Sequences And Series – JEE Main MathematicsPractice Questions & PYQs

Modular arithmetic provides an efficient way to find the remainder of a high-exponent expression by reducing the base to a value congruent to the divisor. Recognizing that $7^2 = 49$, and that 49 is two units less than the nearest multiple of 17, which is $3 \times 17 = 51$, we can effectively work with the equivalent value of $-2$ modulo 17.

Writing the original expression as $7 \cdot (7^2)^{51}$, we substitute $-2$ for $7^2$ to get $7 \cdot (-2)^{51}$, which simplifies to $-7 \cdot 2^{51}$. To evaluate $2^{51}$ efficiently, we observe that $2^4 = 16$, which is $-1$ modulo 17, allowing us to express $2^{51}$ as $(2^4)^{12} \cdot 2^3$. This yields $(-1)^{12} \cdot 8 = 8$, so the overall expression becomes $-7 \cdot 8 = -56$. Finally, adding 68 to -56 brings the value into the positive range, resulting in a remainder of 12.

Geometric progressions are defined by the nth term formula $T_n = ar^{n-1}$ and the sum to infinity formula $S_\infty = \frac{a}{1-r}$. Given two sequences $\{a_k\}$ and $\{b_k\}$ starting at $4$, the combined sequence $c_k = a_k + b_k$ corresponds to $c_k = 4r_1^{k-1} + 4r_2^{k-1}$. Substituting the given values $c_2 = 4r_1 + 4r_2 = 5$ and $c_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4}$ simplifies to $r_1 + r_2 = \frac{5}{4}$ and $r_1^2 + r_2^2 = \frac{13}{16}$.

The identity $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2$ allows us to find the product of the ratios, where $2r_1r_2 = (\frac{5}{4})^2 - \frac{13}{16} = \frac{25}{16} - \frac{13}{16} = \frac{12}{16} = \frac{3}{4}$, resulting in $r_1r_2 = \frac{3}{8}$. These ratios are the roots of the quadratic equation $t^2 - \frac{5}{4}t + \frac{3}{8} = 0$, which factors to $(t - \frac{1}{2})(t - \frac{3}{4}) = 0$. Given the condition $r_1 < r_2$, we determine $r_1 = \frac{1}{2}$ and $r_2 = \frac{3}{4}$.

The sum of the infinite series $\sum_{k=1}^{\infty } c_k$ is calculated as the sum of two individual infinite geometric series, $\frac{4}{1 - 1/2} + \frac{4}{1 - 3/4} = 8 + 16 = 24$. Calculating the individual terms $12a_6 = 12 \cdot 4(\frac{1}{2})^5 = \frac{48}{32} = \frac{3}{2}$ and $8b_4 = 8 \cdot 4(\frac{3}{4})^3 = 32(\frac{27}{64}) = \frac{27}{2}$, the total deduction is $\frac{3}{2} + \frac{27}{2} = 15$. Subtracting this from the infinite sum yields $24 - 15 = 9$.

When several arithmetic progressions share common terms, those terms themselves form a new arithmetic progression characterized by a common difference equal to the least common multiple of the differences of the original sequences. Given the common differences are 4, 3, and 5, calculating $\operatorname{LCM}(4, 3, 5)$ results in 60, which serves as the common difference for our new sequence. By observing the initial terms, we find that the first value common to all three progressions is 47. We generate subsequent common terms by repeatedly adding 60 to this starting value, resulting in 47, 107, and 167. Each common term must also fall within the range of every original progression, meaning no term can exceed the smallest upper boundary of the three given sets, which is 197. Since 167 is the last term below this threshold-as the next term would be 227-we stop there. Summing these three values, $47 + 107 + 167$, yields 321.

Subtracting the sum of terms that meet a specific condition from the total sum of an arithmetic progression yields the sum of terms that do not meet that condition. The series $3, 8, 13, \dots, 373$ has a first term $a = 3$, a common difference $d = 5$, and a last term $l = 373$. Using the formula $l = a + (n - 1)d$, the total number of terms is determined by $373 = 3 + (n - 1)5$, which results in $n = 75$. The total sum of these terms is:
$S = \frac{75}{2}(3 + 373) = 14100$
The terms within this progression that are divisible by $3$ form a new sequence: $3, 18, 33, \dots, 363$. This sub-sequence has a first term of $3$ and a common difference of $15$. Solving $363 = 3 + (k - 1)15$ shows there are $k = 25$ such terms. The sum of these divisible terms is:
$S_{div} = \frac{25}{2}(3 + 363) = 4575$
The final result is obtained by subtracting the sum of the divisible terms from the total sum, resulting in $14100 - 4575 = 9525$.

