Contents / विषय-सूची (8)
  1. 1. Core Concepts & Key Formulas 1. खास Concepts और Formulas
  2. 2. Interactive Sequence & Progression Explorer 2. Interactive Sequence और Progression Explorer
  3. 3. In-Text Proofs & Conceptual Foundations 3. In-Text Proofs और खास विचार
  4. 4. Exercise Set 8.1 (NCERT p. 179) 4. Exercise Set 8.1 (NCERT p. 179)
  5. 5. Exercise Set 8.2 (NCERT p. 185) 5. Exercise Set 8.2 (NCERT p. 185)
  6. 6. Exercise Set 8.3 (NCERT pp. 193–194) 6. Exercise Set 8.3 (NCERT pp. 193–194)
  7. 7. End-of-Chapter Exercises (NCERT pp. 194–195) 7. End-of-Chapter Exercises (NCERT pp. 194–195)
  8. 8. CBSE Board Examination Questions (PYQs) 8. CBSE Board परीक्षा प्रश्न (PYQs)
RK Institute · Class 9 Mathematics

Chapter 08: Predicting What Comes Next — Exploring Sequences and Progressions

Arithmetic Progressions (AP), Geometric Progressions (GP), Explicit & Recursive Rules, Gauss Summation, and Fractals
NCERT Class 9 Mathematics (Ganita Manjari) · Chapter 08 · pp. 174–196

Chapter 08: Predicting What Comes Next — Sequences और Progressions की खोज

Arithmetic Progressions (AP), Geometric Progressions (GP), Explicit और Recursive Rules, Gauss Summation, और Fractals
NCERT Class 9 Mathematics (Ganita Manjari) · Chapter 08 · pp. 174–196

1. Core Concepts & Key Formulas 1. खास Concepts और Formulas

Section Goal: Master the algebraic and geometric foundations of number patterns: formulate explicit formulas $t_n$ and recursive rules $t_n = f(t_{n-1})$, analyze linear growth in Arithmetic Progressions ($d = \text{const}$), analyze exponential scaling in Geometric Progressions ($r = \text{const}$), and evaluate finite sums and fractal series.
Section Goal: Number patterns के algebraic और geometric base को समझें: Explicit formula $t_n$ और recursive rule $t_n = f(t_{n-1})$ बनाना, Arithmetic Progressions में linear growth ($d = \text{const}$) और Geometric Progressions में exponential scaling ($r = \text{const}$) का विश्लेषण करना, और partial sums और fractals को हल करना।

1. Triangular & Square Numbers

$$T_n = \frac{n(n+1)}{2}, \quad S_n = n^2$$

• Consecutive sum: $T_{n-1} + T_n = n^2$
• Odd numbers sum: $1 + 3 + 5 + \dots + (2n-1) = n^2$

2. Explicit vs Recursive Rules

$$t_n = f(n) \quad \text{vs} \quad t_n = f(t_{n-1})$$

• Explicit: Direct evaluation from index $n$
• Recursive: Step-by-step evaluation from previous term

3. Arithmetic Progression (AP)

$$t_n = a + (n-1)d, \quad d = t_k - t_{k-1}$$

Linear progression. Plotting $(n, t_n)$ forms points lying strictly on a straight line with slope $d$.

4. Sum of First $n$ Terms of AP

$$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$$

For natural numbers: $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ (Gauss pairing / Āryabhaṭa rule).

5. Geometric Progression (GP)

$$t_n = a r^{n-1}, \quad r = \frac{t_k}{t_{k-1}}$$

Exponential progression. Successive terms grow by constant multiplicative factor $r$.

6. Fractals (Sierpiński Geometry)

$$N_{\text{carpet}}(n) = 8^n, \quad A_{\text{carpet}}(n) = \left(\frac{8}{9}\right)^n A_0$$

Sierpiński Triangle: $N(n) = 3^n$, Area $A(n) = (\frac{3}{4})^n A_0 \to 0$ as $n \to \infty$.

2. Interactive Sequence & Progression Explorer 2. Interactive Sequence और Progression Explorer

📈 Progression Visualizer & Plotter

AP (Linear) vs GP (Exponential)
23
nth Term ($t_n$)
100
Sum of n Terms ($S_n$)
Linear ($d=3$)
Progression Nature

