### Related Formula textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right)$\text{Sum of an A.P.: } S_k = \frac{k}{2} \left( 2a + (k-1)d \right)$ textGeneral term of an A.P.: a_k = a_1 + (k-1)d$\text{General term of an A.P.: } a_k = a_1 + (k-1)d$ ### Core Logic Let the A.P. have n$n$ terms (where n$n$ is even). The terms are divided into n/2$n/2$ odd-indexed terms and n/2$n/2$ even-indexed terms. ### Step 1: Set up the even and odd sums Sum of even terms: a_2 + a_4 + dots + a_n = 30 quad text--- (1)$a_2 + a_4 + \dots + a_n = 30 \quad \text{--- (1)}$ Sum of odd terms: a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2)$a_1 + a_3 + \dots + a_{n-1} = 24 \quad \text{--- (2)}$ Subtracting equation (2) from (1): (a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6$(a_2 - a_1) + (a_4 - a_3) + \dots + (a_n - a_{n-1}) = 30 - 24 = 6$ Since there are n/2$n/2$ such pairs, and the difference of adjacent terms is the common difference d$d$: fracn2 d = 6 implies n d = 12 quad text--- (3)$\frac{n}{2} d = 6 \implies n d = 12 \quad \text{--- (3)}$ ### Step 2: Solve for n and d We are given that the last term exceeds the first by frac212$\frac{21}{2}$: a_n - a_1 = (n-1)d = frac212$a_n - a_1 = (n-1)d = \frac{21}{2}$ n d - d = 10.5$n d - d = 10.5$ Substitute nd = 12$nd = 12$ from (3): 12 - d = 10.5 implies d = 1.5 = frac32$12 - d = 10.5 \implies d = 1.5 = \frac{3}{2}$ Using this in (3): n left(frac32right) = 12 implies n = 8$n \left(\frac{3}{2}\right) = 12 \implies n = 8$ ### Step 3: Solve for the first term The sum of the odd terms is: S_textodd = frac42 left[ 2a_1 + (4-1)(2d) right] = 24$S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1)(2d) \right] = 24$ 2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32$2 \left[ 2a_1 + 3(3) \right] = 24 \implies 2a_1 + 9 = 12 \implies a_1 = 1.5 = \frac{3}{2}$ Thus, the terms are: frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12$\frac{3}{2}, \, 3, \, \frac{9}{2}, \, 6, \, \frac{15}{2}, \, 9, \, \frac{21}{2}, \, 12$ The terms that are integers are 3, 6, 9, 12$3, 6, 9, 12$. The total number of integer terms is 4. ### Pattern Recognition Sum of even terms minus sum of odd terms in any A.P. with an even number of terms n$n$ is always equal to fracn2 d$\frac{n}{2} d$. This is an extremely useful relation to remember. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula textSophie Germain's algebraic factorization: 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)$\text{Sophie Germain's algebraic factorization: } 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)$ textTelescoping Series representation: T_r = f(r) - f(r+1)$\text{Telescoping Series representation: } T_r = f(r) - f(r+1)$ ### Core Logic This is a telescoping series sum. We expand the denominator using Sophie Germain's algebraic identity to write the general term as a difference of two consecutive rational expressions. ### Step 1: Write down the general term and factor The general term T_r$T_r$ of the series is: T_r = frac4r1 + 4r^4$T_r = \frac{4r}{1 + 4r^4}$ Using the factorization of 1+4r^4$1+4r^4$: T_r = frac4r(2r^2 - 2r + 1)(2r^2 + 2r + 1)$T_r = \frac{4r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$ Notice that the numerator 4r$4r$ is the exact difference of the two quadratic factors: (2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$(2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$ ### Step 2: Split the fraction into telescoping terms Rewrite the