In any binomial expansion $(1+x)^N$, the ratio of consecutive coefficients is governed by the property $\frac{{}^{N}C_{k}}{{}^{N}C_{k-1}} = \frac{N-k+1}{k}$. Given the expansion $(1+x)^{n+2}$, we denote the three consecutive coefficients as ${}^{n+2}C_{r-1}$, ${}^{n+2}C_{r}$, and ${}^{n+2}C_{r+1}$ with the ratio $1 : 3 : 5$.

By setting the ratio of the first two terms as $\frac{{}^{n+2}C_{r}}{{}^{n+2}C_{r-1}} = 3$, we apply the property to get $\frac{(n+2)-r+1}{r} = 3$, which simplifies to $n-r+3 = 3r$, or $n = 4r-3$. For the second and third terms, the ratio $\frac{{}^{n+2}C_{r+1}}{{}^{n+2}C_{r}} = \frac{5}{3}$ yields $\frac{(n+2)-(r+1)+1}{r+1} = \frac{5}{3}$, which simplifies to $\frac{n+2-r}{r+1} = \frac{5}{3}$, or $3n = 8r-1$. Substituting $n = 4r-3$ into the latter equation results in $3(4r-3) = 8r-1$, which simplifies to $12r-9 = 8r-1$, ultimately giving $r = 2$ and $n = 5$. With the expansion now defined as $(1+x)^7$, the three consecutive coefficients are ${}^7C_1$, ${}^7C_2$, and ${}^7C_3$. Calculating these individual values results in $7$, $21$, and $35$, respectively, and their total sum is $7 + 21 + 35 = 63$.

Identifying the pattern in the series $5, 11, 19, 29, 41, \dots$ reveals that the first differences are $6, 8, 10, 12$, and the second differences are constant at $2$. Since the second difference is constant, the general term $T_n$ follows a quadratic expression, which determines to be $T_n = n^2 + 3n + 1$.

Calculating the sum of the first 20 terms is achieved by applying summation properties to this expression, represented as $\sum_{n=1}^{20} (n^2 + 3n + 1)$. Breaking this down gives $\sum_{n=1}^{20} n^2 + 3 \sum_{n=1}^{20} n + \sum_{n=1}^{20} 1$, which equates to $\frac{20 \cdot 21 \cdot 41}{6} + 3 \cdot \frac{20 \cdot 21}{2} + 20$. Evaluating these individual components results in $2870 + 630 + 20$, which produces a final total of $3520$.

Any Arithmetic Progression is fully defined by its initial term and common difference, which allows the sum of any subset of its terms to be expressed as a linear algebraic equation. Given the first term is 8, the sum of the initial four terms is written as $4a_1 + 6d = 50$. Substituting the known first term results in $32 + 6d = 50$, which simplifies to $6d = 18$, establishing a common difference of $d = 3$.

The total number of terms in the progression is found by summing the last four terms, represented as $a_{n-3} + a_{n-2} + a_{n-1} + a_n = 170$. Using the general formula for the $k$-th term, $a_k = a_1 + (k-1)d$, this summation translates to $4a_1 + (4n - 10)d = 170$. Inserting $a_1 = 8$ and $d = 3$ leads to $32 + (4n - 10) \times 3 = 170$, which simplifies to $32 + 12n - 30 = 170$, or $12n = 168$, confirming the sequence contains 14 terms.

Identifying the middle values involves locating the 7th and 8th terms of this 14-term sequence. Calculating these gives $a_7 = 8 + 6 \times 3 = 26$ and $a_8 = 8 + 7 \times 3 = 29$. Multiplying these middle values produces a final result of 754.

Decomposing the general term into partial fractions and utilizing the properties of a telescoping series allows one to calculate the sum of such a sequence efficiently. Factoring the denominator of the expression $a_n = \frac{-2}{4n^2-16n+15}$ as $(2n-3)(2n-5)$ permits us to rewrite each term as the difference of two fractions: $\frac{1}{2n-3} - \frac{1}{2n-5}$.