3. In-Text Proofs & Conceptual Foundations 3. In-Text Proofs और खास विचार

Think & Reflect · p.175 Prove that the sum of any two consecutive triangular numbers is a square number: $T_{n-1} + T_n = n^2$. [NCERT p.175] Think & Reflect · p.175 prove करें कि दो लगातार triangular numbers का sum एक square number होता है: $T_{n-1} + T_n = n^2$। [NCERT p.175]
1. Formulas for Triangular Numbers $$T_n = \frac{n(n+1)}{2}, \quad T_{n-1} = \frac{(n-1)n}{2}$$
2. Algebraic Sum $$T_{n-1} + T_n = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = \frac{n[(n-1) + (n+1)]}{2} = \frac{n(2n)}{2} = \mathbf{n^2}$$
3. Geometric Verification A triangular staircase of $T_{n-1}$ dots inverts and fits seamlessly against a staircase of $T_n$ dots to form an $n \times n$ square array of $n^2$ dots (Figure 8.1 & 8.2).
🎯 Final Answer: $T_{n-1} + T_n = \mathbf{n^2}$ (proved algebraically and geometrically).
1. formula $T_n = \frac{n(n+1)}{2}$ और $T_{n-1} = \frac{(n-1)n}{2}$।
2. sum करने पर $\frac{n(n-1) + n(n+1)}{2} = \frac{2n^2}{2} = \mathbf{n^2}$।
3. Geometric रूप दो staircases मिलकर $n \times n$ square array बनाते हैं।
🎯 final answer: $T_{n-1} + T_n = \mathbf{n^2}$ proved होता है।
Think & Reflect · p.175 Show that the sum of the first $n$ odd natural numbers is always $n^2$: $1 + 3 + 5 + \dots + (2n-1) = n^2$. [NCERT p.175] Think & Reflect · p.175 दिखाएं कि पहले $n$ odd natural numbers का sum हमेशा $n^2$ होता है: $1 + 3 + 5 + \dots + (2n-1) = n^2$। [NCERT p.175]
1. L-Shaped Gnomon Dissection • $n=1: 1 = 1^2$
• $n=2: 1 + 3 = 4 = 2^2$
• $n=3: 1 + 3 + 5 = 9 = 3^2$
• $n=4: 1 + 3 + 5 + 7 = 16 = 4^2$
2. General Inductive Step Adding an L-shaped border of $(2k+1)$ dots around a $k \times k$ square yields a $(k+1) \times (k+1)$ square: $$k^2 + (2k+1) = (k+1)^2$$
🎯 Final Answer: $\sum_{k=1}^n (2k-1) = \mathbf{n^2}$.
1. Pattern $1=1^2, 1+3=2^2, 1+3+5=3^2$।
2. Geometric rule $k^2$ square में $(2k+1)$ border जोड़ने पर $(k+1)^2$ बनता है।
🎯 final answer: odd numbers का sum $= \mathbf{n^2}$ proved होता है।
Think & Reflect · p.181 Explain why plotting the points $(n, t_n)$ of an Arithmetic Progression on a Cartesian plane always forms a straight line. [NCERT p.181] Think & Reflect · p.181 समझाएं कि Cartesian plane पर Arithmetic Progression के बिंदुओं $(n, t_n)$ को दर्शाने पर हमेशा एक सीधी रेखा (straight line) क्यों बनती है। [NCERT p.181]
1. General Term as a Linear Function The explicit formula for an AP is: $$t_n = a + (n-1)d = d \cdot n + (a - d)$$
2. Comparison with Line Equation $y = mx + c$ Replacing index $n$ with continuous variable $x$ gives $y = d x + (a - d)$, which is a linear equation in two variables with constant slope $m = d$ and $y$-intercept $(a-d)$.
3. Slope Invariance The rate of change $\frac{\Delta t_n}{\Delta n} = \frac{t_{n+1} - t_n}{1} = d$ is constant everywhere, ensuring collinearity.
🎯 Final Answer: Points of an AP are collinear because $t_n = d \cdot n + (a-d)$ is a linear polynomial in $n$ of degree 1.
1. Linear equation $t_n = d \cdot n + (a - d)$ एक linear polynomial है।
2. $y = mx + c$ से तुलना इसकी slope $m = d$ हमेशा constant रहती है।
🎯 final answer: Constant slope $d$ होने के कारण सभी बिंदु एक straight line पर स्थित होते हैं।
Think & Reflect · p.183 Derive Gauss's formula for the sum of the first $n$ natural numbers $S = 1 + 2 + \dots + n = \frac{n(n+1)}{2}$ using the reversal and pairing method. [NCERT p.183] Think & Reflect · p.183 Reversal और pairing विधि का उपsum करके पहले $n$ natural numbers के sum $S = \frac{n(n+1)}{2}$ का Gauss formula find करें। [NCERT p.183]
1. Writing Sum in Forward and Reverse Order $$S = 1 + 2 + 3 + \dots + (n-1) + n$$ $$S = n + (n-1) + (n-2) + \dots + 2 + 1$$
2. Adding the Two Equations Column-Wise $$2S = (1+n) + (2+n-1) + (3+n-2) + \dots + (n+1) = \underbrace{(n+1) + (n+1) + \dots + (n+1)}_{n \text{ times}}$$
3. Solving for $S$ $$2S = n(n+1) \implies \mathbf{S = \frac{n(n+1)}{2}}$$
🎯 Final Answer: $\sum_{k=1}^n k = \mathbf{\frac{n(n+1)}{2}}$.
1. सीधे व उल्टे order में लिखना $S = 1 + 2 + \dots + n$ और $S = n + \dots + 1$।
2. दोनों को जोड़ने पर $2S = n(n+1)$।
3. हल $S = \mathbf{\frac{n(n+1)}{2}}$।
🎯 final answer: $S = \mathbf{\frac{n(n+1)}{2}}$ proved होता है।
Think & Reflect · p.189 In the Sierpiński triangle fractal (Figure 8.7), why does the number of black triangles form a GP with $r=3$, while the total shaded area forms a GP with $r=\frac{3}{4}$? [NCERT p.189] Think & Reflect · p.189 Sierpiński triangle fractal (Figure 8.7) में काले triangles की number $r=3$ वाली GP और कुल shaded area $r=\frac{3}{4}$ वाली GP क्यों बनाता है? [NCERT p.189]
1. Triangle Count Tripling At each stage, the central quarter triangle is removed from every existing triangle, leaving 3 smaller congruent triangles $\implies N_n = 3 \times N_{n-1} \implies N_n = 3^n$ (GP with $a=1, r=3$).
2. Shaded Area Shrinking Each remaining triangle retains $\frac{3}{4}$ of the area of the previous stage $\implies A_n = \frac{3}{4} A_{n-1} \implies A_n = \left(\frac{3}{4}\right)^n A_0$ (GP with common ratio $r = \frac{3}{4}$).
3. Infinite Limit As $n \to \infty$, $N_n \to \infty$ while $A_n \to 0$ (infinitely many triangles enclosing zero area!).
🎯 Final Answer: Count $N_n = \mathbf{3^n}$, Area $A_n = \mathbf{\left(\frac{3}{4}\right)^n A_0}$.
1. number में वृद्धि हर stage में 3 गुना triangle बनते हैं $\implies N_n = \mathbf{3^n}$।
2. Area में कमी हर stage में पिछला $\frac{3}{4}$ भाग बचता है $\implies A_n = \mathbf{\left(\frac{3}{4}\right)^n A_0}$।
🎯 final answer: triangle number $3^n$ और Area $\left(\frac{3}{4}\right)^n A_0$ होता है।