general term T_r$T_r$: T_r = frac(2r^2 + 2r + 1) - (2r^2 - 2r + 1)(2r^2 - 2r + 1)(2r^2 + 2r + 1)$T_r = \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$ T_r = frac12r^2 - 2r + 1 - frac12r^2 + 2r + 1 = f(r) - f(r+1)$T_r = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} = f(r) - f(r+1)$ ### Step 3: Expand the sum and solve Sum the first 10 terms: - For r = 1$r = 1$: T_1 = frac11 - frac15$T_1 = \frac{1}{1} - \frac{1}{5}$ - For r = 2$r = 2$: T_2 = frac15 - frac113$T_2 = \frac{1}{5} - \frac{1}{13}$ - ... - For r = 10$r = 10$: T_10 = frac1181 - frac1221$T_{10} = \frac{1}{181} - \frac{1}{221}$ All intermediate terms cancel out: S_10 = 1 - frac1221 = frac220221 = fracmn$S_{10} = 1 - \frac{1}{221} = \frac{220}{221} = \frac{m}{n}$ Since 220$220$ and 221$221$ are coprime (their greatest common divisor is 1): m = 220 quad textand quad n = 221$m = 220 \quad \text{and} \quad n = 221$ m + n = 220 + 221 = 441$m + n = 220 + 221 = 441$ ### Pattern Recognition Sophie Germain identity: The expansion 4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$ is highly common in telescoping series. Spotting this factorization collapses the sum instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an Arithmetic Progression: S_m = fracm2[2a_1 + (m-1)d]$S_m = \frac{m}{2}[2a_1 + (m-1)d]$ ### Core Logic Express the odd terms summation in terms of a_1$a_1$ and d$d$, establish their linear dependency, and then solve for n$n$ where total sum vanishes. ### Step 1: Simplify the Given Summation The summation represents the sum of 12$12$ terms: a_1 + a_3 + a_5 + dots + a_23$a_1 + a_3 + a_5 + \dots + a_{23}$. This is an A.P. with initial term a_1$a_1$ and common difference 2d$2d$. frac122[2a_1 + 11(2d)] = -frac725a_1$\frac{12}{2}[2a_1 + 11(2d)] = -\frac{72}{5}a_1$ 6[2a_1 + 22d] = -frac725a_1 implies 12a_1 + 132d = -frac725a_1$6[2a_1 + 22d] = -\frac{72}{5}a_1 \implies 12a_1 + 132d = -\frac{72}{5}a_1$ ### Step 2: Relate initial term to common difference Multiply through by 5$5$ to eliminate the fraction: 60a_1 + 660d = -72a_1 implies 132a_1 + 660d = 0 implies a_1 = -5d$60a_1 + 660d = -72a_1 \implies 132a_1 + 660d = 0 \implies a_1 = -5d$ ### Step 3: Solve for n Set the general sum of n$n$ terms to 0$0$: fracn2[2a_1 + (n-1)d] = 0 implies 2a_1 + (n-1)d = 0$\frac{n}{2}[2a_1 + (n-1)d] = 0 \implies 2a_1 + (n-1)d = 0$ Substitute a_1 = -5d$a_1 = -5d$: 2(-5d) + (n-1)d = 0 implies -10d + nd - d = 0 implies nd = 11d$2(-5d) + (n-1)d = 0 \implies -10d + nd - d = 0 \implies nd = 11d$ Since a_1 neq 0$a_1 \neq 0$, d neq 0$d \neq 0$, which yields: n = 11$n = 11$ ### Pattern Recognition The condition a_1 = -5d$a_1 = -5d$ means the sequence starts positive/negative and counts down symmetrically. A sum of n$n$ terms equals zero when the middle term or balanced pairs completely wipe each other out, pointing directly to n = 2(5) + 1 = 11$n = 2(5) + 1 = 11$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of first r$r$ odd natural numbers: sum_k=1^r (2k-1) = r^2$\sum_{k=1}^{r} (2k-1) = r^2$ Exponential series expansion: sum_r=0^infty frac1r! = e$\sum_{r=0}^{\infty} \frac{1}{r!} = e$ ### Core Logic Let's find the general r$r$-th term of the series: T_r = frac1 + 3 + 5 + dots + (2r-1)r! = fracr^2r! = fracr(r-1)!$T_r = \frac{1 + 3 + 5 + \dots + (2r-1)}{r!} = \frac{r^2}{r!} = \frac{r}{(r-1)!}$ ### Step 1: Expressing term in terms of sum limits Let's write r = (r-1) + 1$r = (r-1) + 1$: T_r = fracr-1+1(r-1)! = frac1(r-2)! + frac1(r-1)!$T_r = \frac{r-1+1}{(r-1)!