculated as $160 \times \frac{61}{32} = 305$.

Telescoping sums provide an efficient way to evaluate series where each term can be decomposed into a difference of two values, allowing consecutive terms to cancel each other out. Each entry in this sequence follows the algebraic form $\frac{1}{(2r+1)^{2}-1}$, where the variable $r$ takes integer values from $1$ to $100$ to cover the range from the first term, $\frac{1}{3^{2}-1}$, to the last term, $\frac{1}{(201)^{2}-1}$. Expanding the denominator as $(2r+1)^{2}-1 = 4r^{2}+4r+1-1$ simplifies it to $4r(r+1)$, turning the general term into $\frac{1}{4r(r+1)}$.

Applying partial fraction decomposition, the expression $\frac{1}{4r(r+1)}$ becomes $\frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} \right)$. When we sum these fractions from $r=1$ to $100$, the subtraction of each fraction is immediately offset by the addition of that same value in the subsequent term, creating a cascading cancellation throughout the series. This leaves only the first term's initial value, $1$, and the final term's subtracted value, $\frac{1}{101}$. Multiplying the remaining difference of $1 - \frac{1}{101}$ by the common factor of $\frac{1}{4}$ yields $\frac{1}{4} \times \frac{100}{101}$, which simplifies to $\frac{25}{101}$.

The fundamental principle required here is the logarithmic change-of-base property, which states that $\log_{b^k} x = \frac{1}{k} \log_b x$. By applying this identity to transform the base of each term into a power of 9, the series simplifies to $2 \log_9 x + 3 \log_9 x + 4 \log_9 x + \dots$, where the coefficients follow an arithmetic progression.

With 21 terms in the sequence, the coefficients increase from 2 to 22. Factoring out $\log_9 x$ allows us to sum these integers as $\frac{21}{2}(2 + 22) = 252$. Setting the total sum of the series equal to 504 gives the equation $252 \log_9 x = 504$, which simplifies to $\log_9 x = 2$. Solving for $x$ yields $x = 9^2 = 81$.

When dealing with three terms in a geometric progression, representing them as $\frac{a}{r}, a, ar$ is efficient because their product cancels the common ratio. Given that the product is 27, the relationship $a^3 = 27$ holds, which simplifies to $a = 3$. The sum of these three terms is therefore defined by the expression $S = \frac{3}{r} + 3 + 3r$, which rearranges to $S = 3(r + \frac{1}{r}) + 3$.

The constraints on $S$ depend on the sign of $r$. For positive values of $r$, the arithmetic mean-geometric mean inequality confirms that $r + \frac{1}{r} \ge 2$, and substituting this into the sum expression yields $S \ge 3(2) + 3 = 9$. When $r$ is negative, the term $r + \frac{1}{r}$ is equivalent to $-(|r| + \frac{1}{|r|})$. Since the sum of any positive number and its reciprocal is at least 2, it follows that $r + \frac{1}{r} \le -2$, leading to $S \le 3(-2) + 3 = -3$. Consequently, all valid values for $S$ are contained within the interval $(-\infty , -3] \cup [9, \infty )$.

In any geometric progression with common ratio $r$, each term $a_{n+1}$ is defined as $r \cdot a_n$, establishing a direct scalar relationship between sums of terms with consecutive indices. Given that $\sum_{n=1}^{100} a_{2n+1} = 200$ and $\sum_{n=1}^{100} a_{2n} = 100$, we can express the first sum as $\sum_{n=1}^{100} r \cdot a_{2n}$, which factors to $r \sum_{n=1}^{100} a_{2n}$. Equating $200 = r \cdot 100$ determines that the common ratio is 2.

The sum $a_2 + a_3 + \dots + a_{201}$ is equivalent to the sum of the even-indexed terms and the odd-indexed terms, which is $100 + 200 = 300$. Because this sum is also equal to $r \sum_{n=1}^{200} a_n$, we have the relationship $r \sum_{n=1}^{200} a_n = 300$. Substituting $r = 2$ into this equation gives $2 \sum_{n=1}^{200} a_n = 300$, resulting in a final value of 150.