4. Exercise Set 8.1 (NCERT p. 179) 4. Exercise Set 8.1 (NCERT p. 179)

Ex 8.1 · Q1 Find the first five terms of the sequence in which the $n^{\text{th}}$ term is given by: (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, (iii) $t_n = n^2 - 2n + 3$ for $n \ge 1$. [NCERT p.179] Ex 8.1 · Q1 निम्नलिखित अनुorderों के पहले 5 term find करें: (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, (iii) $t_n = n^2 - 2n + 3$ ($n \ge 1$ के लिए)। [NCERT p.179]
1. Part (i) $t_n = 3n - 4$ $t_1 = 3(1)-4 = \mathbf{-1}$, $t_2 = 3(2)-4 = \mathbf{2}$, $t_3 = 3(3)-4 = \mathbf{5}$, $t_4 = 3(4)-4 = \mathbf{8}$, $t_5 = 3(5)-4 = \mathbf{11}$.
2. Part (ii) $t_n = 2 - 5n$ $t_1 = 2-5(1) = \mathbf{-3}$, $t_2 = 2-5(2) = \mathbf{-8}$, $t_3 = 2-5(3) = \mathbf{-13}$, $t_4 = 2-5(4) = \mathbf{-18}$, $t_5 = 2-5(5) = \mathbf{-23}$.
3. Part (iii) $t_n = n^2 - 2n + 3$ $t_1 = 1-2+3 = \mathbf{2}$, $t_2 = 4-4+3 = \mathbf{3}$, $t_3 = 9-6+3 = \mathbf{6}$, $t_4 = 16-8+3 = \mathbf{11}$, $t_5 = 25-10+3 = \mathbf{18}$.
🎯 Final Answer: (i) $\mathbf{-1, 2, 5, 8, 11}$; (ii) $\mathbf{-3, -8, -13, -18, -23}$; (iii) $\mathbf{2, 3, 6, 11, 18}$.
1. भाग (i) $n=1,2,3,4,5$ रखने पर: $\mathbf{-1, 2, 5, 8, 11}$।
2. भाग (ii) $n=1,2,3,4,5$ रखने पर: $\mathbf{-3, -8, -13, -18, -23}$।
3. भाग (iii) $n=1,2,3,4,5$ रखने पर: $\mathbf{2, 3, 6, 11, 18}$।
🎯 final answer: (i) $-1, 2, 5, 8, 11$; (ii) $-3, -8, -13, -18, -23$; (iii) $2, 3, 6, 11, 18$।
Ex 8.1 · Q2 & Q3 (i) Find the $10^{\text{th}}$ and $15^{\text{th}}$ terms of $t_n = 5n - 3$. (ii) Determine whether 402 is a term of the sequence $t_n = 5n - 3$. [NCERT p.179] Ex 8.1 · Q2 & Q3 (i) $t_n = 5n - 3$ के 10वें और 15वें term find करें। (ii) जांचें कि क्या 402 इस sequence का एक term है। [NCERT p.179]
1. Evaluating Specific Terms $$t_{10} = 5(10) - 3 = 50 - 3 = \mathbf{47}$$ $$t_{15} = 5(15) - 3 = 75 - 3 = \mathbf{72}$$
2. Checking if 402 is a Term $$5n - 3 = 402 \implies 5n = 405 \implies n = \frac{405}{5} = \mathbf{81}$$ Since $81 \in \mathbb{N}$ (a positive integer), 402 is indeed the $81^{\text{st}}$ term.
🎯 Final Answer: (i) $t_{10} = \mathbf{47}, t_{15} = \mathbf{72}$; (ii) Yes, 402 is the $81^{\text{st}}$ term.
1. term value $t_{10} = 50 - 3 = \mathbf{47}$, $t_{15} = 75 - 3 = \mathbf{72}$।
2. 402 की जांच $5n - 3 = 402 \implies 5n = 405 \implies n = \mathbf{81}$ (पूर्ण natural number)।
🎯 final answer: (i) $t_{10}=47, t_{15}=72$; (ii) हाँ, 402 इस sequence का 81वां term है।
Ex 8.1 · Q4 & Q5 (i) Which term of $t_n = 5n - 3$ is 607? (ii) A sequence is defined recursively by $a_1 = 2$, $a_n = 2a_{n-1} + 1$ for $n \ge 2$. Find the first 5 terms and determine if 52 is a term. [NCERT p.179] Ex 8.1 · Q4 & Q5 (i) $t_n = 5n - 3$ का कौन सा term 607 है? (ii) Recursive sequence $a_1 = 2$, $a_n = 2a_{n-1} + 1$ के पहले 5 term find करें और बताएं कि क्या 52 इसका term है। [NCERT p.179]
1. Solving for Index of 607 $$5n - 3 = 607 \implies 5n = 610 \implies n = \mathbf{122}$$
2. Computing Recursive Terms • $a_1 = 2$
• $a_2 = 2(2) + 1 = 5$
• $a_3 = 2(5) + 1 = 11$
• $a_4 = 2(11) + 1 = 23$
• $a_5 = 2(23) + 1 = 47$
• $a_6 = 2(47) + 1 = 95$
3. Checking 52 Since $a_5 = 47$ and $a_6 = 95$, 52 is skipped ($47 < 52 < 95$). Thus, 52 is NOT a term.
🎯 Final Answer: (i) $\mathbf{122^{\text{nd}}\text{ term}}$; (ii) Terms: $\mathbf{2, 5, 11, 23, 47}$; 52 is not a term.
1. 607 का index $5n = 610 \implies n = \mathbf{122}$।
2. Recursive term $a_1=2, a_2=5, a_3=11, a_4=23, a_5=47, a_6=95$।
3. 52 की position 52 value 47 और 95 के बीच छूट जाता है, अतः 52 इसका term नहीं है।
🎯 final answer: (i) 122वां term; (ii) पहले 5 term: $2, 5, 11, 23, 47$; 52 term नहीं है।

5. Exercise Set 8.2 (NCERT p. 185) 5. Exercise Set 8.2 (NCERT p. 185)

Ex 8.2 · Q1 & Q2 (i) Find the $10^{\text{th}}$ and $26^{\text{th}}$ terms of the AP: $3, 8, 13, 18, \dots$. (ii) Which term of the AP: $21, 18, 15, \dots$ is $-81$? Is any term $0$? [NCERT p.185] Ex 8.2 · Q1 & Q2 (i) AP: $3, 8, 13, 18, \dots$ के 10वें और 26वें term find करें। (ii) AP: $21, 18, 15, \dots$ का कौन सा term $-81$ है? क्या कोई term 0 है? [NCERT p.185]
1. Part (i) $a=3, d=5$ $$t_{10} = 3 + (10-1)5 = 3 + 45 = \mathbf{48}$$ $$t_{26} = 3 + (26-1)5 = 3 + 125 = \mathbf{128}$$
2. Part (ii) $a=21, d=-3$ $$t_n = -81 \implies 21 + (n-1)(-3) = -81 \implies -3(n-1) = -102 \implies n-1 = 34 \implies n = \mathbf{35}$$ $$t_n = 0 \implies 21 - 3(n-1) = 0 \implies 3(n-1) = 21 \implies n-1 = 7 \implies n = \mathbf{8}$$
🎯 Final Answer: (i) $t_{10} = \mathbf{48}, t_{26} = \mathbf{128}$; (ii) $-81$ is the $\mathbf{35^{\text{th}}\text{ term}}$, and $0$ is the $\mathbf{8^{\text{th}}\text{ term}}$.
1. भाग (i) $t_{10} = 3 + 45 = \mathbf{48}$, $t_{26} = 3 + 125 = \mathbf{128}$।
2. भाग (ii) $21 - 3(n-1) = -81 \implies n = \mathbf{35}$वां term; और $0$ के लिए $n = \mathbf{8}$वां term।
🎯 final answer: (i) $48$ और $128$; (ii) $-81$ 35वां term है और $0$ 8वां term है।
Ex 8.2 · Q3 & Q4 (i) Find the $n^{\text{th}}$ term and recursive rule for the AP: $11, 8, 5, 2, \dots$. (ii) An AP consists of 50 terms of which $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term. [NCERT p.185] Ex 8.2 · Q3 & Q4 (i) AP: $11, 8, 5, 2, \dots$ का $n$वां term और recursive rule find करें। (ii) 50 terms वाली AP का तीसरा term 12 और final term 106 है। 29वां term find करें। [NCERT p.185]
1. Part (i) $a=11, d=-3$ Explicit rule: $t_n = 11 + (n-1)(-3) = \mathbf{14 - 3n}$.
Recursive rule: $\mathbf{t_1 = 11, \quad t_n = t_{n-1} - 3 \text{ for } n \ge 2}$.
2. Part (ii) Setting up Linear Equations $$t_3 = a + 2d = 12 \quad \text{--- (1)}$$ $$t_{50} = a + 49d = 106 \quad \text{--- (2)}$$ Subtracting (1) from (2): $47d = 94 \implies \mathbf{d = 2}$.
Substituting $d=2$ into (1): $a + 2(2) = 12 \implies \mathbf{a = 8}$.
3. Evaluating $29^{\text{th}}$ Term $$t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}$$
🎯 Final Answer: (i) $t_n = \mathbf{14 - 3n}$, $t_n = t_{n-1} - 3$; (ii) $t_{29} = \mathbf{64}$.
1. भाग (i) Explicit: $t_n = \mathbf{14 - 3n}$; Recursive: $t_1 = 11, t_n = t_{n-1} - 3$।
2. भाग (ii) equation हल $47d = 94 \implies d = 2 \implies a = 8$।
3. 29वां term $t_{29} = 8 + 28(2) = \mathbf{64}$।
🎯 final answer: (i) $t_n = 14 - 3n$; (ii) $t_{29} = \mathbf{64}$।

