### Related Formula - General term of an AP: T_n = a + (n-1)d$T_n = a + (n-1)d$ - Sum of n$n$ terms of an AP: S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ - The common difference of a series of common terms between two APs is given by the least common multiple of their respective common differences: d_textcommon = textLCM(d_1, d_2)$d_{\text{common}} = \text{LCM}(d_1, d_2)$ ### Core Logic Let's analyze both arithmetic progressions: - AP 1: 3, 7, 11, 15, dots, 403 implies$3, 7, 11, 15, \dots, 403 \implies$ First term a_1 = 3$a_1 = 3$, common difference d_1 = 4$d_1 = 4$. - AP 2: 2, 5, 8, 11, dots, 404 implies$2, 5, 8, 11, \dots, 404 \implies$ First term a_2 = 2$a_2 = 2$, common difference d_2 = 3$d_2 = 3$. By observation, the first identical value appearing in both series is 11$11$. Therefore, the new common AP has: - First term a = 11$a = 11$ - Common difference d = textLCM(4, 3) = 12$d = \text{LCM}(4, 3) = 12$ ### Step 1: Determine the Number of Common Terms The last term T_n$T_n$ of the common AP cannot exceed the boundary upper limits of either individual series (i.e., le 403$\le 403$): T_n = 11 + (n-1)12 le 403$T_n = 11 + (n-1)12 \le 403$ 12(n-1) le 392$12(n-1) \le 392$ n-1 le 32.66 implies n = 33$n-1 \le 32.66 \implies n = 33$ ### Step 2: Calculate the Series Sum Using the AP summation formula for 33$33$ terms: S_33 = frac332 left[ 2(11) + (33-1)12 right]$S_{33} = \frac{33}{2} \left[ 2(11) + (33-1)12 \right]$ S_33 = frac332 left[ 22 + 32 times 12 right]$S_{33} = \frac{33}{2} \left[ 22 + 32 \times 12 \right]$ S_33 = frac332 left[ 22 + 384 right]$S_{33} = \frac{33}{2} \left[ 22 + 384 \right]$ S_33 = frac332 times 406 = 33 times 203 = 6699$S_{33} = \frac{33}{2} \times 406 = 33 \times 203 = 6699$ ### Pattern Recognition Sees: Overlapping arithmetic progression series elements. Shortcut: Once the first matching number and the LCM are calculated, the maximum term inequality a + (n-1)d le min(L_1, L_2)$a + (n-1)d \le \min(L_1, L_2)$ directly maps out the total number of terms cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula If x, y, z$x, y, z$ are in A.P., then 2y = x + z$2y = x + z$. ### Core Logic From the first sequence condition: 2log_e b = log_e a + log_e c implies log_e b^2 = log_e(ac) implies b^2 = ac quad dots (i)$2\log_e b = \log_e a + \log_e c \implies \log_e b^2 = \log_e(ac) \implies b^2 = ac \quad \dots (i)$ From the second sequence condition, the components are log_eleft(fraca2bright), log_eleft(frac2b3cright), log_eleft(frac3caright)$\log_e\left(\frac{a}{2b}\right), \log_e\left(\frac{2b}{3c}\right), \log_e\left(\frac{3c}{a}\right)$: 2log_eleft(frac2b3cright) = log_eleft(fraca2bright) + log_eleft(frac3caright)$2\log_e\left(\frac{2b}{3c}\right) = \log_e\left(\frac{a}{2b}\right) + \log_e\left(\frac{3c}{a}\right)$ left(frac2b3cright)^2 = fraca2b times frac3ca = frac3c2b$\left(\frac{2b}{3c}\right)^2 = \frac{a}{2b} \times \frac{3c}{a} = \frac{3c}{2b}$ frac4b^29c^2 = frac3c2b implies 8b^3 = 27c^3 implies fracbc = frac32 quad dots (ii)$\frac{4b^2}{9c^2} = \frac{3c}{2b} \implies 8b^3 = 27c^3 \implies \frac{b}{c} = \frac{3}{2} \quad \dots (ii)$ ### Step 1: Finding Ratios Substituting c = frac2b3$c = \frac{2b}{3}$ into equation (i): b^2 = a left(frac2b3right) implies b = frac2a3 implies fracab = frac32$b^2 = a \left(\frac{2b}{3}\right) \implies b = \frac{2a}{3} \implies \frac{a}{b} = \frac{3}{2}$ Thus, consolidating all parts: a : b = 9 : 6$a : b = 9 : 6$ b : c = 6 : 4$b : c = 6 : 4$ a : b : c = 9 : 6 : 4$a : b : c = 9 : 6 : 4$ ### Pattern Recognition Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2log x = log x^2$2\log x = \log x^2$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula 2a_n = a_n+1 + a_n+2$2a_n = a_{n+1} + a_{n+2}$ ### Core Logic Let the terms of the Geometric Progression have a common ratio r$r$. Substituting the geometric forms into the arithmetic mean relationship: 2(a r^n-1) = a r^n + a r^n+1$2(a r^{n-1}) = a r^n + a r^{n+1}$ Dividing out non-zero fields a r^n-1$a r^{n-1}$: 2 = r + r^2 implies r^2 + r - 2 = 0$2 = r + r^2 \implies r^2 + r - 2 = 0$ (r + 2)(r - 1) = 0$(r + 2)(r - 1) = 0$ Since a_2 neq a_1$a_2 \neq a_1$, we have r neq 1$r \neq 1$. Thus, the common ratio is r = -2$r = -2$. ### Step 1: Evaluating the Target Partial Difference