} = \frac{1}{(r-2)!} + \frac{1}{(r-1)!}$ Our infinite sum is: S = sum_r=1^infty T_r = sum_r=2^infty frac1(r-2)! + sum_r=1^infty frac1(r-1)!$S = \sum_{r=1}^{\infty} T_r = \sum_{r=2}^{\infty} \frac{1}{(r-2)!} + \sum_{r=1}^{\infty} \frac{1}{(r-1)!}$ Both sums are standard representations of the exponential expansion. ### Step 2: Summing the parts - First part: sum_r=2^infty frac1(r-2)! = 1 + frac11! + frac12! + dots = e$\sum_{r=2}^{\infty} \frac{1}{(r-2)!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots = e$ - Second part: sum_r=1^infty frac1(r-1)! = 1 + frac11! + frac12! + dots = e$\sum_{r=1}^{\infty} \frac{1}{(r-1)!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots = e$ textTotal Sum S = e + e = 2e$\text{Total Sum } S = e + e = 2e$ ### Pattern Recognition The general term containing r^2$r^2$ in summation with factorials converges to 2e$2e$. Remember the shortcut: sum fracr^2r! = 2e$\sum \frac{r^2}{r!} = 2e$, sum fracr^3r! = 5e$\sum \frac{r^3}{r!} = 5e$. It is extremely useful to memorize these common limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula For a geometric progression, the terms can be set as a, ar, ar^2, ar^3$a, ar, ar^2, ar^3$. For three terms A, B, C$A, B, C$ to be in arithmetic progression, they must satisfy: 2B = A + C$2B = A + C$ ### Core Logic Let the elements be x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3$x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3$. After the specified subtractions, the sequence becomes: a - 2, quad ar - 7, quad ar^2 - 9, quad ar^3 - 5$a - 2, \quad ar - 7, \quad ar^2 - 9, \quad ar^3 - 5$ Since this sequence is in AP, we form two separate common difference linear linkages: 2(ar - 7) = (a - 2) + (ar^2 - 9) implies 2ar - 14 = ar^2 + a - 11 implies ar^2 - 2ar + a + 3 = 0 quad dots (1)$2(ar - 7) = (a - 2) + (ar^2 - 9) \implies 2ar - 14 = ar^2 + a - 11 \implies ar^2 - 2ar + a + 3 = 0 \quad \dots (1)$ 2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) implies 2ar^2 - 18 = ar^3 + ar - 12 implies ar^3 - 2ar^2 + ar + 6 = 0 quad dots (2)$2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) \implies 2ar^2 - 18 = ar^3 + ar - 12 \implies ar^3 - 2ar^2 + ar + 6 = 0 \quad \dots (2)$ ### Step 1: Solve the Simultaneous Polynomials Multiply equation (1) by r$r$: ar^3 - 2ar^2 + ar + 3r = 0 quad dots (3)$ar^3 - 2ar^2 + ar + 3r = 0 \quad \dots (3)$ Subtract equation (3) from equation (2): (ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0$(ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0$ 6 - 3r = 0 implies 3r = 6 implies r = 2$6 - 3r = 0 \implies 3r = 6 \implies r = 2$ Substitute r = 2$r = 2$ back into equation (1): a(2)^2 - 2a(2) + a + 3 = 0$a(2)^2 - 2a(2) + a + 3 = 0$ 4a - 4a + a + 3 = 0 implies a = -3$4a - 4a + a + 3 = 0 \implies a = -3$ ### Step 2: Find the Continuous Product Value The continuous product term is: mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = a cdot ar cdot ar^2 cdot ar^3 = a^4 r^6$\mathbf{x}_1\mathbf{x}_2\mathbf{x}_3\mathbf{x}_4 = a \cdot ar \cdot ar^2 \cdot ar^3 = a^4 r^6$ mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = (-3)^4 cdot (2)^6 = 81 times 64 = 5184$\mathbf{x}_1\mathbf{x}_2\mathbf{x}_3\mathbf{x}_4 = (-3)^4 \cdot (2)^6 = 81 \times 64 = 5184$ Now divide by 24 as required: frac124(5184) = 216$\frac{1}{24}(5184) = 216$ ### Pattern Recognition Notice that multiplying the first AP condition equation by r$r$ perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series