6. Exercise Set 8.3 (NCERT pp. 193–194) 6. Exercise Set 8.3 (NCERT pp. 193–194)

Ex 8.3 · Q1 & Q2 (i) Find the $12^{\text{th}}$ term of a GP with common ratio $r=2$ whose $8^{\text{th}}$ term is 192. (ii) Find the $10^{\text{th}}$ and $n^{\text{th}}$ terms of the GP: $5, 25, 125, \dots$. [NCERT p.193] Ex 8.3 · Q1 & Q2 (i) $r=2$ वाले GP का 12वां term find करें जिसका 8वां term 192 है। (ii) GP: $5, 25, 125, \dots$ के 10वें और $n$वें term find करें। [NCERT p.193]
1. Part (i) Finding $a$ and $t_{12}$ $$t_8 = a r^7 = 192 \implies a (2^7) = 192 \implies 128 a = 192 \implies a = \frac{192}{128} = \frac{3}{2}$$ $$t_{12} = a r^{11} = \frac{3}{2} (2^{11}) = 3 \times 2^{10} = 3 \times 1024 = \mathbf{3072}$$ Alternative step: $t_{12} = t_8 \times r^4 = 192 \times 2^4 = 192 \times 16 = \mathbf{3072}$.
2. Part (ii) $a=5, r=5$ $$t_n = a r^{n-1} = 5 \times 5^{n-1} = \mathbf{5^n}$$ $$t_{10} = 5^{10} = \mathbf{9765625}$$
🎯 Final Answer: (i) $t_{12} = \mathbf{3072}$; (ii) $t_n = \mathbf{5^n}, t_{10} = \mathbf{9765625}$.
1. भाग (i) $t_{12} = t_8 \times r^4 = 192 \times 16 = \mathbf{3072}$।
2. भाग (ii) $t_n = 5^n$, और $t_{10} = 5^{10} = \mathbf{9765625}$।
🎯 final answer: (i) $3072$; (ii) $t_n = 5^n, t_{10} = 9765625$।
Ex 8.3 · Q3 & Q4 (i) Which term of the sequence $t_n = 3^n + 1$ is 730? (ii) Which term of the GP: $2, 2\sqrt{2}, 4, \dots$ is 128? [NCERT pp. 193–194] Ex 8.3 · Q3 & Q4 (i) Sequence $t_n = 3^n + 1$ का कौन सा term 730 है? (ii) GP: $2, 2\sqrt{2}, 4, \dots$ का कौन सा term 128 है? [NCERT pp. 193–194]
1. Part (i) Solving $3^n + 1 = 730$ $$3^n = 729 = 3^6 \implies n = \mathbf{6\text{ (the } 6^{\text{th}}\text{ term)}}$$
2. Part (ii) GP $a=2, r=\sqrt{2}$ $$t_n = a r^{n-1} = 2 (\sqrt{2})^{n-1} = 128$$ $$(\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12} \implies n - 1 = 12 \implies n = \mathbf{13}$$
🎯 Final Answer: (i) $\mathbf{6^{\text{th}}\text{ term}}$; (ii) $\mathbf{13^{\text{th}}\text{ term}}$.
1. भाग (i) $3^n = 729 = 3^6 \implies n = \mathbf{6}$।
2. भाग (ii) $2(\sqrt{2})^{n-1} = 128 \implies (\sqrt{2})^{n-1} = (\sqrt{2})^{12} \implies n = \mathbf{13}$।
🎯 final answer: (i) 6वां term; (ii) 13वां term।
Ex 8.3 · Q5 A ball is dropped from a height of $24\text{ feet}$. Each time it bounces, it rebounds to $\frac{3}{4}$ of its previous height. (i) Write the GP for heights after each bounce. (ii) What height does it reach on the $5^{\text{th}}$ bounce? (iii) What total distance has the ball travelled by the time it hits the ground for the $6^{\text{th}}$ time? [NCERT p.193] Ex 8.3 · Q5 एक गेंद को 24 फीट की height से गिराया जाता है। हर टप्पे पर यह अपनी पिछली height के $\frac{3}{4}$ भाग तक उछलती है। (i) उछाल की heights की GP लिखें। (ii) 5वें टप्पे पर यह कितनी height पर पहुंचेगी? (iii) 6ठी बार जमीन से टकराने तक गेंद कुल कितनी distance तय कर चुकी होगी? [NCERT p.193]
1. Sequence of Bounce Heights ($a_1 = 18\text{ ft}, r = 0.75$) $$h_1 = 24 \times \frac{3}{4} = 18\text{ ft}, \quad h_2 = 13.5\text{ ft}, \quad h_3 = 10.125\text{ ft}, \quad \dots$$ $$h_n = 24 \left(\frac{3}{4}\right)^n = 18 \left(\frac{3}{4}\right)^{n-1}$$
2. Height on $5^{\text{th}}$ Bounce $$h_5 = 24 \left(\frac{3}{4}\right)^5 = 24 \times \frac{243}{1024} = \frac{729}{128} \approx \mathbf{5.695\text{ feet}}$$
3. Total Distance up to $6^{\text{th}}$ Impact $$\text{Total Distance} = 24 + 2(h_1 + h_2 + h_3 + h_4 + h_5) = 24 + 2 \times 18 \left[\frac{1 - (3/4)^5}{1 - 3/4}\right]$$ $$= 24 + 36 \times 4 \left(1 - \frac{243}{1024}\right) = 24 + 144 \left(\frac{781}{1024}\right) = 24 + 109.83 = \mathbf{133.83\text{ feet}}$$
🎯 Final Answer: (i) $h_n = 18(0.75)^{n-1}$; (ii) $h_5 = \mathbf{5.70\text{ ft}}$; (iii) Total distance $= \mathbf{133.83\text{ feet}}$.
1. heights की GP $18, 13.5, 10.125, \dots$ ($a=18, r=0.75$)।
2. 5वें टप्पे पर height $h_5 = 24(0.75)^5 = \mathbf{5.70\text{ feet}}$।
3. 6ठी टक्कर तक कुल distance $\text{Distance} = 24 + 2(S_5) = \mathbf{133.83\text{ feet}}$।
🎯 final answer: (i) $h_n = 18(0.75)^{n-1}$; (ii) $5.70\text{ ft}$; (iii) कुल distance $= 133.83\text{ ft}$।
Ex 8.3 · Q7 In the Sierpiński square carpet (Figure 8.12): (i) Find the number of red squares in Stages 0 to 5. (ii) Give the explicit and recursive formula for the number of red squares at Stage $n$. (iii) If the Stage 0 square has area $1\text{ sq unit}$, find the area of the red region at Stage $n$. What happens to this area as $n \to \infty$? [NCERT p.194] Ex 8.3 · Q7 Sierpiński square carpet (Figure 8.12) में: (i) Stages 0 से 5 में लाल वर्गों की number find करें। (ii) Stage $n$ के लिए explicit व recursive formula दें। (iii) यदि Stage 0 का area $1$ है, तो Stage $n$ का area find करें। $n \to \infty$ होने पर area का क्या होगा? [NCERT p.194]
1. Number of Squares • Stage 0: $8^0 = \mathbf{1}$
• Stage 1: $8^1 = \mathbf{8}$
• Stage 2: $8^2 = \mathbf{64}$
• Stage 3: $8^3 = \mathbf{512}$
• Stage 4: $8^4 = \mathbf{4096}$
• Stage 5: $8^5 = \mathbf{32768}$
2. Explicit & Recursive Formulas for Count • Explicit: $\mathbf{N(n) = 8^n}$
• Recursive: $\mathbf{N(0) = 1, \quad N(n) = 8 \cdot N(n-1) \text{ for } n \ge 1}$
3. Area of Red Region At each step, $\frac{1}{9}$ is removed and $\frac{8}{9}$ is retained: $$A(n) = \left(\frac{8}{9}\right)^n A_0 = \mathbf{\left(\frac{8}{9}\right)^n}$$ Recursive: $A(0) = 1, A(n) = \frac{8}{9}A(n-1)$. Since $\frac{8}{9} < 1$, as $n \to \infty$, $\lim_{n \to \infty} A(n) = \mathbf{0}$.
🎯 Final Answer: Counts: $1, 8, 64, 512, 4096, 32768$; $N(n) = \mathbf{8^n}$; Area $A(n) = \mathbf{(\frac{8}{9})^n} \to \mathbf{0}$.
1. वर्गों की number $1, 8, 64, 512, 4096, 32768$ ($N_n = 8^n$)।
2. formula Explicit: $N(n) = \mathbf{8^n}$; Recursive: $N(n) = 8 N(n-1)$।
3. Area $A(n) = \mathbf{\left(\frac{8}{9}\right)^n}$। जब $n \to \infty$ तो Area शून्य ($0$) की ओर घटता है।
🎯 final answer: number $= 8^n$; Area $= (\frac{8}{9})^n \to 0$।