When calculating the sum from $n=1$ to $n=25$, the terms cancel in a chain where the subtrahend of one term matches the minuend of the subsequent term. This reduces the entire summation to the difference between the final remaining value and the initial negative value, which is $\frac{1}{47} - \frac{1}{-3}$. Combining these fractions yields $\frac{1}{47} + \frac{1}{3}$, which results in $\frac{50}{141}$.

The average of the first $K$ integers, defined as $S_K$, simplifies to $\frac{K+1}{2}$ when applying the sum formula $\frac{K(K+1)}{2}$. Squaring each term gives $S_j^2 = \frac{j^2+2j+1}{4}$. Summing this expression from $j = 1$ to $n$ yields $\frac{1}{4} \sum_{j=1}^n (j^2+2j+1)$, which expands to $\frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + n(n+1) + n \right]$. Simplifying the bracketed expression and factoring out $\frac{n}{24}$ results in $\frac{n}{24} (2n^2 + 9n + 13)$. By equating this to the given expression $\frac{n}{A} (B n^2 + C n + D)$, the values are determined as $A = 24$, $B = 2$, $C = 9$, and $D = 13$. Testing the options with these results reveals that $A+B = 26$ is divisible by $D = 13$, confirming the valid relationship.

Optimizing the product of terms in an arithmetic progression involves expressing the product as a quadratic function of the common difference, then applying the derivative to locate the minimum value. With the sequence parameters fixed, the number of terms $n$ is found by solving the sum equation $S_n=0$, allowing for the final arithmetic evaluation.

Given the seventh term $a+6d=3$, we write the first term as $a=3-6d$. The product $a_1 a_4 = a(a+3d)$ becomes $(3-6d)(3-3d)$, expanding to $18d^2-27d+9$. Setting the derivative $36d-27$ equal to zero reveals $d=\frac{3}{4}$, which implies $a=3-6(\frac{3}{4})=-\frac{3}{2}$. Applying these values to the sum formula $S_n = \frac{n}{2}[2a+(n-1)d]=0$, we get $-3+(n-1)\frac{3}{4}=0$, which determines that $n=5$. Calculating the final expression $n!-4a_{n(n+2)}$ where $n=5$ requires finding the 35th term $a_{35}=a+34d$. Substituting the known values gives $120-4(-\frac{3}{2}+34(\frac{3}{4}))$, which simplifies to $120-4(24)$, resulting in 24.

Determining the fractional part of a value $\frac{N}{D}$ relies on identifying the remainder left when the numerator $N$ is divided by the denominator $D$. Any fraction of this form can be expressed as an integer plus $\frac{r}{D}$, where $r$ is that specific remainder. For the expression $\frac{4^{2022}}{15}$, the goal is to find the remainder when $4^{2022}$ is divided by $15$.

Powers of $4$ behave predictably under modulo $15$ arithmetic. Specifically, $4^1 = 4$, and $4^2 = 16$, which is $1 \pm od{15}$ because $16$ divided by $15$ leaves a remainder of $1$. Since $4^{2022}$ can be written as $(4^2)^{1011}$, it is equivalent to $16^{1011}$. Because $16$ is congruent to $1 \pm od{15}$, raising $16$ to any integer power results in a value that is also $1 \pm od{15}$. Therefore, $4^{2022}$ leaves a remainder of $1$ when divided by $15$, making the resulting fractional part $\frac{1}{15}$.

The core concept for solving this series relies on the property of telescoping sums, where the general term is decomposed into the difference of two values, causing intermediate terms to cancel out. The denominator of the general term, $r^4+r^2+1$, can be factored into $(r^2+r+1)(r^2-r+1)$. By writing the numerator $r$ as half the difference between these two factors, specifically $\frac{1}{2}[(r^2+r+1)-(r^2-r+1)]$, the general term $T_r$ transforms into a form suitable for telescoping:

$T_r = \frac{1}{2} \left[ \frac{1}{r^2-r+1} - \frac{1}{r^2+r+1} \right]$

Substituting values from $r=1$ up to $r=10$ into this decomposed expression creates a chain of subtractions. The first term $T_1$ becomes $\frac{1}{2}(1 - \frac{1}{3})$, the second term $T_2$ is $\frac{1}{2}(\frac{1}{3} - \frac{1}{7})$, and this pattern continues sequentially. By the time we reach the tenth term, $T_{10}$ is expressed as $\frac{1}{2}(\frac{1}{91} - \frac{1}{111})$.