We need to evaluate: S_20 - S_18 = T_19 + T_20$S_{20} - S_{18} = T_{19} + T_{20}$ T_19 + T_20 = a r^18 + a r^19 = a r^18(1 + r)$T_{19} + T_{20} = a r^{18} + a r^{19} = a r^{18}(1 + r)$ Substituting a = frac18$a = \frac{1}{8}$ and r = -2$r = -2$: T_19 + T_20 = frac18 (-2)^18 (1 - 2) = frac12^3 cdot 2^18 cdot (-1) = -2^15$T_{19} + T_{20} = \frac{1}{8} (-2)^{18} (1 - 2) = \frac{1}{2^3} \cdot 2^{18} \cdot (-1) = -2^{15}$ ### Pattern Recognition Partial sum differences simplify into standard standalone term values (S_n - S_n-2 = T_n + T_n-1$S_n - S_{n-2} = T_n + T_{n-1}$). This eliminates the need to apply long fraction sum formats. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula T_n = a + (n-1)d$T_n = a + (n-1)d$ D_textcommon = textLCM(d_1, d_2)$D_{\text{common}} = \text{LCM}(d_1, d_2)$ ### Core Logic First Progression (S_1$S_1$): 4, 9, 14, 19, dots$4, 9, 14, 19, \dots$ Common difference d_1 = 5$d_1 = 5$. Last term (T_25$T_{25}$) = 4 + (25-1)5 = 4 + 120 = 124$= 4 + (25-1)5 = 4 + 120 = 124$. Second Progression (S_2$S_2$): 3, 6, 9, 12, dots$3, 6, 9, 12, \dots$ Common difference d_2 = 3$d_2 = 3$. Last term (T_37$T_{37}$) = 3 + (37-1)3 = 3 + 108 = 111$= 3 + (37-1)3 = 3 + 108 = 111$. ### Step 1: Forming the Common AP By inspecting the sequences, the first common term (a_textcommon$a_{\text{common}}$) is 9$9$. The common difference of the new series is the LCM of the original differences: D_textcommon = textLCM(5, 3) = 15$D_{\text{common}} = \text{LCM}(5, 3) = 15$ Thus, the common terms form a new AP: 9, 24, 39, 54, dots$9, 24, 39, 54, \dots$ ### Step 2: Bounding the Sequence The last term of the common AP must be less than or equal to the smallest maximum limit of the two series. Here, min(124, 111) = 111$\min(124, 111) = 111$. So, the n$n$-th term of the common sequence is bounded by 111: 9 + (n-1)15 le 111$9 + (n-1)15 \le 111$ 15(n-1) le 102$15(n-1) \le 102$ (n-1) le frac10215 = 6.8$(n-1) \le \frac{102}{15} = 6.8$ n le 7.8$n \le 7.8$ Since n$n$ must be an integer, n = 7$n = 7$. ### Pattern Recognition The common terms of two APs always form a new AP. Its common difference is the LCM of the original differences. Find the first common term manually, then cap the n$n$-th term inequality with the smallest end-boundary of the original sets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula S_infty = fraca1-r + fracdr(1-r)^2$S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ (Sum of an infinite Arithmetico-Geometric Progression, where a$a$ is the first AP term, d$d$ is common difference, and r$r$ is geometric ratio). ### Core Logic The series given is an AGP. However, let's look at it explicitly. Let S = 8$S = 8$. 8 = 3 + frac3+p4 + frac3+2p4^2 + dots$8 = 3 + \frac{3+p}{4} + \frac{3+2p}{4^2} + \dots$ Multiply the entire equation by the geometric ratio (1/4$1/4$): frac84 = frac34 + frac3+p4^2 + frac3+2p4^3 + dots$\frac{8}{4} = \frac{3}{4} + \frac{3+p}{4^2} + \frac{3+2p}{4^3} + \dots$ ### Step 1: Shift and Subtract Subtract the shifted series from the original series: 8 - frac84 = 3 + left(frac3+p4 - frac34right) + left(frac3+2p4^2 - frac3+p4^2right) + dots$8 - \frac{8}{4} = 3 + \left(\frac{3+p}{4} - \frac{3}{4}\right) + \left(\frac{3+2p}{4^2} - \frac{3+p}{4^2}\right) + \dots$ 8 - 2 = 3 + fracp4 + fracp4^2 + fracp4^3 + dots$8 - 2 = 3 + \frac{p}{4} + \frac{p}{4^2} + \frac{p}{4^3} + \dots$ 6 = 3 + fracp4 left( 1 + frac14 + frac14^2 + dots right)$6 = 3 + \frac{p}{4} \left( 1 + \frac{1}{4} + \frac{1}{4^2} + \dots \right)$ ### Step 2: Summing the pure Infinite GP The term in parentheses is an infinite geometric series with a=1$a=1$ and r=1/4$r=1/4$. Sum = frac11 - 1/4 = frac13/4 = frac43$= \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$ ### Step 3: Final Output Evaluation Substitute this sum back: 6 = 3 + fracp4 times frac43$6 = 3 + \frac{p}{4} \times \frac{4}{3}$ 6 - 3 = fracp3$6 - 3 = \frac{p}{3}$ 3 = fracp3 Rightarrow p = 9$3 = \frac{p}{3} \Rightarrow p = 9$ ### Pattern Recognition The shift-and-subtract technique natively nullifies the arithmetic growth leaving behind a uniform geometric progression. Using the AGP direct formula S = a/(1-r) + dr/(1-r)^2$S = a/(1-r) + dr/(1-r)^2$ works perfectly here as well. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series