Chapter 5 — ARITHMETIC PROGRESSIONS

Concept Used (Sum of Arithmetic Progression)

An Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant.

Sum of first n terms is given by:

\[\boxed{S_n = \frac{n}{2}\left[2a + (n-1)d\right]}\]

Where:
a = first term
d = common difference
n = number of terms

Visual Understanding of AP Growth

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Solution Roadmap

• Step 1: Identify first term (a)
• Step 2: Find common difference (d)
• Step 3: Identify number of terms (n)
• Step 4: Substitute in sum formula
• Step 5: Solve carefully step-by-step

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Solution (i)

AP: 2, 7, 12, .... (10 terms)

$$\begin{aligned} a &= 2 \\ d &= 7 - 2 = 5 \\ n &= 10 \end{aligned}$$

Using formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$ $$\begin{aligned} S_{10} &= \frac{10}{2}[2(2) + (10-1)(5)] \\ &= 5[4 + 9 \times 5] \\ &= 5[4 + 45] \\ &= 5 \times 49 \\ &= 245 \end{aligned}$$

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Solution (ii)

AP: -37, -33, -29, .... (12 terms)

$$\begin{aligned} a &= -37 \\ d &= -33 - (-37) = 4 \\ n &= 12 \end{aligned}$$ $$\begin{aligned} S_{12} &= \frac{12}{2}[2(-37) + (12-1)(4)] \\ &= 6[-74 + 11 \times 4] \\ &= 6[-74 + 44] \\ &= 6(-30) \\ &= -180 \end{aligned}$$

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Solution (iii)

AP: 0.6, 1.7, 2.8, .... (100 terms)

$$\begin{aligned} a &= 0.6 \\ d &= 1.7 - 0.6 = 1.1 \\ n &= 100 \end{aligned}$$ $$\begin{aligned} S_{100} &= \frac{100}{2}[2(0.6) + (100-1)(1.1)] \\ &= 50[1.2 + 99 \times 1.1] \\ &= 50[1.2 + 108.9] \\ &= 50[110.1] \\ &= 5505 \end{aligned}$$

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Solution (iv)

AP: \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, ....\) (11 terms)

$$\begin{aligned} a &= \frac{1}{15} \\ d &= \frac{1}{12} - \frac{1}{15} \end{aligned}$$

Take LCM of 12 and 15 = 60

$$\begin{aligned} d &= \frac{5 - 4}{60} = \frac{1}{60} \\ n &= 11 \end{aligned}$$ $$\begin{aligned} S_{11} &= \frac{11}{2}\left[2\left(\frac{1}{15}\right) + (11-1)\left(\frac{1}{60}\right)\right] \\ &= \frac{11}{2}\left[\frac{2}{15} + \frac{10}{60}\right] \\ &= \frac{11}{2}\left[\frac{2}{15} + \frac{1}{6}\right] \end{aligned}$$

Take LCM of 15 and 6 = 30

$$\begin{aligned} &= \frac{11}{2}\left[\frac{4 + 5}{30}\right] \\ &= \frac{11}{2} \times \frac{9}{30} \\ &= \frac{11}{2} \times \frac{3}{10} \\ &= \frac{33}{20} \end{aligned}$$

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Exam Significance

• Direct formula-based questions (very high probability in CBSE boards)
• Common in 2–3 mark questions
• Forms base for higher concepts like sum to n terms problems
• Frequently used in competitive exams (NTSE, Olympiads, SSC, Banking)
• Tests accuracy in substitution and fraction handling

Concept Used

When first term and last term are given, we follow two steps:

• Find number of terms using: \(a_n = a + (n-1)d\)
• Then use sum formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)

Visual Flow of Calculation

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Solution Roadmap

• Step 1: Convert mixed numbers (if any)
• Step 2: Identify \(a\), \(d\), and last term
• Step 3: Use \(a_n = a + (n-1)d\) to find n
• Step 4: Substitute in sum formula
• Step 5: Solve carefully step-by-step

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Solution (i)

AP: \(7,\ 10\frac{1}{2},\ 14,\ \ldots,\ 84\)

$$\begin{aligned} a &= 7 \\ d &= 10\frac{1}{2} - 7 \\ &= \frac{21}{2} - 7 \\ &= \frac{21 - 14}{2} \\ &= \frac{7}{2} \\ a_n &= 84 \end{aligned}$$

Finding number of terms:

$$\begin{aligned} a_n &= a + (n-1)d \\ 84 &= 7 + (n-1)\frac{7}{2} \\ 84 - 7 &= \frac{7(n-1)}{2} \\ 77 &= \frac{7(n-1)}{2} \\ 77 \times 2 &= 7(n-1) \\ 154 &= 7(n-1) \\ n-1 &= \frac{154}{7} \\ n-1 &= 22 \\ n &= 23 \end{aligned}$$

Now calculate sum:

$$\begin{aligned} S_{23} &= \frac{23}{2}\left[2(7) + (23-1)\frac{7}{2}\right] \\ &= \frac{23}{2}\left[14 + \frac{22 \times 7}{2}\right] \\ &= \frac{23}{2}\left[14 + \frac{154}{2}\right] \\ &= \frac{23}{2}(14 + 77) \\ &= \frac{23}{2} \times 91 \\ &= \frac{2093}{2} \\ &= 1046\frac{1}{2} \end{aligned}$$

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Solution (ii)

AP: \(34,\ 32,\ 30,\ \ldots,\ 10\)

$$\begin{aligned} a &= 34 \\\ d &= 32 - 34 = -2 \\\ a_n &= 10 \end{aligned}$$ $$\begin{aligned} 10 &= 34 + (n-1)(-2) \\\ 10 - 34 &= (n-1)(-2) \\\ -24 &= -2(n-1) \\\ n-1 &= \frac{24}{2} \\\ n-1 &= 12 \\\ n &= 13 \end{aligned}$$ $$\begin{aligned} S_{13} &= \frac{13}{2}[2(34) + (13-1)(-2)] \\\ &= \frac{13}{2}[68 + 12(-2)] \\\ &= \frac{13}{2}[68 - 24] \\\ &= \frac{13}{2}(44) \\\ &= 13 \times 22 \\\ &= 286 \end{aligned}$$