7. End-of-Chapter Exercises (NCERT pp. 194–195) 7. End-of-Chapter Exercises (NCERT pp. 194–195)

End Ex · Q1 & Q2 (i) Find the $31^{\text{st}}$ term of an AP whose $11^{\text{th}}$ term is 38 and $16^{\text{th}}$ term is 73. (ii) Determine the AP whose $3^{\text{rd}}$ term is 16 and $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12. [NCERT p.194] End Ex · Q1 & Q2 (i) AP का 31वां term find करें जिसका 11वां term 38 और 16वां term 73 है। (ii) AP निर्धारित करें जिसका तीसरा term 16 है और 7वां term 5वें term से 12 अधिक है। [NCERT p.194]
1. Solving Q1 System $$t_{11} = a + 10d = 38, \quad t_{16} = a + 15d = 73$$ Subtracting: $5d = 35 \implies \mathbf{d = 7} \implies a = 38 - 70 = \mathbf{-32}$. $$t_{31} = a + 30d = -32 + 30(7) = -32 + 210 = \mathbf{178}$$
2. Solving Q2 System $$t_7 - t_5 = 12 \implies (a+6d) - (a+4d) = 12 \implies 2d = 12 \implies \mathbf{d = 6}$$ $$t_3 = a + 2d = 16 \implies a + 2(6) = 16 \implies \mathbf{a = 4}$$ The AP is: $4, 4+6, 4+12, \dots \implies \mathbf{4, 10, 16, 22, \dots}$.
🎯 Final Answer: Q1: $t_{31} = \mathbf{178}$; Q2: AP is $\mathbf{4, 10, 16, 22, \dots}$.
1. Q1 हल $5d = 35 \implies d=7, a=-32 \implies t_{31} = -32 + 210 = \mathbf{178}$।
2. Q2 हल $2d = 12 \implies d=6, a=4 \implies \text{AP: } \mathbf{4, 10, 16, 22, \dots}$।
🎯 final answer: Q1: $t_{31} = 178$; Q2: $4, 10, 16, 22, \dots$।
End Ex · Q3 & Q4 (i) How many three-digit numbers are divisible by 7? (ii) How many multiples of 4 lie between 10 and 250? [NCERT p.195] End Ex · Q3 & Q4 (i) 7 से divisible होने वाली 3-अंकीय numbers कितनी हैं? (ii) 10 और 250 के बीच 4 के कितने multiples (multiples) हैं? [NCERT p.195]
1. Q3 Three-Digit Numbers Divisible by 7 Smallest 3-digit number divisible by 7 is $105$; largest is $994$. $$105, 112, 119, \dots, 994 \quad (a=105, d=7, t_n=994)$$ $$105 + (n-1)7 = 994 \implies 7(n-1) = 889 \implies n-1 = 127 \implies \mathbf{n = 128}$$
2. Q4 Multiples of 4 between 10 and 250 Smallest multiple $> 10$ is $12$; largest $< 250$ is $248$. $$12, 16, 20, \dots, 248 \quad (a=12, d=4, t_n=248)$$ $$12 + (n-1)4 = 248 \implies 4(n-1) = 236 \implies n-1 = 59 \implies \mathbf{n = 60}$$
🎯 Final Answer: (i) $\mathbf{128\text{ numbers}}$; (ii) $\mathbf{60\text{ multiples}}$.
1. Q3 हल AP: $105, 112, \dots, 994 \implies 7(n-1) = 889 \implies n = \mathbf{128}$।
2. Q4 हल AP: $12, 16, \dots, 248 \implies 4(n-1) = 236 \implies n = \mathbf{60}$।
🎯 final answer: (i) 128 numbers; (ii) 60 multiples।
End Ex · Q5 & Q7 (i) Find a GP for which the sum of the first two terms is $-4$ and the $5^{\text{th}}$ term is 4 times the $3^{\text{rd}}$ term. (ii) The number of bacteria doubles every hour. If 30 bacteria were present originally, find the population after 2 hours, 4 hours, and $n$ hours. [NCERT p.195] End Ex · Q5 & Q7 (i) वह GP find करें जिसके पहले दो terms का sum $-4$ है और 5वां term तीसरे term का 4 गुना है। (ii) बैक्टीरिया की number प्रति घंटा double होती है। यदि शुरुआत में 30 बैक्टीरिया थे, तो 2, 4 और $n$ घंटे बाद जनnumber find करें। [NCERT p.195]
1. Q5 GP Determination $$t_5 = 4 t_3 \implies a r^4 = 4 a r^2 \implies r^2 = 4 \implies r = \pm 2$$ • Case 1 ($r=2$): $a + ar = -4 \implies a(1+2) = -4 \implies a = -\frac{4}{3} \implies \mathbf{-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots}$
• Case 2 ($r=-2$): $a(1-2) = -4 \implies -a = -4 \implies a = 4 \implies \mathbf{4, -8, 16, -32, \dots}$
2. Q7 Bacteria Doubling Model Initial $P_0 = 30$, common ratio $r=2$: $$P(t) = 30 \times 2^t$$ • After 2 hours: $30 \times 2^2 = 30 \times 4 = \mathbf{120}$
• After 4 hours: $30 \times 2^4 = 30 \times 16 = \mathbf{480}$
• After $n$ hours: $\mathbf{30 \times 2^n}$
🎯 Final Answer: Q5: $\mathbf{-\frac{4}{3}, -\frac{8}{3}, \dots}$ or $\mathbf{4, -8, 16, \dots}$; Q7: 2h: $\mathbf{120}$, 4h: $\mathbf{480}$, $n$h: $\mathbf{30 \times 2^n}$.
1. Q5 हल $r^2 = 4 \implies r = \pm 2$। GP: $-\frac{4}{3}, -\frac{8}{3}, \dots$ या $4, -8, 16, \dots$।
2. Q7 Bacteria $P(n) = 30 \times 2^n$ $\implies 2\text{h: } \mathbf{120}, 4\text{h: } \mathbf{480}, n\text{h: } \mathbf{30 \times 2^n}$।
🎯 final answer: Q5: GP प्राप्त होती है; Q7: $120, 480, 30 \times 2^n$।
