Consider an A.P. of positive integers, whose \sum of the first three terms is 54 and the \sum of the first twenty terms lies between 1600 and 1800. Then its 11^textth$11^{\text{th}}$ term is:

### 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

### 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 n^textth text term of a GP: T_n = a r^n-1$$n^{\text{th}} \text{ term of a GP: } T_n = a r^{n-1}$$ ### Core Logic For the first Geometric Progression (GP): First term t_1 = a$t_1 = a$ Third term t_3 = b = a r_1^2 Rightarrow r_1^2 = fracba$t_3 = b = a r_1^2 \Rightarrow r_1^2 = \frac{b}{a}$ The 11^textth$11^{\text{th}}$ term is: t_11 = a r_1^10 = a (r_1^2)^5 = a left(fracbaright)^5$$t_{11} = a r_1^{10} = a (r_1^2)^5 = a \left(\frac{b}{a}\right)^5$$ For the second Geometric Progression (GP): First term T_1 = a$T_1 = a$ Fifth term T_5 = a r_2^4 = b Rightarrow r_2^4 = fracba Rightarrow r_2 = left(fracbaright)^1/4$T_5 = a r_2^4 = b \Rightarrow r_2^4 = \frac{b}{a} \Rightarrow r_2 = \left(\frac{b}{a}\right)^{1/4}$ ### Step 1: Equating the Terms The p^textth$p^{\text{th}}$ term of the second GP is: T_p = a r_2^p-1 = a left(left(fracbaright)^1/4right)^p-1 = a left(fracbaright)^fracp-14$$T_p = a r_2^{p-1} = a \left(\left(\frac{b}{a}\right)^{1/4}\right)^{p-1} = a \left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$ Given that t_11 = T_p$t_{11} = T_p$: a left(fracbaright)^5 = a left(fracbaright)^fracp-14$$a \left(\frac{b}{a}\right)^5 = a \left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$ ### Step 2: Solving for p Since a$a$ and b$b$ are distinct positive real numbers, fracba neq 1$\frac{b}{a} \neq 1$. Therefore, we can equate the exponents: 5 = fracp - 14$$5 = \frac{p - 1}{4}$$ 20 = p - 1 Rightarrow p = 21$$20 = p - 1 \Rightarrow p = 21$$ ### Pattern Recognition Express the common ratios strictly in terms of powers of (b/a)$(b/a)$ to bypass isolated radical tracking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series