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Solution (iii)

AP: \(-5,\ -8,\ -11,\ \ldots,\ -230\)

$$\begin{aligned} a &= -5 \\\ d &= -8 - (-5) = -3 \\\ a_n &= -230 \end{aligned}$$ $$\begin{aligned} -230 &= -5 + (n-1)(-3) \\\ -230 + 5 &= -3(n-1) \\\ -225 &= -3(n-1) \\\ n-1 &= \frac{225}{3} \\\ n-1 &= 75 \\\ n &= 76 \end{aligned}$$ $$\begin{aligned} S_{76} &= \frac{76}{2}[2(-5) + (76-1)(-3)] \\\ &= 38[-10 + 75(-3)] \\\ &= 38[-10 - 225] \\\ &= 38(-235) \\\ &= -8930 \end{aligned}$$

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Exam Significance

• Frequently asked in CBSE board exams (3–4 marks)
• Tests ability to find number of terms correctly
• Mixed fraction handling is a common trap
• Negative APs appear in competitive exams
• Strong base for higher-level AP applications

Concept Framework

In Arithmetic Progression problems, we mainly use:

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

Strategy Flow

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Solution (i)

\[50 = 5 + (n-1)3\] $$\begin{aligned} 45 &= 3(n-1) \\ n-1 &= 15 \\ n &= 16 \end{aligned}$$ $$\begin{aligned} S_{16} &= \frac{16}{2}[2(5) + 15(3)] \\ &= 8[10 + 45] \\ &= 8 \times 55 \\ &= 440 \end{aligned}$$

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Solution: ii

\(\text{given }a = 7, a_{13} = 35,\text{ find }d\text{ and }S_{13}.\) $$\begin{aligned}a_{13}&=a_{1}+\left( 13-1\right) d\\ 35&=7+12d\\ \Rightarrow 12d&=28\\ d&=\dfrac{28}{12}\\ d&=\dfrac{7}{3}\end{aligned}$$ Sum of series $$\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{13}&=\dfrac{13}{2}\left[ 2\times 7+\left( 13-1\right) \times \dfrac{7}{3}\right] \\ &=\dfrac{13}{2}\left[ 14+\dfrac{12\times 7}{3}\right] \\ &=\dfrac{13}{2}\left( 14+28\right) \\ &=\dfrac{13}{2}\times 42\\ &=13\times 21\\ &=273\end{aligned}$$ $$\begin{aligned}\boxed{d=\dfrac{7}{3},\ S_{13}=273}\end{aligned}$$

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Solution: iii

Given: \(a_{12}=37,d=3\)
To find \(S_{12}\)

$$\begin{aligned}a_{12}&=a+11d\\ 37&=a+11\times 3\\ a&=37-33\\ a&=4\end{aligned}$$ $$a=4,d=3$$ $$\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{12}&=\dfrac{12}{2}\left[ 2\times 4+\left( 12-1\right) 3\right] \\ &=6\left[ 8+11\times 3\right] \\ &=6\left[ 8+33\right] \\ &=6\left( 41\right) \\ &=246\end{aligned}$$

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Solution: iv

\(\text{given } a_3 = 15, S_{10}= 125,\text{ find }d\text{ and }a_{10}.\) $$\begin{align}a_{3}&=15\\ S_{10}&=125\\ a_{3}&=a+2d\\ 15&=a+2d\\ \Rightarrow 2d&=15-a\\ d&=\dfrac{15-a}{2}\tag{1}\end{align}$$ $$\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 125&=\dfrac{10}{2}\left[ 2a+9d\right] \\ 12&5=5\left[ 2a+\dfrac{9\left( 15-a\right) }{2}\right] \\ 25&=\dfrac{4a+135-9a}{2}\\ 50&=135-5a\\ 5a&=135-50\\ 5a&=85\\ a&=\dfrac{85}{5}\\ a&=17\end{aligned}$$ Substituting value of a in equation-(1) $$\begin{aligned}d&=\dfrac{15-\left( 17\right) }{2}\\ &=\dfrac{-2}{2}\\ &=-1\end{aligned}$$ $$\begin{aligned}a_{10}&=a_{1}+9d\\ &=17+9\left( -1\right) \\ &=17-9\\ &=8\end{aligned}$$ \[\boxed{d=-1,\ a_{10}=8}\]

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Solution (v)

$$\begin{aligned} 75 &= \frac{9}{2}[2a + 8 \times 5] \\ 75 &= \frac{9}{2}[2a + 40] \\ 150 &= 9(2a + 40) \\ 150 &= 18a + 360 \\ 18a &= -210 \\ a &= -\frac{35}{3} \end{aligned}$$ $$\begin{aligned} a_9 &= a + 8d \\ &= -\frac{35}{3} + 40 \\ &= \frac{85}{3} \end{aligned}$$

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Solution (vi))

$$\begin{aligned} 90 &= \frac{n}{2}[4 + (n-1)8] \\ 180 &= n(8n - 4) \\ 180 &= 8n^2 - 4n \\ 8n^2 - 4n - 180 &= 0 \\ 2n^2 - n - 45 &= 0 \end{aligned}$$ $$\begin{aligned} (2n+9)(n-5)=0 \end{aligned}$$ \[n=5 \quad (\text{valid})\] \[a_5 = 2 + 4 \times 8 = 34\]

