### Related Formula Summation Identities: sum r = fracn(n+1)2, quad sum r^2 = fracn(n+1)(2n+1)6$\sum r = \frac{n(n+1)}{2}, \quad \sum r^2 = \frac{n(n+1)(2n+1)}{6}$ ### Core Logic Split the 40-term series into two sub-series of 20 terms each: Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd): 1^2 + 5^2 + 9^2 + dots$1^2 + 5^2 + 9^2 + \dots$ upto 20 terms. General term T_r = (4r - 3)^2$T_r = (4r - 3)^2$. Series 2 (linear terms at positions 2, 4, 6...): 3 + 7 + 11 + dots$3 + 7 + 11 + \dots$ upto 20 terms. General term t_r = (4r - 1)$t_r = (4r - 1)$. ### Step 1: Formulate Total Sigma Expression textSum = sum_r=1^20 left[ (4r - 3)^2 + (4r - 1) right]$\text{Sum} = \sum_{r=1}^{20} \left[ (4r - 3)^2 + (4r - 1) \right]$ textSum = sum_r=1^20 (16r^2 - 24r + 9 + 4r - 1) = sum_r=1^20 (16r^2 - 20r + 8)$\text{Sum} = \sum_{r=1}^{20} (16r^2 - 24r + 9 + 4r - 1) = \sum_{r=1}^{20} (16r^2 - 20r + 8)$ textSum = 16sum_r=1^20 r^2 - 20sum_r=1^20 r + 8sum_r=1^20 1$\text{Sum} = 16\sum_{r=1}^{20} r^2 - 20\sum_{r=1}^{20} r + 8\sum_{r=1}^{20} 1$ ### Step 2: Arithmetic Evaluation sum_r=1^20 r^2 = frac20 times 21 times 416 = 2870$\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870$ sum_r=1^20 r = frac20 times 212 = 210$\sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$ textSum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$\text{Sum} = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$ ### Pattern Recognition When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequence and Series

### Related Formula Sum of first n$n$ terms of an AP is given by: S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ ### Core Logic Given specifications: 1) a_6 = 7 implies a + 5d = 7 quad dots text(ii)$a_6 = 7 \implies a + 5d = 7 \quad \dots \text{(ii)}$ 2) S_7 = 7 implies frac72(2a + 6d) = 7 implies a + 3d = 1 quad dots text(iii)$S_7 = 7 \implies \frac{7}{2}(2a + 6d) = 7 \implies a + 3d = 1 \quad \dots \text{(iii)}$ Subtracting (iii) from (ii): 2d = 6 implies d = 3$2d = 6 \implies d = 3$ Substituting d=3$d=3$ into (iii): a + 3(3) = 1 implies a = -8$a + 3(3) = 1 \implies a = -8$ ### Step 1: Find n from Sn = 700 Substitute a = -8$a = -8$ and d = 3$d = 3$ into the equation for S_n = 700$S_n = 700$: 700 = fracn2[2(-8) + (n-1)3]$700 = \frac{n}{2}[2(-8) + (n-1)3]$ 1400 = n[-16 + 3n - 3]$1400 = n[-16 + 3n - 3]$ 3n^2 - 19n - 1400 = 0$3n^2 - 19n - 1400 = 0$ Factoring the quadratic equation: (3n + 56)(n - 25) = 0$(3n + 56)(n - 25) = 0$ Since n$n$ must be a positive integer, n = 25$n = 25$. ### Step 2: Determine standard term value We need to find a_25$a_{25}$ corresponding to index n=25$n=25$: a_25 = a + 24d$a_{25} = a + 24d$ a_25 = -8 + 24(3) = -8 + 72 = 64$a_{25} = -8 + 24(3) = -8 + 72 = 64$ ### Pattern Recognition When S_n$S_n$ and specific terms are given, prioritize finding the first term a$a$ and common difference d$d$ through simple elimination headers before targeting the value of n$n$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of first n$n$ terms of a Geometric Progression (GP) is: S_n = fraca(r^n - 1)r - 1$S_n = \frac{a(r^n - 1)}{r - 1}$ ### Core Logic Let the first term be a$a$ and common ratio be r$r$. Given: 1) ar + ar^3 + ar^5 = 21 implies ar(1 + r^2 + r^4) = 21 quad dots text(1)$ar + ar^3 + ar^5 = 21 \implies ar(1 + r^2 + r^4) = 21 \quad \dots \text{(1)}$ 2) ar^7 + ar^9 + ar^11 = 15309 implies ar^7(1 + r^2 + r^4) = 15309 quad dots text(2)$ar^7 + ar^9 + ar^{11} = 15309 \implies ar^7(1 + r^2 + r^4) = 15309 \quad \dots \text{(2)}$ Dividing equation (2) by equation (1): fracar^7ar = frac1530921 implies r^6 = 729 implies r = 3$\frac{ar^7}{ar} = \frac{15309}{21} \implies r^6 = 729 \implies r = 3$ ### Step 1: Solve for a Substitute r = 3$r = 3$ into equation (1): a(3)(1 + 9 + 81) = 21$a(3)(1 + 9 + 81) = 21$ 3a(91) = 21 implies a = frac791 = frac113$3a(91) = 21 \implies a = \frac{7}{91} = \frac{1}{13}$ ### Step 2: Find Sum of 9 terms Evaluating S_9$S_9$: S_9 = fraca(r^9 - 1)r - 1 = fracfrac113(3^9 - 1)3 - 1 = frac19683 - 126 = frac1968226 = 757$S_9 = \frac{a(r^9 - 1)}{r - 1} = \frac{\frac{1}{13}(3^9 - 1)}{3 - 1} = \frac{19683 - 1}{26} = \frac{19682}{26} = 757$ ### Pattern Recognition Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio r^k$r^k$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of first n$n$ terms of an Arithmetic Progression: S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ ### Core Logic Set up linear expressions for the given sums : S_40 = frac402[2a + 39d] = 1030 Rightarrow 2a + 39d = 51.5$S_{40} = \frac{40}{2}[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5$ S_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5$S_{12} = \frac{12}{2}[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5$ ### Step 1: Solve for a$a$ and d$d$ Subtract the second equation from the first : (2a + 39d) - (2a + 11d) = 51.5 - 9.5$(2a + 39d) - (2a + 11d) = 51.5 - 9.5$ 28d = 42 Rightarrow d = frac4228 = frac32 = 1.5$28d = 42 \Rightarrow d = \frac{42}{28} = \frac{3}{2} = 1.5$ Substitute d = 1.5$d = 1.5$ back to find a$a$: 2a + 11(1.5) = 9.5 Rightarrow 2a + 16.5 = 9.5 Rightarrow 2a = -7 Rightarrow a = -3.5$2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5$ ### Step 2: Evaluate S_30 - S_10$S_{30} - S_{10}$ Write out the formula for the target subtraction : S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d]$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d]$ = 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$ [cite: 3964, 3965] Substitute the values of a$a$ and d$d$ : = 20(-3.5) + 390(1.5) = -70 + 585 = 515$= 20(-3.5) + 390(1.5) = -70 + 585 = 515$ ### Pattern Recognition Notice that S_30 - S_10$S_{30} - S_{10}$ represents the \sum of terms from T_11$T_{11}$ to T_30$T_{30}$, which can also be formulated as 20 times A_20.5$20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an infinite geometric progression: S_infty = fraca1-r quad textfor |r| < 1$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$ ### Core Logic The given expression is an infinite Arithmetico-Geometric Progression (AGP) : S = 5 + frac5+alpha7 + frac5+2alpha7^2 + frac5+3alpha7^3 + dots infty$S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots \infty$ ### Step 1: Shift and Subtract Multiply the equation by the common ratio frac17$\frac{1}{7}$ and shift it by one position : frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty$\frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots \infty$ Subtract this from the original equation: S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots$S - \frac{1}{7}S = 5 + \left(\frac{5+\alpha-5}{7}\right) + \left(\frac{5+2\alpha-(5+\alpha)}{7^2}\right) + \dots$ frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots$\frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$ ### Step 2: Sum the Infinite Geometric Series Apply the infinite GP formula to the terms involving alpha$\alpha$ : frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6$\frac{6}{7}S = 5 + \frac{\alpha}{7}\left(\frac{1}{1 - \frac{1}{7}}\right) = 5 + \frac{\alpha}{7}\left(\frac{7}{6}\right) = 5 + \frac{\alpha}{6}$ Given that S = 7$S = 7$ : frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6$\frac{6}{7}(7) = 5 + \frac{\alpha}{6} \Rightarrow 6 = 5 + \frac{\alpha}{6}$ 1 = fracalpha6 Rightarrow alpha = 6$1 = \frac{\alpha}{6} \Rightarrow \alpha = 6$ ### Pattern Recognition Standard trick for infinite AGPs: Multiply by the common ratio r$r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series