End Ex · Q8 & Q9 (i) The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is 24, and the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44. Find the first three terms. (ii) Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than 1000. [NCERT p.195] End Ex · Q8 & Q9 (i) AP के 4थे व 8वें terms का sum 24 और 6ठे व 10वें terms का sum 44 है। पहले तीन term find करें। (ii) $n$ का वह minimum value find करें जिसके लिए पहले $n$ natural numbers का sum 1000 से अधिक हो। [NCERT p.195]
1. Q8 AP System $$t_4 + t_8 = (a+3d) + (a+7d) = 2a + 10d = 24 \implies a + 5d = 12 \quad \text{--- (1)}$$ $$t_6 + t_{10} = (a+5d) + (a+9d) = 2a + 14d = 44 \implies a + 7d = 22 \quad \text{--- (2)}$$ Subtracting: $2d = 10 \implies \mathbf{d = 5} \implies a = 12 - 25 = \mathbf{-13}$. First 3 terms: $\mathbf{-13, -8, -3}$.
2. Q9 Smallest $n$ for $S_n > 1000$ $$\frac{n(n+1)}{2} > 1000 \implies n(n+1) > 2000$$ Since $44 \times 45 = 1980 < 2000$ and $45 \times 46 = 2070 > 2000$, the smallest integer is $\mathbf{n = 45}$ ($S_{45} = 1035$).
🎯 Final Answer: Q8: First three terms are $\mathbf{-13, -8, -3}$; Q9: Smallest $n = \mathbf{45}$.
1. Q8 हल $2d = 10 \implies d = 5, a = -13 \implies \text{पहले तीन term: } \mathbf{-13, -8, -3}$।
2. Q9 हल $n(n+1) > 2000 \implies 44 \times 45 = 1980, 45 \times 46 = 2070 \implies n = \mathbf{45}$।
🎯 final answer: Q8: $-13, -8, -3$; Q9: $n = \mathbf{45}$।
End Ex · Q10 & Q12 (i) Which term of the GP: $2, 8, 32, \dots$ is 131072? Write explicit and recursive formulas. (ii) If the $4^{\text{th}}, 10^{\text{th}}$, and $16^{\text{th}}$ terms of a GP are $x, y, z$, prove that $x, y, z$ are in GP. [NCERT p.195] End Ex · Q10 & Q12 (i) GP: $2, 8, 32, \dots$ का कौन सा term 131072 है? Explicit व recursive formula लिखें। (ii) यदि GP के चौथे, 10वें और 16वें term $x, y, z$ हैं, तो prove करें कि $x, y, z$ GP में हैं। [NCERT p.195]
1. Q10 Solving $a=2, r=4$ $$t_n = 2 \times 4^{n-1} = 131072 \implies 4^{n-1} = 65536 = 4^8 \implies n-1 = 8 \implies \mathbf{n = 9}$$ • Explicit: $\mathbf{t_n = 2 \cdot 4^{n-1}}$
• Recursive: $\mathbf{t_1 = 2, \quad t_n = 4 t_{n-1} \text{ for } n \ge 2}$
2. Q12 Proof of GP for $x, y, z$ Let GP have first term $a$ and common ratio $r$: $$x = t_4 = a r^3, \quad y = t_{10} = a r^9, \quad z = t_{16} = a r^{15}$$ $$y^2 = (a r^9)^2 = a^2 r^{18}$$ $$x \cdot z = (a r^3)(a r^{15}) = a^2 r^{18}$$ Since $y^2 = xz$, $x, y, z$ strictly form a geometric progression with common ratio $r^6$.
🎯 Final Answer: Q10: $\mathbf{9^{\text{th}}\text{ term}}$, $t_n = 2 \cdot 4^{n-1}$; Q12: $y^2 = xz = a^2 r^{18}$ (proved).
1. Q10 हल $4^{n-1} = 65536 = 4^8 \implies n = \mathbf{9}$वां term। Explicit: $t_n = 2 \cdot 4^{n-1}$।
2. Q12 हल $x=ar^3, y=ar^9, z=ar^{15} \implies y^2 = a^2 r^{18} = xz \implies x, y, z$ GP में हैं।
🎯 final answer: Q10: 9वां term; Q12: $y^2 = xz$ proved होता है।
End Ex · Q14 & Q15 (i) Given $P_1=1, P_2=2$, and $P_n = P_1 + \dots + P_{n-1} + 1$ for $n > 2$. Find $P_1$ to $P_8$ and give the explicit formula. (ii) Given $W_1=1, W_2=2$, and $W_n = W_1 + \dots + W_{n-2} + 2$ for $n > 2$. Find $W_1$ to $W_8$ and identify the sequence. [NCERT p.195] End Ex · Q14 & Q15 (i) $P_1=1, P_2=2$, और $P_n = P_1 + \dots + P_{n-1} + 1$ ($n>2$) के लिए $P_1$ से $P_8$ और explicit formula find करें। (ii) $W_1=1, W_2=2$, और $W_n = W_1 + \dots + W_{n-2} + 2$ ($n>2$) के लिए $W_1$ से $W_8$ find करें। [NCERT p.195]
1. Q14 Sequence Evaluation • $P_1 = 1$
• $P_2 = 2$
• $P_3 = 1+2+1 = 4$
• $P_4 = 1+2+4+1 = 8$
• $P_5 = 16, P_6 = 32, P_7 = 64, P_8 = 128$
Simpler recursive rule: $P_n = 2 P_{n-1}$ for $n \ge 3$. Explicit formula: $\mathbf{P_n = 2^{n-1}}$.
2. Q15 Sequence Evaluation • $W_1 = 1$
• $W_2 = 2$
• $W_3 = W_1 + 2 = 1 + 2 = 3$
• $W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$
• $W_5 = 1 + 2 + 3 + 2 = 8$
• $W_6 = 1 + 2 + 3 + 5 + 2 = 13$
• $W_7 = 21, W_8 = 34$
This is the famous Fibonacci Sequence ($F_1=1, F_2=2, F_3=3, F_4=5, F_5=8, F_6=13, F_7=21, F_8=34$).
🎯 Final Answer: Q14: Powers of two $P_n = \mathbf{2^{n-1}}$ ($1, 2, 4, 8, 16, 32, 64, 128$); Q15: Fibonacci Sequence ($1, 2, 3, 5, 8, 13, 21, 34$).
1. Q14 हल $P_1=1, P_2=2, P_3=4, P_4=8, P_5=16, P_6=32, P_7=64, P_8=128$ $\implies P_n = \mathbf{2^{n-1}}$।
2. Q15 हल $1, 2, 3, 5, 8, 13, 21, 34$ $\implies$ यह प्रproved Fibonacci Sequence है।
🎯 final answer: Q14: $P_n = 2^{n-1}$; Q15: Fibonacci sequence ($1, 2, 3, 5, 8, 13, 21, 34$)।