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Solution (viii)

\[4 = a + 2(n-1)\] \[n = \frac{6-a}{2}\] Substitute in sum: $$\begin{aligned} -14 &= \frac{n}{2}[2a + (n-1)2] \\ -14 &= \frac{6-a}{4}[2a + 6-a-2] \\ -56 &= (6-a)(a+4) \end{aligned}$$ $$\begin{aligned} -56 &= 6a + 24 - a^2 - 4a \\ -56 &= 2a + 24 - a^2 \\ a^2 - 2a - 80 &= 0 \end{aligned}$$ $$\begin{aligned} (a-10)(a+8)=0 \end{aligned}$$ \[a = -8 \quad (\text{valid})\] \[n = \frac{6-(-8)}{2} = 7\]

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Solution: viii

\(\text{given } a_n= 4, d = 2, S_n= –14,\text{ find }n\text{ and }a.\)

To find \(n\) and \(a\)

$$\begin{align}a_{n}&=a+\left( n-1\right) d\\ a_{n}&=a+2\left( n-1\right) \\ 4&=a+2\left( n-1\right) \\ 2\left( n-1\right) &=4-a\\ n-1&=\dfrac{4-a}{2}\\ n&=\dfrac{4-a}{2}+1\\ &=\dfrac{4-a+2}{2}\\ n&=\dfrac{6-a}{2}\tag{1}\end{align}$$ $$\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ -14&=\dfrac{6-a}{2\times 2}\left[ 2a+\left( \dfrac{6-a}{2}-1\right) \cdot 2\right] \\ -56&=6-a\left[ 2a+6-a-2\right] \\ -56&=\left( 6-a\right) \left( a+4\right) \\ -5&6=6a+24-a^{2}-4a\\ -56&=2a+24-a^{2}\\ \Rightarrow &a^{2}-2a-24-56=0\\ \Rightarrow &a^{2}-2a-80=0\\ \Rightarrow &a^{2}-10a+8a-80=0\\ &a\left( a-10\right) +8\left( a-10\right) =0\\ &\left( a-10\right) \left( a+8\right) =0\\ &a-10=0\\ &a=10\\ &a+8=0\\ &a=-8\end{aligned}$$

\(a=-8\) or \(a = 10\)

\[n=\dfrac{6-a}{2}\]

when \(a =-8\)

$$\begin{aligned}n&=\dfrac{6-\left( -8\right) }{2}\\ &=\dfrac{14}{2}\\ &=7\end{aligned}$$

when \(a = 10\)

$$\begin{aligned}n&=\dfrac{6-10}{2}\\ &=\dfrac{-4}{2}\\ &=-2\end{aligned}$$

\(n\) can not be negative
therefore,

\[\boxed{n=7,\; a=-8}\]

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Solution (ix)

\[192 = 4[6 + 7d]\] \[48 = 6 + 7d\] \[7d = 42\] \[d = 6\]

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Solution (x)

\[144 = \frac{9}{2}(a + 28)\] \[288 = 9a + 252\] \[9a = 36\] \[a = 4\]

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Exam Significance

• Most important mixed-type AP question pattern (CBSE favorite)
• Tests multi-formula application
• Quadratic formation (very high exam probability)
• Concept integration (aₙ + Sₙ combined)
• Common in NTSE, Olympiad, SSC, Banking

Concept Used

When sum of AP is given and number of terms is unknown, we:

• Use sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\)
• Form a quadratic equation in \(n\)
• Solve and reject negative/non-integer values

Visual Idea of Increasing Sum

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Solution Roadmap

• Step 1: Identify \(a\) and \(d\)
• Step 2: Substitute into sum formula
• Step 3: Simplify to quadratic equation
• Step 4: Solve and validate value of \(n\)

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Solution

AP: \(9,\ 17,\ 25,\ \ldots\)

$$\begin{aligned} a &= 9 \\ d &= 17 - 9 = 8 \end{aligned}$$

Given: \(S_n = 636\)

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} 636 &= \frac{n}{2}[2(9) + (n-1)(8)] \\ &= \frac{n}{2}[18 + 8n - 8] \\ &= \frac{n}{2}[8n + 10] \end{aligned}$$

Multiply both sides by 2 to remove denominator:

$$\begin{aligned} 1272 &= n(8n + 10) \end{aligned}$$

Expand RHS:

$$\begin{aligned} 1272 &= 8n^2 + 10n \end{aligned}$$

Bring all terms to one side:

$$\begin{aligned} 8n^2 + 10n - 1272 &= 0 \end{aligned}$$

Divide entire equation by 2:

$$\begin{aligned} 4n^2 + 5n - 636 &= 0 \end{aligned}$$

Factorisation:

$$\begin{aligned} 4n^2 + 53n - 48n - 636 &= 0 \\\ (4n^2 + 53n) - (48n + 636) &= 0 \\\ n(4n + 53) - 12(4n + 53) &= 0 \\\ (4n + 53)(n - 12) &= 0 \end{aligned}$$ $$\begin{aligned} n - 12 &= 0 \Rightarrow n = 12 \\\ 4n + 53 &= 0 \Rightarrow n = -\frac{53}{4} \end{aligned}$$

Number of terms cannot be negative or fractional, hence:

\[\boxed{n = 12}\]

Therefore, 12 terms must be taken to get the sum 636.