When these terms are added together, the negative portion of each term cancels out the positive portion of the subsequent term. This leaves only the first value from the first term and the final value from the last term. Consequently, the entire sum simplifies to $\frac{1}{2}(1 - \frac{1}{111})$, which resolves to $\frac{55}{111}$.

Representing four positive consecutive terms of a geometric progression as $a, ar, ar^2, ar^3$ allows us to utilize the properties of products and sums. Given the product $a^4r^6 = 1296$, taking the square root reveals $a^2r^3 = 36$, which implies $a = 6r^{-3/2}$. Substituting this expression into the sum $a(1+r+r^2+r^3) = 126$ leads to $\frac{6}{r^{3/2}}(1+r+r^2+r^3) = 126$, which simplifies to $r^{-3/2} + r^{-1/2} + r^{1/2} + r^{3/2} = 21$.

Grouping the terms as $(r^{3/2} + r^{-3/2}) + (r^{1/2} + r^{-1/2}) = 21$ allows for a substitution. By letting $x = r^{1/2} + r^{-1/2}$, we note that $x^3 = r^{3/2} + 3(r^{1/2} + r^{-1/2}) + r^{-3/2}$, which means $r^{3/2} + r^{-3/2} = x^3 - 3x$. Substituting these into the simplified sum equation gives $(x^3 - 3x) + x = 21$, or $x^3 - 2x = 21$. Testing integer values reveals that $x=3$ is the valid solution.

Solving the equation $\sqrt{r} + \frac{1}{\sqrt{r}} = 3$ requires multiplying by $\sqrt{r}$ to obtain $r + 1 = 3\sqrt{r}$. Squaring both sides produces $(r+1)^2 = 9r$, which expands to $r^2 + 2r + 1 = 9r$, or $r^2 - 7r + 1 = 0$. Since the common ratios are the roots of this quadratic equation, their sum is determined by the coefficient of the linear term, resulting in a value of 7.

The fundamental property relating the sum of a geometric series to its individual terms is that the difference $S_n - S_{n-1}$ is equal to the $n^{\text{th}}$ term, denoted by $T_n$. Given the $4^{\text{th}}$ term $T_4 = 500$ and common ratio $r = \frac{1}{m}$, the general term is expressed as $T_n = 500 \cdot \left(\frac{1}{m}\right)^{n-4}$. The condition $S_6 > S_5 + 1$ is equivalent to $T_6 > 1$, which leads to the inequality $\frac{500}{m^2} > 1$, or $m^2 < 500$. Similarly, the condition $S_7 < S_6 + \frac{1}{2}$ is equivalent to $T_7 < \frac{1}{2}$, resulting in $\frac{500}{m^3} < \frac{1}{2}$, or $m^3 > 1000$. Solving these constraints for the integer $m$ shows that $m$ must be greater than 10 and less than or equal to 22, as the square of 22 is 484 and the square of 23 is 529. Consequently, $m$ can be any integer from 11 to 22, giving 12 possible values.

The expression relies on the binomial identity $\frac{1}{k+1} \binom{n}{k} = \frac{1}{n+1} \binom{n+1}{k+1}$, which allows us to simplify the entire sum by factoring out $\frac{1}{n+1}$. By summing the binomial coefficients $\binom{n+1}{k+1}$ from $k=0$ to $n$, we obtain the sum of the $(n+1)$-th row of Pascal's triangle excluding the first term, which results in the expression $\frac{2^{n+1}-1}{n+1}$. Comparing this simplified form to the fraction $\frac{1023}{10}$ clearly establishes that $n+1$ must equal 10, leading to a final value of $n = 9$.

Any sequence of common terms shared by two separate arithmetic progressions constitutes its own arithmetic progression, where the common difference is the least common multiple of the individual series' differences. The first series increases by 4 and the second by 5, meaning the combined sequence progresses by $LCM(4, 5) = 20$.

Identifying the first shared value reveals that 11 is the initial common term for both lists. Using the general term formula $T_n = a + (n-1)d$ with the starting term $a = 11$ and common difference $d = 20$, the 8th term is calculated as $T_8 = 11 + (8-1) \times 20$. Completing this arithmetic results in $11 + 140$, which gives 151.

