### Related Formula S_n = fraca(1 - r^n)1 - r$S_n = \frac{a(1 - r^n)}{1 - r}$ where S_n$S_n$ is the sum of n$n$ terms, a$a$ is the first term, and r$r$ is the common ratio. ### Core Logic Let the terms of the G.P. be a, ar, ar^2, ar^3, dots, ar^63$a, ar, ar^2, ar^3, \dots, ar^{63}$. The sum of all 64 terms is given by: S_textall = a + ar + ar^2 + dots + ar^63 = fraca(1 - r^64)1 - r$S_{\text{all}} = a + ar + ar^2 + \dots + ar^{63} = \frac{a(1 - r^{64})}{1 - r}$ The odd terms are a, ar^2, ar^4, dots, ar^62$a, ar^2, ar^4, \dots, ar^{62}$. This forms another G.P. with 32 terms and a common ratio of r^2$r^2$. The sum of the odd terms is: S_textodd = a + ar^2 + ar^4 + dots + ar^62 = fraca(1 - (r^2)^32)1 - r^2 = fraca(1 - r^64)1 - r^2$S_{\text{odd}} = a + ar^2 + ar^4 + \dots + ar^{62} = \frac{a(1 - (r^2)^{32})}{1 - r^2} = \frac{a(1 - r^{64})}{1 - r^2}$ ### Step 1: Equate and Solve for r We are given that S_textall = 7 cdot S_textodd$S_{\text{all}} = 7 \cdot S_{\text{odd}}$. Substituting our formulas: fraca(1 - r^64)1 - r = 7 cdot fraca(1 - r^64)1 - r^2$\frac{a(1 - r^{64})}{1 - r} = 7 \cdot \frac{a(1 - r^{64})}{1 - r^2}$ Assuming a neq 0$a \neq 0$ and r neq 1$r \neq 1$, we can cancel the common terms dots a(1 - r^64) dots$\dots a(1 - r^{64}) \dots$ from both sides: frac11 - r = frac71 - r^2$\frac{1}{1 - r} = \frac{7}{1 - r^2}$ Since 1 - r^2 = (1 - r)(1 + r)$1 - r^2 = (1 - r)(1 + r)$, we have: frac11 - r = frac7(1 - r)(1 + r)$\frac{1}{1 - r} = \frac{7}{(1 - r)(1 + r)}$ 1 + r = 7$1 + r = 7$ r = 6$r = 6$ ### Pattern Recognition Shortcut: In any G.P. with an even number of terms, the ratio of the total sum to the sum of the odd-positioned terms is exactly 1 + r$1 + r$. Thus, 1 + r = 7 Rightarrow r = 6$1 + r = 7 \Rightarrow r = 6$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula a_n = a + (n-1)d$a_n = a + (n-1)d$ For finding extrema of a polynomial function f(x)$f(x)$, we set its derivative f'(x) = 0$f'(x) = 0$. ### Core Logic Given the 6th term of the A.P. is a_6 = 2$a_6 = 2$. a + 5d = 2 Rightarrow a = 2 - 5d$a + 5d = 2 \Rightarrow a = 2 - 5d$ We need to maximize the product P = a_1 a_4 a_5$P = a_1 a_4 a_5$. P = a(a + 3d)(a + 4d)$P = a(a + 3d)(a + 4d)$ Substituting a = 2 - 5d$a = 2 - 5d$ into the expression for P$P$: P = (2 - 5d)(2 - 5d + 3d)(2 - 5d + 4d)$P = (2 - 5d)(2 - 5d + 3d)(2 - 5d + 4d)$ P = (2 - 5d)(2 - 2d)(2 - d)$P = (2 - 5d)(2 - 2d)(2 - d)$ ### Step 1: Expand and Differentiate Let's expand P$P$ as a function of d$d$, f(d)$f(d)$: f(d) = (2 - 5d)(4 - 6d + 2d^2)$f(d) = (2 - 5d)(4 - 6d + 2d^2)$ f(d) = 8 - 12d + 4d^2 - 20d + 30d^2 - 10d^3$f(d) = 8 - 12d + 4d^2 - 20d + 30d^2 - 10d^3$ f(d) = -10d^3 + 34d^2 - 32d + 8$f(d) = -10d^3 + 34d^2 - 32d + 8$ To find the maximum, we differentiate f(d)$f(d)$ with respect to d$d$ and equate to zero: f'(d) = -30d^2 + 68d - 32 = 0$f'(d) = -30d^2 + 68d - 32 = 0$ 15d^2 - 34d + 16 = 0$15d^2 - 34d + 16 = 0$ Factoring the quadratic: 15d^2 - 24d - 10d + 16 = 0$15d^2 - 24d - 10d + 16 = 0$ 3d(5d - 8) - 2(5d - 8) = 0$3d(5d - 8) - 2(5d - 8) = 0$ (5d - 8)(3d - 2) = 0$(5d - 8)(3d - 2) = 0$ This gives critical points d = frac85$d = \frac{8}{5}$ and d = frac23$d = \frac{2}{3}$. ### Step 2: Check for Maximum We check the second derivative to confirm a maximum: f''(d) = -60d + 68$f''(d) = -60d + 68$ At d = frac85$d = \frac{8}{5}$: f''left(frac85right) = -60left(frac85right) + 68 = -96 + 68 = -28 lt 0 quad (textMaximum)$f''\left(\frac{8}{5}\right) = -60\left(\frac{8}{5}\right) + 68 = -96 + 68 = -28 \lt 0 \quad (\text{Maximum})$ At d = frac23$d = \frac{2}{3}$: f''left(frac23right) = -60left(frac23right) + 68 = -40 + 68 = 28 gt 0 quad (textMinimum)$f''\left(\frac{2}{3}\right) = -60\left(\frac{2}{3}\right) + 68 = -40 + 68 = 28 \gt 0 \quad (\text{Minimum})$ Therefore, the greatest product occurs at d = frac85$d = \frac{8}{5}$. ### Pattern Recognition When asked to maximize a product of A.P. terms with a known constant term, express all terms strictly in d$d$, build the cubic, and use standard calculus f'(x)=0$f'(x)=0$ checking roots against the 2nd derivative test (Wavy Curve method works beautifully here). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Application of Derivatives

### Related Formula textSum of AP: S_k = frack2 [2a + (k - 1)d]$\text{Sum of AP: } S_k = \frac{k}{2} [2a + (k - 1)d]$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ sum_k=1^n k = fracn(n+1)2$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ ### Core Logic For the arithmetic progression 3, 7, 11, dots$3, 7, 11, \dots$ First term a = 3$a = 3$, Common difference d = 4$d = 4$. The sum of the first k$k$ terms is: S_k = frack2 (2(3) + (k - 1)4) = frack2 (6 + 4k - 4) = frack2 (4k + 2) = 2k^2 + k$S_k = \frac{k}{2} (2(3) + (k - 1)4) = \frac{k}{2} (6 + 4k - 4) = \frac{k}{2} (4k + 2) = 2k^2 + k$ ### Step 1: Finding the Sum of Sums Now compute the sum sum_k=1^n S_k$\sum_{k=1}^{n} S_k$: sum_k=1^n S_k = sum_k=1^n (2k^2 + k) = 2sum_k=1^n k^2 + sum_k=1^n k$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$ = 2 left( fracn(n+1)(2n+1)6 right) + fracn(n+1)2$= 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2}$ = n(n+1) left[ frac2(2n+1)6 + frac12 right] = n(n+1) left[ frac2n+13 + frac12 right]$= n(n+1) \left[ \frac{2(2n+1)}{6} + \frac{1}{2} \right] = n(n+1) \left[ \frac{2n+1}{3} + \frac{1}{2} \right]$ = n(n+1) left[ frac4n + 2 + 36 right] = fracn(n+1)(4n + 5)6$= n(n+1) \left[ \frac{4n + 2 + 3}{6} \right] = \frac{n(n+1)(4n + 5)}{6}$ ### Step 2: Resolving the Inequality Substitute this sum into the given expression: frac6n(n+1) sum_k=1^n S_k = frac6n(n+1) cdot fracn(n+1)(4n+5)6 = 4n + 5$\frac{6}{n(n+1)} \sum_{k=1}^{n} S_k = \frac{6}{n(n+1)} \cdot \frac{n(n+1)(4n+5)}{6} = 4n + 5$ We are given the bounds: 40 lt 4n + 5 lt 42$40 \lt 4n + 5 \lt 42$ 35 lt 4n lt 37$35 \lt 4n \lt 37$ 8.75 lt n lt 9.25$8.75 \lt n \lt 9.25$ Since n$n$ must be an integer (representing the number of terms), the only valid integer is n = 9$n = 9$. ### Pattern Recognition Evaluating a 'sum of sums' for an AP effectively requires applying the Sigma k^2$\Sigma k^2$ and Sigma k$\Sigma k$ standard formulas to the generic S_n$S_n$ quadratic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ ### Core Logic Using the sum formula for an AP: S_20 = frac202[2a + 19d] = 790$S_{20} = \frac{20}{2}[2a + 19d] = 790$ 10[2a + 19d] = 790$10[2a + 19d] = 790$ 2a + 19d = 79 quad dots (1)$2a + 19d = 79 \quad \dots (1)$ S_10 = frac102[2a + 9d] = 145$S_{10} = \frac{10}{2}[2a + 9d] = 145$ 5[2a + 9d] = 145$5[2a + 9d] = 145$ 2a + 9d = 29 quad dots (2)$2a + 9d = 29 \quad \dots (2)$ ### Step 1: Solving for a and d Subtracting (2) from (1): 10d = 50 Rightarrow d = 5$10d = 50 \Rightarrow d = 5$ Substituting d=5$d=5$ into (2): 2a + 9(5) = 29 Rightarrow 2a = 29 - 45 = -16$2a + 9(5) = 29 \Rightarrow 2a = 29 - 45 = -16$ a = -8$a = -8$ ### Step 2: Evaluating the required expression We need to find S_15 - S_5$S_{15} - S_{5}$: S_15 - S_5 = frac152[2a + 14d] - frac52[2a + 4d]$S_{15} - S_5 = \frac{15}{2}[2a + 14d] - \frac{5}{2}[2a + 4d]$ Substituting 2a = -16$2a = -16$ and d = 5$d = 5$: = frac152[-16 + 70] - frac52[-16 + 20]$= \frac{15}{2}[-16 + 70] - \frac{5}{2}[-16 + 20]$ = frac152[54] - frac52[4]$= \frac{15}{2}[54] - \frac{5}{2}[4]$ = 15 times 27 - 5 times 2$= 15 \times 27 - 5 \times 2$ = 405 - 10 = 395$= 405 - 10 = 395$ ### Pattern Recognition When two sums of an AP are given, immediately set up the linear equations in terms of a$a$ and d$d$. Solve for them, and substitute directly into the target expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series