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Exam Significance

• Very common CBSE 3–4 mark question
• Tests quadratic formation from AP
• Checks algebra accuracy under pressure
• Frequently appears in NTSE & Olympiads
• Important for higher AP applications

Concept Used

When first term \(a\), last term \(l\), and sum \(S\) are given:

• Use: \(S_n = \frac{n}{2}(a + l)\) to find \(n\)
• Then use: \(a_n = a + (n-1)d\) to find \(d\)

Visual Interpretation

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Solution Roadmap

• Step 1: Use sum formula with \(a\) and \(l\) to find \(n\)
• Step 2: Use nth term formula to find \(d\)
• Step 3: Solve step-by-step carefully

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Solution

Given:
First term \(a = 5\)
Last term \(l = 45\)
Sum \(S = 400\)

Using sum formula:

\[S_n = \frac{n}{2}(a + l)\] $$\begin{aligned} 400 &= \frac{n}{2}(5 + 45) \\ &= \frac{n}{2}(50) \end{aligned}$$ $$\begin{aligned} 400 &= 25n \end{aligned}$$ $$\begin{aligned} n &= \frac{400}{25} \\ n &= 16 \end{aligned}$$

Now use nth term formula to find \(d\):

$$a_n = a + (n-1)d$$ $$\begin{aligned} 45 &= 5 + (16 - 1)d \\ 45 &= 5 + 15d \end{aligned}$$ $$\begin{aligned} 45 - 5 &= 15d \\ 40 &= 15d \end{aligned}$$ $$\begin{aligned} d &= \frac{40}{15} \\ &= \frac{8}{3} \end{aligned}$$ \[\boxed{n = 16,\; d = \frac{8}{3}}\]

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Exam Significance

• Very common CBSE board question pattern (3–4 marks)
• Tests ability to switch between formulas
• Important for case-study questions
• Frequently appears in competitive exams
• Concept base for advanced AP problems

Concept Used

When first term \(a\), last term \(l\), and common difference \(d\) are given:

• Use \(a_n = a + (n-1)d\) to find number of terms \(n\)
• Then use \(S_n = \frac{n}{2}(a + l)\) to find sum

Visual Understanding

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Solution Roadmap

• Step 1: Use nth term formula to find \(n\)
• Step 2: Substitute into sum formula
• Step 3: Simplify carefully

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Solution

Given:
First term \(a = 17\)
Last term \(l = 350\)
Common difference \(d = 9\)

Using nth term formula:

$$a_n = a + (n-1)d$$ $$\begin{aligned} 350 &= 17 + (n-1) \cdot 9 \end{aligned}$$

Expand RHS:

$$\begin{aligned} 350 &= 17 + 9n - 9 \\ 350 &= 8 + 9n \end{aligned}$$

Bring constant to LHS:

$$\begin{aligned} 350 - 8 &= 9n \\ 342 &= 9n \end{aligned}$$ $$\begin{aligned} n &= \frac{342}{9} \\ n &= 38 \end{aligned}$$

Now calculate sum:

$$S_n = \frac{n}{2}(a + l)$$ $$\begin{aligned} S_{38} &= \frac{38}{2}(17 + 350) \\ &= 19 \times 367 \end{aligned}$$

Final multiplication:

$$\begin{aligned} 19 \times 367 &= 19 \times (300 + 60 + 7) \\ &= 5700 + 1140 + 133 \\ &= 6973 \end{aligned}$$ $$\boxed{n = 38,\; S = 6973}$$

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Exam Significance

• Very common CBSE board 3–4 mark question
• Tests connection between \(a_n\) and \(S_n\)
• Frequently appears in case-study questions
• Important for competitive exams (NTSE, SSC)
• Builds strong algebra + AP linkage

Concept Used

When a specific term and common difference are given:

• Use \(a_n = a + (n-1)d\) to find first term \(a\)
• Then use \(S_n = \frac{n}{2}[2a + (n-1)d]\) to find sum

Visual Idea

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Solution Roadmap

• Step 1: Use \(a_n\) formula to find \(a\)
• Step 2: Substitute into sum formula
• Step 3: Simplify step-by-step

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Solution

Given:
Common difference \(d = 7\)
\(22^{\text{nd}}\) term \(a_{22} = 149\)

Using nth term formula:

$$a_n = a + (n-1)d$$ $$\begin{aligned} 149 &= a + (22 - 1)\times 7 \\ &= a + 21 \times 7 \\ &= a + 147 \end{aligned}$$

Find first term:

$$\begin{aligned} a &= 149 - 147 \\ a &= 2 \end{aligned}$$

Now calculate sum of first 22 terms:

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} S_{22} &= \frac{22}{2}[2(2) + (22-1)\times 7] \\ &= 11[4 + 21 \times 7] \\ &= 11[4 + 147] \\ &= 11 \times 151 \end{aligned}$$

Final multiplication:

$$\begin{aligned} 11 \times 151 &= 11 \times (100 + 50 + 1) \\ &= 1100 + 550 + 11 \\ &= 1661 \end{aligned}$$ \[\boxed{S_{22} = 1661}\]

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Exam Significance

• Very common CBSE 3–4 mark question
• Tests reverse thinking (finding \(a\) first)
• Important for case-study problems
• Frequently appears in competitive exams
• Builds strong link between term formula and sum formula

Concept Used

When middle terms are given:

• Use relation \(a_2 = a + d\), \(a_3 = a + 2d\)
• First find common difference \(d\)
• Then find first term \(a\)
• Finally apply sum formula

Visual Understanding

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Solution Roadmap

• Step 1: Find \(d = a_3 - a_2\)
• Step 2: Use \(a_2 = a + d\) to find \(a\)
• Step 3: Apply sum formula

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Solution

Given:
Second term \(a_2 = 14\)
Third term \(a_3 = 18\)

Find common difference:

$$\begin{aligned} d &= a_3 - a_2 \\ &= 18 - 14 \\ &= 4 \end{aligned}$$

Now find first term using \(a_2 = a + d\):

$$\begin{aligned} a + d &= 14 \\ a + 4 &= 14 \\ a &= 14 - 4 \\ a &= 10 \end{aligned}$$

Now calculate sum of first 51 terms:

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} S_{51} &= \frac{51}{2}[2(10) + (51-1)\times 4] \\ &= \frac{51}{2}[20 + 50 \times 4] \\ &= \frac{51}{2}[20 + 200] \\ &= \frac{51}{2}(220) \\ &= 51 \times 110 \end{aligned}$$