Determining the value of $2^7(2S-T)$ requires analyzing the recursive definitions of $a_n$ and $b_n$ to identify the sequence terms and then applying series transformation techniques. The sequence $a_n$ begins with $a_1=1$ and progresses as $a_n = a_{n-1} + (n-1)$, yielding the sequence $1, 2, 4, 7, 11, 16, 22, 29, 37$. Given $b_1=1$ and the relation $b_n - b_{n-1} = a_{n-1}$, we calculate $b_{10}$ by summing $b_1$ and the terms of $a_n$ from $a_1$ to $a_9$, which results in $b_{10} = 1 + (1+2+4+7+11+16+22+29+37) = 130$.

The series $S = \sum_{n=1}^{10} \frac{b_n}{2^n}$ and $T = \sum_{n=1}^{8} \frac{n}{2^{n+1}}$ allow for a simplification using the method of differences by subtracting $S/2$ from $S$. This operation establishes the relationship $2S = 2(a_1+b_1) - \frac{b_{10}+2a_9}{2^9} + T$, which rearranges directly into the form needed for the expression $2^7(2S-T)$. Substituting the values into this derived formula gives $2^7(2S-T) = 2^8(a_1+b_1) - \frac{b_{10}+2a_9}{4}$.

Using the established sequence values, $a_1=1$ and $a_9=37$, we compute $2^8(1+1) - \frac{130+2(37)}{4}$. This simplifies to $512 - \frac{204}{4}$, which leads to $512 - 51$, ultimately equaling 461.

Grouping terms based on their signs allows the sum to be split into two manageable components. The positive terms follow the pattern $\sum_{n=1}^{8} (2n-1)(4n-3)^2 = 29856$. The negative terms follow the pattern $\sum_{n=1}^{7} (2n)(4n-1)^2 = 22904$. Subtracting the latter from the former, $29856 - 22904 = 6952$, provides the final result.

The differences between consecutive terms of the sequence $5, 8, 14, 23, 35, 50, \dots$ form an arithmetic progression of $3, 6, 9, 12, 15, \dots$, indicating that the general term $a_n$ follows a quadratic pattern. Expressing $a_n$ as the sum of its first term and the cumulative differences, we obtain the general formula:
$a_n = 5 + \sum_{k=1}^{n-1} 3k = \frac{3n^2 - 3n + 10}{2}$
Substituting $n = 40$ into this expression yields $a_{40} = \frac{3(40)^2 - 3(40) + 10}{2} = 2345$.

The sum of the first 30 terms is defined as $S_{30} = \sum_{n=1}^{30} \frac{3n^2 - 3n + 10}{2}$, which decomposes into:
$S_{30} = \frac{3}{2} \sum_{n=1}^{30} n^2 - \frac{3}{2} \sum_{n=1}^{30} n + \sum_{n=1}^{30} 5$
Using the standard summation identities for the sum of squares and the sum of the first $n$ integers, this expression simplifies to $13635$. Calculating the final value requested gives $13635 - 2345 = 11290$.

Expanding a trinomial of the form $(a+bx+cx^2)^{10}$ involves determining the coefficients of each power of $x$ by accounting for all combinations of terms that multiply to produce that specific degree. The linear term $p_1 x$ arises solely from the selection of the first power of the binomial $(bx)$ paired with the constant $a$ chosen nine times, expressed as $\binom{10}{1} a^9 (bx) = 10 a^9 b x$. Setting $10 a^9 b = 20$ simplifies to $a^9 b = 2$, and because $a$ and $b$ must be natural numbers, this relationship is satisfied only when $a=1$ and $b=2$.

To find the coefficient $p_2$ of $x^2$, we must combine the contributions from the $cx^2$ term, which is $10 a^9 (cx^2)$, and the square of the $bx$ term, which is $\binom{10}{2} a^8 (bx)^2 = 45 a^8 b^2 x^2$. Substituting the identified values of $a=1$ and $b=2$ into the equation $10 a^9 c + 45 a^8 b^2 = 210$ leads to $10(1)^9 c + 45(1)^8(2)^2 = 210$. This expression simplifies to $10c + 180 = 210$, revealing that $10c = 30$, or $c=3$. Evaluating $2(a+b+c)$ with these constants results in $2(1+2+3)$, which equals 12.