8. CBSE Board Examination Questions (PYQs) 8. CBSE Board परीक्षा प्रश्न (PYQs)

CBSE 2023 · 1 Mark If the $n^{\text{th}}$ term of an AP is $t_n = 7 - 4n$, then its common difference $d$ is: (A) $7$, (B) $-4$, (C) $4$, (D) $3$. [CBSE Class 9/10 Board Standard] CBSE 2023 · 1 Mark यदि किसी AP का $n$वां term $t_n = 7 - 4n$ है, तो इसका common difference $d$ होगा: (A) $7$, (B) $-4$, (C) $4$, (D) $3$। [CBSE Class 9/10 Board Standard]
1. Method 1 (Coefficient of $n$) For any AP with $t_n = An + B$, the common difference is the linear coefficient of $n \implies d = \mathbf{-4}$.
2. Method 2 (Consecutive Terms) $t_1 = 7 - 4(1) = 3$, $t_2 = 7 - 4(2) = -1 \implies d = t_2 - t_1 = -1 - 3 = \mathbf{-4}$.
Marking Scheme:
• [0.5M] Stating $d = t_2 - t_1$ or identifying linear slope
• [0.5M] Correct answer option (B) $-4$
🎯 Final Answer: Correct Option: (B) $-4$.
1. calculation $t_1 = 3, t_2 = -1 \implies d = -1 - 3 = \mathbf{-4}$।
🎯 final answer: सही विकल्प: (B) $-4$।
CBSE 2022 · 2 Marks Find the sum of all two-digit odd positive natural numbers. [CBSE Board Standard] CBSE 2022 · 2 Marks सभी 2-अंकीय odd धनात्मक natural numbers का sum find करें। [CBSE Board Standard]
1. Identifying the AP The sequence of 2-digit odd numbers is: $11, 13, 15, \dots, 99$. Here $a = 11, d = 2, l = t_n = 99$.
2. Finding Number of Terms ($n$) $$11 + (n-1)2 = 99 \implies 2(n-1) = 88 \implies n-1 = 44 \implies \mathbf{n = 45}$$
3. Sum Calculation $$S_{45} = \frac{n}{2}(a + l) = \frac{45}{2}(11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = \mathbf{2475}$$
Marking Scheme:
• [1.0M] Identifying AP and calculating $n = 45$
• [1.0M] Applying $S_n = \frac{n}{2}(a+l)$ and evaluating $2475$
🎯 Final Answer: Sum of 2-digit odd numbers $= \mathbf{2475}$.
1. AP $11, 13, \dots, 99 \implies a=11, d=2, l=99$।
2. number of terms $11 + 2(n-1) = 99 \implies n = 45$।
3. sum $S_{45} = \frac{45}{2}(11 + 99) = 45 \times 55 = \mathbf{2475}$।
🎯 final answer: कुल sum $= \mathbf{2475}$ है।
CBSE 2024 · 3 Marks The $8^{\text{th}}$ term of an AP is zero. Prove that its $38^{\text{th}}$ term is triple its $18^{\text{th}}$ term. [CBSE Board Standard] CBSE 2024 · 3 Marks किसी AP का 8वां term शून्य है। prove करें कि इसका 38वां term इसके 18वें term का तीन गुना होगा। [CBSE Board Standard]
1. Given Condition $$t_8 = a + 7d = 0 \implies \mathbf{a = -7d}$$
2. Expressing $t_{18}$ in terms of $d$ $$t_{18} = a + 17d = (-7d) + 17d = \mathbf{10d}$$
3. Expressing $t_{38}$ in terms of $d$ $$t_{38} = a + 37d = (-7d) + 37d = \mathbf{30d}$$
4. Ratio Comparison $$\frac{t_{38}}{t_{18}} = \frac{30d}{10d} = 3 \implies \mathbf{t_{38} = 3 \times t_{18}}$$
Marking Scheme:
• [1.0M] Establishing $a = -7d$ from $t_8 = 0$
• [1.0M] Calculating $t_{18} = 10d$ and $t_{38} = 30d$
• [1.0M] Concluding $t_{38} = 3 t_{18}$
🎯 Final Answer: $t_{38} = 30d = 3(10d) = \mathbf{3 \times t_{18}}$ (proved).
1. दिया गया value $t_8 = a + 7d = 0 \implies a = -7d$।
2. $t_{18}$ और $t_{38}$ $t_{18} = -7d + 17d = 10d$, और $t_{38} = -7d + 37d = 30d$।
3. संबंध $t_{38} = 30d = 3 \times (10d) = \mathbf{3 \times t_{18}}$ proved.
🎯 final answer: $t_{38} = 3 \times t_{18}$ proved होता है।
CBSE 2024 · 4 Marks (Case Study) Industrial Equipment Depreciation & Output Modelling: A manufacturing machine produces 5000 units in its first year. Due to wear, its production decreases by 200 units every subsequent year (forming an AP). Simultaneously, the machine's market value depreciates by $10\%$ every year (forming a GP) from an initial purchase price of ₹10,00,000. (i) How many units will the machine produce in its $8^{\text{th}}$ year? (ii) What is the total cumulative production over the first 10 years? (iii) What will be the machine's market value at the end of the $2^{\text{nd}}$ year? (iv) In which year will the annual production drop to 3200 units? [CBSE Class 9/10 Case Study Benchmark] CBSE 2024 · 4 Marks (Case Study) Industrial Equipment Depreciation & Output Modelling: एक मशीन पहले वर्ष 5000 इकाइयों का उत्पादन करती है और हर वर्ष उत्पादन 200 इकाइयां घटता है (AP)। साथ ही, ₹10,00,000 के खरीद मूल्य से इसका बाजार मूल्य हर वर्ष $10\%$ घटता है (GP)। (i) 8वें वर्ष में उत्पादन कितना होगा? (ii) पहले 10 वर्षों में कुल उत्पादन कितना होगा? (iii) 2 वर्ष बाद मशीन का मूल्य क्या होगा? (iv) किस वर्ष उत्पादन घटकर 3200 इकाइयां रह जाएगा? [CBSE Class 9/10 Case Study Benchmark]