Final multiplication:

$$\begin{aligned} 51 \times 110 &= 51 \times (100 + 10) \\ &= 5100 + 510 \\ &= 5610 \end{aligned}$$ \[\boxed{S_{51} = 5610}\]

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Exam Significance

• Tests reverse concept of AP (finding \(a\) from middle terms)
• Very common CBSE 3–4 mark question
• Important for case-study questions
• Frequently appears in Olympiad and NTSE
• Builds conceptual depth (not direct formula use)

Concept Used

When sums for two different values of \(n\) are given:

• Form two equations using \(S_n = \frac{n}{2}[2a + (n-1)d]\)
• Solve to find \(a\) and \(d\)
• Substitute into general sum formula

Visual Insight

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Solution Roadmap

• Step 1: Form equation using \(S_7\)
• Step 2: Form equation using \(S_{17}\)
• Step 3: Solve simultaneous equations
• Step 4: Substitute into general formula

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Solution

Given:
\(S_7 = 49\)
\(S_{17} = 289\)

Using sum formula:

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

Step 1: Use \(S_7\)

$$\begin{aligned} S_7 &= \frac{7}{2}[2a + 6d] \\ 49 &= \frac{7}{2}[2a + 6d] \end{aligned}$$ Multiply both sides by 2: $$\begin{aligned} 98 &= 7[2a + 6d] \\ 14 &= 2a + 6d \end{aligned}$$ Divide by 2: $$\begin{align} 7 &= a + 3d \tag{1} \end{align}$$

Step 2: Use \(S_{17}\)

$$\begin{aligned} S_{17} &= \frac{17}{2}[2a + 16d] \\ 289 &= \frac{17}{2}[2a + 16d] \end{aligned}$$ Multiply both sides by 2: $$\begin{aligned} 578 &= 17[2a + 16d] \\ 34 &= 2a + 16d \end{aligned}$$ Divide by 2: $$\begin{align} 17 &= a + 8d \tag{2} \end{align}$$

Step 3: Solve equations

Subtract (1) from (2): $$\begin{aligned} (a + 8d) - (a + 3d) &= 17 - 7 \\ 5d &= 10 \\ d &= 2 \end{aligned}$$ Substitute \(d = 2\) into (1): $$\begin{aligned} 7 &= a + 3(2) \\ 7 &= a + 6 \\ a &= 1 \end{aligned}$$

Step 4: General Sum

$$\begin{aligned} S_n &= \frac{n}{2}[2a + (n-1)d] \\ &= \frac{n}{2}[2(1) + (n-1)(2)] \\ &= \frac{n}{2}[2 + 2n - 2] \\ &= \frac{n}{2}(2n) \\ &= n^2 \end{aligned}$$ \[\boxed{S_n = n^2}\]

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Exam Significance

• Very high probability CBSE question (derivation-based)
• Tests simultaneous equations + AP concepts
• Important for case-study pattern
• Frequently appears in Olympiads
• Builds strong algebraic reasoning

Concept Used

A sequence is an AP if:

• \(a_{n+1} - a_n = \text{constant}\)
• OR consecutive differences are equal

After proving AP, use:

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

Visual Insight

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Solution (i) \(a_n = 3 + 4n\)

Find first few terms:

$$\begin{aligned} a_1 &= 3 + 4(1) = 7 \\ a_2 &= 3 + 4(2) = 11 \\ a_3 &= 3 + 4(3) = 15 \end{aligned}$$

Sequence: \(7,\ 11,\ 15,\ \ldots\)

Check common difference:

$$\begin{aligned} a_2 - a_1 &= 11 - 7 = 4 \\ a_3 - a_2 &= 15 - 11 = 4 \end{aligned}$$

Since the difference is constant, the sequence is an AP with:

\[a = 7,\quad d = 4\]

Sum of first 15 terms:

$$\begin{aligned} S_{15} &= \frac{15}{2}[2(7) + (15-1)(4)] \\ &= \frac{15}{2}[14 + 14 \times 4] \\ &= \frac{15}{2}[14 + 56] \\ &= \frac{15}{2}(70) \\ &= 15 \times 35 \\ &= 525 \end{aligned}$$ $$\boxed{S_{15} = 525}$$

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Solution (ii) \(a_n = 9 - 5n\)

Find first few terms:

$$\begin{aligned} a_1 &= 9 - 5(1) = 4 \\ a_2 &= 9 - 5(2) = -1 \\ a_3 &= 9 - 5(3) = -6 \\ a_4 &= 9 - 5(4) = -11 \end{aligned}$$

Sequence: \(4,\ -1,\ -6,\ -11,\ \ldots\)

Check common difference:

$$\begin{aligned} a_2 - a_1 &= -1 - 4 = -5 \\ a_3 - a_2 &= -6 - (-1) = -5 \\ a_4 - a_3 &= -11 - (-6) = -5 \end{aligned}$$

Hence, it is an AP with:

\[a = 4,\quad d = -5\]

Sum of first 15 terms:

$$\begin{aligned} S_{15} &= \frac{15}{2}[2(4) + (15-1)(-5)] \\ &= \frac{15}{2}[8 + 14(-5)] \\ &= \frac{15}{2}[8 - 70] \\ &= \frac{15}{2}(-62) \\ &= -15 \times 31 \\ &= -465 \end{aligned}$$ \[\boxed{S_{15} = -465}\]

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Exam Significance

• Tests fundamental definition of AP (difference method)
• Very common CBSE proof-based question
• Combines concept + computation
• Frequently appears in Olympiad & NTSE
• Builds deep understanding of sequence behavior

Concept Used

If sum of first \(n\) terms is given, then:

• \(a_n = S_n - S_{n-1}\)
• \(a_1 = S_1\)

Visual Insight

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Solution Roadmap

• Step 1: Use \(S_1\) to find first term
• Step 2: Use \(S_2\) to find second term
• Step 3: Use \(a_n = S_n - S_{n-1}\)
• Step 4: Derive general term