In any geometric progression, terms are related by a constant ratio, and grouping consecutive terms helps simplify complex expressions into manageable equations. We are given the sum of the second, third, and fourth terms as $ar + ar^2 + ar^3 = 3$, and the sum of the sixth, seventh, and eighth terms as $ar^5 + ar^6 + ar^7 = 243$. By factoring $r^4$ out of the second expression, we observe that $r^4(ar + ar^2 + ar^3) = 243$. Substituting the value of the first sum into this equality gives $3r^4 = 243$, which simplifies to $r^4 = 81$, yielding a positive common ratio of $r = 3$.

With the ratio established, we substitute $r = 3$ into the first equation to find $a(3 + 9 + 27) = 3$, or $39a = 3$, which simplifies to $a = \frac{1}{13}$. Using the sum formula for a geometric progression, $S_n = \frac{a(r^n - 1)}{r - 1}$, for $n = 50$, we calculate $S_{50} = \frac{\frac{1}{13}(3^{50} - 1)}{3 - 1}$. This leads directly to the result of $\frac{1}{26}(3^{50} - 1)$.

When dealing with five numbers in an arithmetic progression, it is most efficient to define them symmetrically as $a-2d$, $a-d$, $a$, $a+d$, and $a+2d$, where $a$ represents the middle term and $d$ is the common difference. Summing these terms immediately yields $5a = 25$, confirming the middle term is 5. Utilizing the product condition, we calculate $(a-2d)(a+2d)(a-d)(a+d)a = 2520$. Substituting $a=5$ into this product results in $(25-4d^2)(25-d^2) \times 5 = 2520$, which simplifies to the equation $(25-4d^2)(25-d^2) = 504$.

Expanding the expression leads to the quadratic equation in terms of $d^2$, specifically $4d^4 - 125d^2 + 121 = 0$. Factoring this equation reveals two possible scenarios for the square of the common difference: $d^2 = 1$ or $d^2 = \frac{121}{4}$. Testing the first case, $d = \pm 1$ generates sequences where no term equals $-\frac{1}{2}$, effectively eliminating it from consideration. The second case, where $d = \pm \frac{11}{2}$, yields the arithmetic progression $-6$, $-\frac{1}{2}$, 5, $\frac{21}{2}$, 16. Comparing these values, we identify 16 as the largest number.

The behavior of an Arithmetic Progression is entirely governed by its first term and common difference, where any term at index $n$ is calculated using the general formula
$a_n = a + (n-1)d$
We determine the parameters of the first sequence by comparing the fortieth and hundredth terms. Subtracting the equation $a + 39d = -159$ from $a + 99d = -399$ results in $60d = -240$, confirming that the common difference is $d = -4$. Plugging this value into the expression for the fortieth term, $a + 39(-4) = -159$, reveals the first term is $a = -3$.

With the first sequence defined, we find the seventieth term by calculating $a + 69d = -3 + 69(-4)$, which equals $-279$. This term is equivalent to the hundredth term of the second progression, $b_{100}$. Because the second progression has a common difference two units larger than the first, its common difference is $-4 + 2 = -2$. Using the general term formula for the hundredth term of the second sequence, we set up the equation
$b_1 + 99(-2) = -279$
Solving this equation leads to $b_1 - 198 = -279$, which indicates that $b_1 = -81$.

Recognizing that each set of parentheses follows the algebraic identity for the sum of a geometric progression, the infinite series can be written as the sum of terms where the $n$-th term is $\frac{x^{n+1}-y^{n+1}}{x-y}$. Factoring out the constant $\frac{1}{x-y}$, the expression transforms into:
$\frac{1}{x-y} \left( \sum_{n=1}^{\infty } x^{n+1} - \sum_{n=1}^{\infty } y^{n+1} \right)$
Because the conditions $|x|<1$ and $|y|<1$ are met, both infinite sums converge using the standard formula $\sum_{k=2}^{\infty } r^k = \frac{r^2}{1-r}$, yielding $\frac{1}{x-y} \left( \frac{x^2}{1-x} - \frac{y^2}{1-y} \right)$.

