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 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