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Solution

Given: \(S_n = 4n - n^2\)

Step 1: First term

$$\begin{aligned} S_1 &= 4(1) - 1^2 \\ &= 4 - 1 \\ &= 3 \end{aligned}$$ \[a_1 = S_1 = 3\]

Step 2: Sum of first two terms

$$\begin{aligned} S_2 &= 4(2) - 2^2 \\ &= 8 - 4 \\ &= 4 \end{aligned}$$

Step 3: Second term

$$\begin{aligned} a_2 &= S_2 - S_1 \\ &= 4 - 3 \\ &= 1 \end{aligned}$$

Step 4: Third term

$$\begin{aligned} S_3 &= 4(3) - 3^2 \\ &= 12 - 9 \\ &= 3 \end{aligned}$$ $$\begin{aligned} a_3 &= S_3 - S_2 \\ &= 3 - 4 \\ &= -1 \end{aligned}$$

Step 5: Tenth term

$$\begin{aligned} S_{10} &= 4(10) - 10^2 \\ &= 40 - 100 \\ &= -60 \end{aligned}$$ $$\begin{aligned} S_9 &= 4(9) - 9^2 \\ &= 36 - 81 \\ &= -45 \end{aligned}$$ $$\begin{aligned} a_{10} &= S_{10} - S_9 \\ &= -60 - (-45) \\ &= -60 + 45 \\ &= -15 \end{aligned}$$

Step 6: General term

$$\begin{aligned} S_n &= 4n - n^2 \\ S_{n-1} &= 4(n-1) - (n-1)^2 \end{aligned}$$ Expand \(S_{n-1}\): $$\begin{aligned} S_{n-1} &= 4n - 4 - (n^2 - 2n + 1) \\ &= 4n - 4 - n^2 + 2n - 1 \\ &= 6n - 5 - n^2 \end{aligned}$$ Now, $$\begin{aligned} a_n &= S_n - S_{n-1} \\ &= (4n - n^2) - (6n - 5 - n^2) \\ &= 4n - n^2 - 6n + 5 + n^2 \\ &= -2n + 5 \end{aligned}$$ \[\boxed{a_n = 5 - 2n}\]

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Final Answers

• \(a_1 = 3\)
• \(S_2 = 4\)
• \(a_2 = 1\)
• \(a_3 = -1\)
• \(a_{10} = -15\)
• \(a_n = 5 - 2n\)

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Exam Significance

• Very important CBSE pattern question
• Tests concept: \(a_n = S_n - S_{n-1}\)
• Common in case-study and HOTS
• Appears in NTSE, Olympiad
• Builds deep conceptual understanding of AP structure

Concept Used

Numbers divisible by 6 form an AP:

• Sequence: \(6, 12, 18, 24, \ldots\)
• General term: multiples of 6 → \(6n\)

This is an AP with:

\[a = 6,\quad d = 6\]

Visual Understanding

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Solution Roadmap

• Step 1: Identify AP parameters \(a, d, n\)
• Step 2: Apply sum formula
• Step 3: Simplify step-by-step

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Solution

First 40 positive integers divisible by 6 form the AP:
\(6,\ 12,\ 18,\ \ldots\)

$$\begin{aligned} a &= 6 \\ d &= 12 - 6 = 6 \\ n &= 40 \end{aligned}$$

Using sum formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$ $$\begin{aligned} S_{40} &= \frac{40}{2}[2(6) + (40-1)(6)] \\ &= 20[12 + 39 \times 6] \end{aligned}$$

Calculate inside bracket:

$$\begin{aligned} 39 \times 6 &= 234 \\ 12 + 234 &= 246 \end{aligned}$$ $$\begin{aligned} S_{40} &= 20 \times 246 \\ &= 4920 \end{aligned}$$ \[\boxed{S_{40} = 4920}\]

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Shortcut Insight (Very Important)

Since numbers are \(6, 12, 18, \ldots\), we can write:

\[\text{Sum} = 6(1 + 2 + 3 + \cdots + 40)\] $$\begin{aligned} &= 6 \times \frac{40 \times 41}{2} \\ &= 6 \times 820 \\ &= 4920 \end{aligned}$$

This shortcut is extremely useful in competitive exams.

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Exam Significance

• Very common CBSE board question
• Tests identification of AP in real context
• Shortcut method highly useful in exams
• Appears in NTSE, SSC, Banking
• Strengthens link between AP and number patterns

Concept Used

Multiples of a number form an Arithmetic Progression (AP).

• Multiples of 8: \(8, 16, 24, 32, \ldots\)
• This is an AP with constant difference

\[a = 8,\quad d = 8\]

Visual Understanding

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Solution Roadmap

• Step 1: Identify \(a, d, n\)
• Step 2: Apply sum formula
• Step 3: Simplify step-by-step

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Solution

First 15 multiples of 8 form the AP:
\(8,\ 16,\ 24,\ 32,\ \ldots\)

$$\begin{aligned} a &= 8 \\ d &= 8 \\ n &= 15 \end{aligned}$$

Using sum formula:

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} S_{15} &= \frac{15}{2}[2(8) + (15-1)(8)] \\ &= \frac{15}{2}[16 + 14 \times 8] \end{aligned}$$

Calculate inside bracket:

$$\begin{aligned} 14 \times 8 &= 112 \\ 16 + 112 &= 128 \end{aligned}$$ $$\begin{aligned} S_{15} &= \frac{15}{2} \times 128 \\ &= 15 \times 64 \\ &= 960 \end{aligned}$$ \[\boxed{S_{15} = 960}\]

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Shortcut Insight (Highly Important)

Multiples of 8 can be written as:

$$8(1 + 2 + 3 + \cdots + 15)$$ $$\begin{aligned} &= 8 \times \frac{15 \times 16}{2} \\ &= 8 \times 120 \\ &= 960 \end{aligned}$$

This method is much faster in exams.