Simplifying the numerator within the parentheses requires finding a common denominator to reach $\frac{x^2(1-y) - y^2(1-x)}{(1-x)(1-y)}$, which expands to $\frac{x^2-x^2y-y^2+xy^2}{(1-x)(1-y)}$. Rearranging the numerator terms as $(x^2-y^2) - xy(x-y)$ and using the difference of squares identity allows us to factor out $(x-y)$ from the numerator. Once this factor cancels with the $(x-y)$ in the denominator, the entire expression simplifies to $\frac{x+y-xy}{(1-x)(1-y)}$.

Recognizing that the series consists of alternating terms reveals two distinct arithmetic progressions hidden within. By grouping every other number, we create one sequence starting at 3 and another starting at 4, each with a common difference of 5, which allows us to calculate their totals separately and simplify the summation process.

Since there are 40 terms in total, each separate progression contains 20 terms. Applying the arithmetic series formula $S = \frac{n}{2}[2a + (n-1)d]$ to the first sequence with $a = 3$ and $d = 5$ gives a total of $10(6 + 95)$, while the second sequence with $a = 4$ and $d = 5$ results in $10(8 + 95)$. Combining these sums yields $10(101 + 103)$, which simplifies to $10 \times 204$, or 2040. Equating this result to the expression $(102)m$ allows us to solve for $m$ by calculating $\frac{2040}{102}$, resulting in $m = 20$.

Since the three expressions are in an arithmetic progression, the middle term $f(x)$ is the arithmetic mean of the other two, leading to the relation:

$f(x) = \frac{2(2^x + 2^{-x}) + (3^x + 3^{-x})}{2}$

The arithmetic mean-geometric mean inequality guarantees that for any positive base $a$, the sum $a^x + a^{-x}$ is at least 2 for all real numbers, with the equality achieved precisely at $x = 0$. Evaluating the function at $x = 0$, we substitute the minimum value of 2 into the numerator to obtain 3 as the result.

An arithmetic progression is defined by two fundamental variables: the first term $a$ and the common difference $d$. The terms $T_{10} = a + 9d = \frac{1}{20}$ and $T_{20} = a + 19d = \frac{1}{10}$ establish a system of linear equations that allow us to isolate these parameters. Subtracting the first equation from the second eliminates $a$, leaving $10d = \frac{1}{10} - \frac{1}{20}$, which simplifies to $10d = \frac{1}{20}$, meaning $d = \frac{1}{200}$. Plugging this value of $d$ back into the equation for the 10th term results in $a + \frac{9}{200} = \frac{10}{200}$, confirming that $a = \frac{1}{200}$. With $a$ and $d$ determined, we apply the summation formula $S_n = \frac{n}{2}[2a + (n-1)d]$ for the first 200 terms. This yields $S_{200} = \frac{200}{2} \left[ \frac{2}{200} + \frac{199}{200} \right]$, which simplifies to $100 \times \frac{201}{200}$, resulting in a final value of $100\frac{1}{2}$.

The $n$-th term of an arithmetic progression is defined by the relationship $a_n = a_1 + (n-1)d$. By substituting the provided values $a_1 = 1$ and $a_n = 300$, we obtain the equation $(n-1)d = 299$. Because the common difference $d$ must be an integer, $n-1$ must be a factor of $299$, which breaks down into the prime factors $13 \times 23$. Given the restriction that $15 \le n \le 50$, we evaluate the divisors of $299$ to find a suitable $n$. Setting $n-1 = 23$ results in $n = 24$, which fits within the required interval and implies a common difference of $d = 13$.

With the number of terms established as $n = 24$, the requested values become $a_{20}$ and $S_{20}$. Calculating the specific term $a_{20}$ involves applying the formula $a_{20} = a_1 + 19d$, which yields $1 + 19(13) = 1 + 247 = 248$. To find the sum of these terms, we use the formula $S_{k} = \frac{k}{2}(a_1 + a_k)$ for $k=20$, resulting in $S_{20} = \frac{20}{2}(1 + 248) = 10 \times 249 = 2490$. Combining these results provides the final ordered pair $(2490, 248)$.