1. Part (i) Production in $8^{\text{th}}$ Year ($a=5000, d=-200$) $$t_8 = 5000 + (8-1)(-200) = 5000 - 1400 = \mathbf{3600\text{ units}}$$
2. Part (ii) Total 10-Year Production $$S_{10} = \frac{10}{2}[2(5000) + 9(-200)] = 5[10000 - 1800] = 5 \times 8200 = \mathbf{41000\text{ units}}$$
3. Part (iii) Value after 2 Years ($V_0 = 1000000, r = 0.90$) $$V_2 = 1000000 \times (0.90)^2 = 1000000 \times 0.81 = \mathbf{₹8,10,000}$$
4. Part (iv) Year for 3200 Units $$5000 - 200(n-1) = 3200 \implies 200(n-1) = 1800 \implies n-1 = 9 \implies \mathbf{n = 10^{\text{th}}\text{ year}}$$
Marking Scheme:
• [1.0M] Part (i) $t_8 = 3600$
• [1.0M] Part (ii) $S_{10} = 41000$
• [1.0M] Part (iii) $V_2 = ₹8,10,000$
• [1.0M] Part (iv) $n = 10$
🎯 Final Answer: (i) $\mathbf{3600\text{ units}}$, (ii) $\mathbf{41000\text{ units}}$, (iii) $\mathbf{₹8,10,000}$, (iv) $\mathbf{10^{\text{th}}\text{ year}}$.
1. भाग (i) 8वां वर्ष $t_8 = 5000 - 1400 = \mathbf{3600\text{ units}}$।
2. भाग (ii) 10 वर्ष का कुल उत्पादन $S_{10} = 5(10000 - 1800) = \mathbf{41000\text{ units}}$।
3. भाग (iii) 2 वर्ष बाद मूल्य $1000000 \times 0.81 = \mathbf{₹8,10,000}$।
4. भाग (iv) 3200 का वर्ष $200(n-1) = 1800 \implies n = \mathbf{10}$वां वर्ष।
🎯 final answer: (i) 3600; (ii) 41000; (iii) ₹8,10,000; (iv) 10वां वर्ष।
CBSE 2023 · 5 Marks (i) The ratio of the $11^{\text{th}}$ term to the $18^{\text{th}}$ term of an AP is $2:3$. Find the ratio of the $5^{\text{th}}$ term to the $21^{\text{st}}$ term, and the ratio of the sum of the first 5 terms to the sum of the first 21 terms. (ii) If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find its $n^{\text{th}}$ term and determine which term is 164. [CBSE Board Standard] CBSE 2023 · 5 Marks (i) किसी AP के 11वें और 18वें terms का ratio $2:3$ है। इसके 5वें और 21वें terms का ratio, और पहले 5 terms के sum और पहले 21 terms के sum का ratio find करें। (ii) यदि AP के पहले $n$ terms का sum $S_n = 3n^2 + 5n$ है, तो इसका $n$वां term find करें और बताएं कि कौन सा term 164 है। [CBSE Board Standard]
1. Part (i) Solving Ratio $t_{11} : t_{18} = 2 : 3$ $$\frac{a + 10d}{a + 17d} = \frac{2}{3} \implies 3(a + 10d) = 2(a + 17d) \implies 3a + 30d = 2a + 34d \implies \mathbf{a = 4d}$$
2. Ratio of $5^{\text{th}}$ to $21^{\text{st}}$ Term $$\frac{t_5}{t_{21}} = \frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \mathbf{\frac{1}{3} \quad (1:3)}$$
3. Ratio of Sums $S_5 : S_{21}$ $$\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} = \frac{5[2(4d) + 4d]}{21[2(4d) + 20d]} = \frac{5[12d]}{21[28d]} = \frac{60}{588} = \mathbf{\frac{5}{49} \quad (5:49)}$$
4. Part (ii) Finding $t_n$ from $S_n = 3n^2 + 5n$ $$t_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5]$$ $$= 3n^2 + 5n - [3n^2 - n - 2] = \mathbf{6n + 2}$$ $$6n + 2 = 164 \implies 6n = 162 \implies \mathbf{n = 27}$$
Marking Scheme:
• [1.5M] Proving $a = 4d$ and $t_5 : t_{21} = 1:3$
• [1.5M] Proving $S_5 : S_{21} = 5:49$
• [1.0M] Deriving $t_n = S_n - S_{n-1} = 6n+2$
• [1.0M] Solving $6n+2 = 164 \implies n=27$
🎯 Final Answer: (i) $t_5 : t_{21} = \mathbf{1:3}$, $S_5 : S_{21} = \mathbf{5:49}$; (ii) $t_n = \mathbf{6n+2}$, $164$ is the $\mathbf{27^{\text{th}}\text{ term}}$.
1. भाग (i) $a=4d$ संबंध $3(a+10d) = 2(a+17d) \implies a = 4d$।
2. $t_5 : t_{21}$ $\frac{4d+4d}{4d+20d} = \frac{8}{24} = \mathbf{1:3}$।
3. $S_5 : S_{21}$ $\frac{5(12d)}{21(28d)} = \mathbf{5:49}$।
4. भाग (ii) $t_n$ व 164 $t_n = S_n - S_{n-1} = \mathbf{6n+2} \implies 6n+2=164 \implies n=\mathbf{27}$।
🎯 final answer: (i) $1:3$ और $5:49$; (ii) $t_n = 6n+2$, $n = \mathbf{27}$वां term।