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Exam Significance

• Very common CBSE board question
• Tests conversion of real-world pattern → AP
• Shortcut method saves time in exams
• Frequently appears in SSC, Banking, NTSE
• Builds strong number pattern understanding

Concept Used

Odd numbers form an Arithmetic Progression (AP):

• Sequence: \(1, 3, 5, 7, \ldots\)
• Each term increases by 2

$$a = 1,\quad d = 2$$

Visual Understanding

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Solution Roadmap

• Step 1: Identify AP parameters \(a, d, l\)
• Step 2: Find number of terms using nth term formula
• Step 3: Apply sum formula

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Solution

Odd numbers between 0 and 50 are:
\(1,\ 3,\ 5,\ 7,\ \ldots,\ 49\)

$$\begin{aligned} a &= 1 \\ d &= 2 \\ a_n &= 49 \end{aligned}$$

Find number of terms:

$$\begin{aligned} a_n &= a + (n-1)d \\ 49 &= 1 + (n-1)\times 2 \\ 49 &= 1 + 2n - 2 \\ 49 &= 2n - 1 \end{aligned}$$ $$\begin{aligned} 49 + 1 &= 2n \\ 50 &= 2n \\ n &= 25 \end{aligned}$$

Now calculate sum:

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} S_{25} &= \frac{25}{2}[2(1) + (25-1)\times 2] \\ &= \frac{25}{2}[2 + 24 \times 2] \\ &= \frac{25}{2}[2 + 48] \\ &= \frac{25}{2}(50) \\ &= 25 \times 25 \\ &= 625 \end{aligned}$$ \[\boxed{S_{25} = 625}\]

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Shortcut Insight (Very Powerful)

Sum of first \(n\) odd numbers = \(n^2\)

$$\begin{aligned} \text{Here } n = 25 \\ \Rightarrow \text{Sum} = 25^2 = 625 \end{aligned}$$

This is one of the most important identities in mathematics.

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Exam Significance

• Very common CBSE question
• Tests recognition of number patterns
• Shortcut saves huge time in exams
• Frequently used in Olympiad & NTSE
• Builds strong conceptual number sense

Concept Used

The penalties increase uniformly each day → this forms an Arithmetic Progression (AP).

• First day penalty = ₹200
• Increase per day = ₹50
• Total days = 30

\[a = 200,\quad d = 50,\quad n = 30\]

Visual Interpretation

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Solution Roadmap

• Step 1: Identify AP parameters \(a, d, n\)
• Step 2: Apply sum formula
• Step 3: Simplify step-by-step

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Solution

Penalties form the AP:
₹200, ₹250, ₹300, …

$$\begin{aligned} a &= 200 \\ d &= 250 - 200 = 50 \\ n &= 30 \end{aligned}$$

Using sum formula:

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} S_{30} &= \frac{30}{2}[2(200) + (30-1)\times 50] \\ &= 15[400 + 29 \times 50] \end{aligned}$$

Calculate inside bracket:

$$\begin{aligned} 29 \times 50 &= 1450 \\ 400 + 1450 &= 1850 \end{aligned}$$ $$\begin{aligned} S_{30} &= 15 \times 1850 \end{aligned}$$

Final multiplication:

$$\begin{aligned} 15 \times 1850 &= 15 \times (1000 + 800 + 50) \\ &= 15000 + 12000 + 750 \\ &= 27750 \end{aligned}$$ \[\boxed{\text{Total Penalty} = \text{₹ }27,750}\]

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Shortcut Insight (Exam Gold)

You can also use:

\[S_n = \frac{n}{2}(a + l)\]

Find last term:

$$\begin{aligned} l &= a + (n-1)d \\ &= 200 + 29 \times 50 \\ &= 200 + 1450 \\ &= 1650 \end{aligned}$$ $$\begin{aligned} S_{30} &= \frac{30}{2}(200 + 1650) \\ &= 15 \times 1850 \\ &= 27750 \end{aligned}$$

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Exam Significance

• Very important CBSE case-study type question
• Tests real-life modelling using AP
• Frequently appears in board exams (4–5 marks)
• Important for SSC, Banking, Olympiads
• Builds application-based thinking

Concept Used

The prizes decrease uniformly → this forms an Arithmetic Progression (AP) with negative common difference.

• Let highest prize = \(x\)
• Each next prize decreases by ₹20
• Total number of prizes = 7

\[a = x,\quad d = -20,\quad n = 7\]

Visual Understanding

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Solution Roadmap

• Step 1: Form AP expression
• Step 2: Use sum formula
• Step 3: Solve for \(x\)
• Step 4: Write all terms

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Solution

Let highest prize = \(x\)

Then prizes are:
\(x,\ x-20,\ x-40,\ \ldots\)

$$\begin{aligned} a &= x \\ d &= -20 \\ n &= 7 \end{aligned}$$

Find last term:

$$\begin{aligned} a_7 &= a + (n-1)d \\ &= x + 6(-20) \\ &= x - 120 \end{aligned}$$

Using sum formula:

$$S_n = \frac{n}{2}(a + l)$$ $$\begin{aligned} 700 &= \frac{7}{2}[x + (x - 120)] \\ &= \frac{7}{2}(2x - 120) \end{aligned}$$

Multiply both sides by 2:

$$\begin{aligned} 1400 &= 7(2x - 120) \end{aligned}$$

Divide by 7:

$$\begin{aligned} 200 &= 2x - 120 \end{aligned}$$

Solve for \(x\):

$$\begin{aligned} 2x &= 200 + 120 \\ 2x &= 320 \\ x &= 160 \end{aligned}$$

Hence, the prizes are:

\[\boxed{ \text{₹ }160,\ \text{₹ }140,\ \text{₹ }120,\ \text{₹ }100,\ \text{₹ }80,\ \text{₹ }60,\ \text{₹ }40 }\]