The logarithmic identity $\log_{a^n} b = \frac{1}{n} \log_a b$ simplifies each term of the series by moving the exponent of the base to the front as a multiplier. Applying this property to the first 20 terms transforms the expression $\log_{7^{1/2}} x + \log_{7^{1/3}} x + \dots + \log_{7^{1/21}} x$ into a sum of $2 \log_7 x + 3 \log_7 x + \dots + 21 \log_7 x$. Factoring out $\log_7 x$, the expression becomes the product of $\log_7 x$ and the sum of integers from 2 to 21, which must equal 460. Since the sum of these integers is 230, the relationship is expressed as follows:

$230 \log_7 x = 460$

Dividing both sides by 230 results in $\log_7 x = 2$, which means $x = 7^2$, or 49.

This problem is solved by recognizing that the series consists of an initial constant term followed by ten distinct terms that can be aggregated using summation notation. There are eleven ones in the expression, contributing to a constant value of 11, while the remaining parts form a sum of products of squares and linear factors. By grouping these as $11 - \sum_{r=1}^{10} (2r)^2(2r-1)$, we can evaluate the summation systematically.

Factoring out a constant 4 leaves us with $4 \sum_{r=1}^{10} (2r^3 - r^2)$, and utilizing standard summation formulas for $n=10$ provides $4(2 \times 3025 - 385)$. This simplifies to $4(5665)$, resulting in 22660, so the expression becomes $11 - 22660$. Matching this to $\alpha - 220 \beta$ and dividing 22660 by 220 confirms that the ordered pair $(\alpha, \beta)$ is $(11, 103)$.

The arithmetic mean-geometric mean inequality provides a direct method for establishing the minimum value of expressions involving sums of positive terms, asserting that the average of two quantities is at least their geometric mean. Applying this to the given function yields $\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$, which simplifies to $2^{\sin x} + 2^{\cos x} \geq 2^{1 + \frac{\sin x + \cos x}{2}}$. Since $\sin x + \cos x$ has a minimum value of $-\sqrt{2}$, the exponent reaches a minimum value of $-\frac{1}{\sqrt{2}}$. Substituting this into the inequality, we find the overall minimum value of the expression is $2^{1 - \frac{1}{\sqrt{2}}}$.

In a geometric progression, each term is the product of the previous term and a common ratio $r$, which implies that any pair of consecutive terms $a_{n} + a_{n+1}$ relates to the subsequent pair $a_{n+2} + a_{n+3}$ by the factor $r^2$. Given the equations $a_1 + a_2 = 4$ and $a_3 + a_4 = 16$, we can express the second equation as $r^2(a_1 + a_2) = 16$. Substituting the known sum, we get $r^2(4) = 16$, which simplifies to $r^2 = 4$, giving $r = 2$ or $r = -2$. To determine the correct ratio, we use the first expression $a_1(1 + r) = 4$. If $r$ were 2, then $a_1$ would be $4/3$, which contradicts the condition that $a_1 < 0$. Therefore, we must have $r = -2$, which leads to $a_1(1 - 2) = 4$, or $a_1 = -4$.

To find the value of $\lambda$, calculate the sum of the first nine terms using the formula $S_n = a_1 \frac{r^n - 1}{r - 1}$. Substituting the determined values results in $S_9 = -4 \frac{(-2)^9 - 1}{-2 - 1}$. This simplifies to $S_9 = -4 \frac{-512 - 1}{-3}$, which is $-4 \frac{-513}{-3}$. Dividing the terms gives $S_9 = -4(171)$, which equals -684. Given that this sum is defined as $4\lambda$, we have $4\lambda = -684$, which yields $\lambda = -171$.

The series $2^{10} + 2^9 \cdot 3^1 + 2^8 \cdot 3^2 + \dots + 3^{10}$ follows the pattern of a geometric progression where the first term is $2^{10}$ and the common ratio is $3/2$. By observing the exponents of the base 2 decreasing from 10 to 0 while the base 3 increases from 0 to 10, we can determine there are 11 terms in total.

Applying the standard sum formula for a geometric progression, $S_n = a \frac{r^n - 1}{r - 1}$, we calculate the sum as:

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

Simplifying the denominator to $1/2$ causes the $2^{10}$ term to be multiplied by 2, becoming $2^{11}$. The expression then expands to:

$2^{11} \left( \frac{3^{11}}{2^{11}} - 1 \right) = 3^{11} - 2^{11}$

Since this sum is equivalent to the expression $S - 2^{11}$ provided in the problem, equating the two results clearly identifies $S$ as $3^{11}$.