The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by frac212$\frac{21}{2}$. Then the number of terms which are integers in the A.P. is :

### 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

### Related Formula textTotal Sum = alpha + beta$$\text{Total Sum} = \alpha + \beta$$ ### Core Logic Factor out common fractions from the even terms component (beta$\beta$) to represent it as a scalar multiple of the universal sum sequence. ### Step 1: Simplify the Even Terms Series beta = frac12^4 + frac14^4 + frac16^4 + dots = frac12^4 left( frac11^4 + frac12^4 + frac13^4 + dots right)$$\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots = \frac{1}{2^4} \left( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots \right)$$ beta = frac116 left( fracpi^490 right)$$\beta = \frac{1}{16} \left( \frac{\pi^4}{90} \right)$$ ### Step 2: Express Alpha by Remainder Deduction Since total sum equals alpha + beta$\alpha + \beta$: alpha = textTotal Sum - beta = fracpi^490 - frac116 left( fracpi^490 right) = frac1516 left( fracpi^490 right)$$\alpha = \text{Total Sum} - \beta = \frac{\pi^4}{90} - \frac{1}{16} \left( \frac{\pi^4}{90} \right) = \frac{15}{16} \left( \frac{\pi^4}{90} \right)$$ ### Step 3: Compute the Relative Ratio fracalphabeta = fracfrac1516 left( fracpi^490 right)frac116 left( fracpi^490 right) = 15$\frac{\alpha}{\beta} = \frac{\frac{15}{16} \left( \frac{\pi^4}{90} \right)}{\frac{1}{16} \left( \frac{\pi^4}{90} \right)} = 15$ ### Pattern Recognition For alternating p-series powers like sum n^-p$\sum n^{-p}$, the even component fractions always condense via factor steps to 2^-p cdot S_texttotal$2^{-p} \cdot S_{\text{total}}$, decoupling power values cleanly from final simple scalar quotients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an infinite geometric progression with |r| < 1$|r| < 1$: S_infty = fraca1 - r$$S_{\infty} = \frac{a}{1 - r}$$ ### Core Logic From the definition of the relation: log_e y = x log_eleft(frac25 ight) implies log_e y = log_eleft(frac25right)^x implies y = left(frac25 ight)^x$$\log_e y = x \log_e\left(\frac{2}{5} ight) \implies \log_e y = \log_e\left(\frac{2}{5}\right)^x \implies y = \left(\frac{2}{5} ight)^x$$

Since x in S = \0, 1, 2, 3, dots\$x \in S = \{0, 1, 2, 3, \dots\}$, the output values of y$y$ represent elements of the range. ### Step 1: Compute Infinite Sum Generating elements by plugging in values of x$x$: For x = 0 implies y = 1$x = 0 \implies y = 1$ For x = 1 implies y = frac25$x = 1 \implies y = \frac{2}{5}$ For x = 2 implies y = left(frac25 ight)^2$x = 2 \implies y = \left(\frac{2}{5} ight)^2$ Sum of elements in the range: textSum = 1 + left(frac25 ight)^1 + left(frac25 ight)^2 + dots = frac11 - frac25 = frac53$$\text{Sum} = 1 + \left(\frac{2}{5} ight)^1 + \left(\frac{2}{5} ight)^2 + \dots = \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}$$ ### Pattern Recognition Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions

### Related Formula Symmetry identity rule inside Arithmetic Progressions: a_k + a_n-k+1 = a_1 + a_n$$a_k + a_{n-k+1} = a_1 + a_n$$ ### Core Logic Group matching paired steps equidistant from sequence boundary ends: a_1 + a_2024 = a_5 + a_2020 = a_10 + a_2015 = dots$$a_1 + a_{2024} = a_5 + a_{2020} = a_{10} + a_{2015} = \dots$$ The sequence of inner indices follows an AP tracking loop: 5, 10, 15, dots, 2020$$5, 10, 15, \dots, 2020$$ Calculate internal block element count N$N$: 2020 = 5 + (N-1)5 implies 2015 = 5(N-1) implies N - 1 = 403 implies N = 404 text terms$$2020 = 5 + (N-1)5 \implies 2015 = 5(N-1) \implies N - 1 = 403 \implies N = 404 \text{ terms}$$ ### Step 1: Simplify Expression Equations Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding a_1$a_1$ and a_2024$a_{2024}$ introduces one more pair, resulting in 203 identical sum blocks: 203(a_1 + a_2024) = 2233 implies a_1 + a_2024 = frac2233203 = 11$$203(a_1 + a_{2024}) = 2233 \implies a_1 + a_{2024} = \frac{2233}{203} = 11$$ ### Step 2: Evaluate the Total Sum Using the standard AP sum formula: S_2024 = frac20242(a_1 + a_2024) = 1012 times 11 = 11132$$S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) = 1012 \times 11 = 11132$$ ### Pattern Recognition Progressions possess natural positional balance. Grouping symmetrical index pairs (a_k + a_n-k+1$a_k + a_{n-k+1}$) allows factoring out variable steps right away. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series