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Exam Significance

• Very important CBSE case-study type question
• Tests AP modelling with variables
• Negative common difference concept
• Frequently appears in real-life applications
• Important for SSC, Banking, Olympiad exams

Concept Used

Number of trees planted per class forms an Arithmetic Progression (AP):

• Class I → 1 tree
• Class II → 2 trees
• …
• Class XII → 12 trees

\[1,\ 2,\ 3,\ \ldots,\ 12\] \[a = 1,\quad d = 1,\quad n = 12\]

Visual Understanding

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Solution Roadmap

• Step 1: Find total trees planted by one section (AP sum)
• Step 2: Multiply by number of sections (3)

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Solution

Trees planted by one section follow the AP:
\(1, 2, 3, \ldots, 12\)

Sum of trees for one section:

\[S_n = \frac{n}{2}(a + l)\] $$\begin{aligned} S_{12} &= \frac{12}{2}(1 + 12) \\ &= 6 \times 13 \\ &= 78 \end{aligned}$$

Since there are 3 sections in each class:

$$\begin{aligned} \text{Total trees} &= 78 \times 3 \\ &= 234 \end{aligned}$$ \[\boxed{234\ \text{trees}}\]

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Shortcut Insight

Sum of first \(n\) natural numbers:

\[\frac{n(n+1)}{2} = \frac{12 \times 13}{2} = 78\]

Then multiply by 3:

\[78 \times 3 = 234\]

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Exam Significance

• Classic CBSE case-study question
• Tests real-life modelling using AP
• Involves two-step reasoning (sum + multiplication)
• Frequently appears in board exams (4–5 marks)
• Important for NTSE, Olympiad, SSC

Concept Used

Length of a semicircle = \(\pi r\)

Since radii form an AP, their corresponding arc lengths also form an AP.

• Radii: \(0.5,\ 1.0,\ 1.5,\ 2.0,\ \ldots\)
• Each term increases by \(0.5\)

Visual Understanding

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Solution Roadmap

• Step 1: Convert radii into semicircle lengths
• Step 2: Identify AP
• Step 3: Find last term
• Step 4: Apply sum formula

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Solution

Radii form AP:
\(0.5,\ 1.0,\ 1.5,\ 2.0,\ \ldots\)

Length of each semicircle = \(\pi r\)

So arc lengths form AP:

\[0.5\pi,\ \pi,\ 1.5\pi,\ 2\pi,\ \ldots\] $$\begin{aligned} a &= 0.5\pi \\ d &= 0.5\pi \\ n &= 13 \end{aligned}$$

Find last term:

$$\begin{aligned} a_{13} &= a + (n-1)d \\ &= 0.5\pi + 12(0.5\pi) \\ &= 0.5\pi + 6\pi \\ &= 6.5\pi \end{aligned}$$

Now calculate total length:

\[S_n = \frac{n}{2}(a + l)\] $$\begin{aligned} S_{13} &= \frac{13}{2}[0.5\pi + 6.5\pi] \\ &= \frac{13}{2}(7\pi) \end{aligned}$$ $$\begin{aligned} &= \frac{13}{2} \times 7 \times \frac{22}{7} \\ &= \frac{13}{2} \times 22 \\ &= 13 \times 11 \\ &= 143 \end{aligned}$$ \[\boxed{143\ \text{cm}}\]

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Exam Significance

• Very important geometry + AP integration question
• Tests conversion of geometry → algebra
• Frequently appears in case-study questions
• High probability in CBSE boards
• Important for Olympiad level thinking

Conceptual Spiral Representation

Concept Used

The number of logs per row forms a decreasing Arithmetic Progression (AP).

• Bottom row = 20 logs
• Each next row has 1 less log

\[20,\ 19,\ 18,\ \ldots\] \[a = 20,\quad d = -1\]

Visual Understanding

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Solution Roadmap

• Step 1: Use sum formula to form equation
• Step 2: Convert into quadratic equation
• Step 3: Solve and validate value of \(n\)
• Step 4: Find top row using \(a_n\)

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Solution

Given total logs \(S_n = 200\)

\[S_n = \frac{n}{2}[2a + (n-1)d]\] $$\begin{aligned} 200 &= \frac{n}{2}[2(20) + (n-1)(-1)] \\ &= \frac{n}{2}[40 - (n-1)] \end{aligned}$$ $$\begin{aligned} &= \frac{n}{2}[40 - n + 1] \\ &= \frac{n}{2}[41 - n] \end{aligned}$$

Multiply both sides by 2:

$$\begin{aligned} 400 &= n(41 - n) \end{aligned}$$

Expand RHS:

$$\begin{aligned} 400 &= 41n - n^2 \end{aligned}$$

Rearrange:

$$\begin{aligned} n^2 - 41n + 400 &= 0 \end{aligned}$$

Factorisation:

$$\begin{aligned} n^2 - 25n - 16n + 400 &= 0 \\ n(n - 25) - 16(n - 25) &= 0 \\ (n - 25)(n - 16) &= 0 \end{aligned}$$ $$\begin{aligned} n = 25 \quad \text{or} \quad n = 16 \end{aligned}$$

Since number of logs cannot become negative, we reject \(n = 25\).

\[n = 16\]

Now find logs in top row:

$$\begin{aligned} a_n &= a + (n-1)d \\ a_{16} &= 20 + 15(-1) \\ &= 20 - 15 \\ &= 5 \end{aligned}$$ \[\boxed{ \text{Number of rows} = 16,\quad \text{Top row logs} = 5 }\]

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Exam Significance

• Classic AP + quadratic modelling question
• Tests logical rejection of invalid root
• Very common CBSE 4–5 mark question
• Appears in NTSE and Olympiad exams
• Builds strong reasoning with real-life patterns