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 ntext-th term of a GP: T_n = a r^n-1$$n\text{-th term of a GP}: T_n = a r^{n-1}$$ ### Core Logic Let the terms of the GP be a, ar, ar^2, ar^3, dots$a, ar, ar^2, ar^3, \dots$ Given: a_2 cdot a_3 cdot a_4 = 64$a_2 \cdot a_3 \cdot a_4 = 64$ (ar) cdot (ar^2) cdot (ar^3) = 64$$(ar) \cdot (ar^2) \cdot (ar^3) = 64$$ a^3 r^6 = 64 Rightarrow (ar^2)^3 = 64 Rightarrow ar^2 = 4$$a^3 r^6 = 64 \Rightarrow (ar^2)^3 = 64 \Rightarrow ar^2 = 4$$ Since it is a GP of increasing positive terms, r > 1$r > 1$ and a > 0$a > 0$. ### Step 1: Utilize the sum condition Given: a_1 + a_3 + a_5 = frac8137$a_1 + a_3 + a_5 = \frac{813}{7}$ a + ar^2 + ar^4 = frac8137$$a + ar^2 + ar^4 = \frac{813}{7}$$ Extract a$a$ from ar^2 = 4 Rightarrow a = frac4r^2$ar^2 = 4 \Rightarrow a = \frac{4}{r^2}$. Substitute this in: frac4r^2 + 4 + 4r^2 = frac8137$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$ 4left(frac1r^2 + 1 + r^2right) = frac8137$$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$ Let r^2 = t$r^2 = t$: 4left(frac1t + 1 + tright) = frac8137$$4\left(\frac{1}{t} + 1 + t\right) = \frac{813}{7}$$ Wait, there is a much faster method by just scaling the required expression. ### Step 2: Calculate the required expression We need a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$. Notice that ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$. So, required sum = r^2 left(frac8137right)$r^2 \left(\frac{813}{7}\right)$. To find r^2$r^2$, we solve the quadratic in t = r^2$t = r^2$: 4left(fract^2 + t + 1tright) = frac8137$$4\left(\frac{t^2 + t + 1}{t}\right) = \frac{813}{7}$$ 28t^2 + 28t + 28 = 813t$$28t^2 + 28t + 28 = 813t$$ 28t^2 - 785t + 28 = 0$$28t^2 - 785t + 28 = 0$$ The roots are t = 28$t = 28$ and t = frac128$t = \frac{1}{28}$. Since the GP is increasing, r > 1 Rightarrow r^2 = 28$r > 1 \Rightarrow r^2 = 28$. ### Step 3: Final evaluation Alternatively, expand directly: ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ (Note: r^2 times frac8137 = 28 times frac8137 = 4 times 813 = 3252$r^2 \times \frac{813}{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252$) ### Pattern Recognition In GP questions demanding a sum shifted by a fixed index (like a_1+a_3+a_5$a_1+a_3+a_5$ to a_3+a_5+a_7$a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier r^k$r^k$. Here it's a simple scaling by r^2$r^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

### Related Formula Partial fraction decomposition for telescopic summing: frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions.$$\frac{2n^2 - (n+1)^2 + 1 \dots}{\dots} \text{ structures directly cancel in series expansions.}$$ ### Core Logic Given recurrence: a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2$$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$ Rewrite the numerator to split the fraction: n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$ a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2$$a_{n+1} - \frac{1}{2}a_n = \frac{2n^2 - (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} - \frac{1}{n^2}$$ ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1$n=1$: a_2 - frac12a_1 = frac22^2 - frac11^2$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$ For n=2$n=2$: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$ Wait, let's look at a cleaner telescopic scaling: a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right)$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$. ### Step 2: Identify Geometric Progression Let V_n = a_n - frac2n^2$V_n = a_n - \frac{2}{n^2}$. The recurrence gives V_n+1 = frac12 V_n$V_{n+1} = \frac{1}{2} V_n$. This proves V_n$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$. First term V_1 = a_1 - frac21^2 = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$. ### Step 3: Infinite Summation We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$. Since V_n$V_n$ is an infinite GP: S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2$$S_{\infty} = \frac{V_1}{1 - r} = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$ Taking absolute value: |-2| = 2$|-2| = 2$ ### Pattern Recognition When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute V_n = A_n - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: 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