NCERT Class 10 Maths AP Ex 5.2 Solutions

Concept Used

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

nth Term Formula:

\[ a_n = a + (n - 1)d \]

Where:
• \(a\) = first term
• \(d\) = common difference
• \(n\) = term number
• \(a_n\) = nth term

Equal spacing shows constant difference (d)

Solution Roadmap

Step 1: Identify known values (a, d, n, a_n)
Step 2: Use formula \(a_n = a + (n-1)d\)
Step 3: Substitute values carefully
Step 4: Solve algebraically step-by-step
Step 5: Verify final value

Solution (i)

\[ \begin{aligned} a &= 7,\quad d = 3,\quad n = 8\\ a_n &= a + (n-1)d\\ &= 7 + (8-1)\times 3\\ &= 7 + 7\times 3\\ &= 7 + 21\\ &= 28 \end{aligned} \]

Solution (ii)

\[ \begin{aligned} a &= -18,\quad n = 10,\quad a_n = 0\\ a_n &= a + (n-1)d\\ 0 &= -18 + (10-1)d\\ 0 &= -18 + 9d\\ 9d &= 18\\ d &= \frac{18}{9}\\ d &= 2 \end{aligned} \]

Solution (iii)

\[ \begin{aligned} d &= -3,\quad n = 18,\quad a_n = -5\ a_n &= a + (n-1)d\\ -5 &= a + (18-1)(-3)\\ -5 &= a + 17(-3)\\ -5 &= a - 51\\ a &= -5 + 51\\ a &= 46 \end{aligned} \]

Solution (iv)

\[ \begin{aligned} a &= -18.9,\quad d = 2.5,\quad a_n = 3.6\\\ a_n &= a + (n-1)d\\\ 3.6 &= -18.9 + (n-1)\times 2.5\\\ 3.6 + 18.9 &= (n-1)\times 2.5\\\ 22.5 &= (n-1)\times 2.5\\\ n-1 &= \frac{22.5}{2.5}\\\ n-1 &= 9\\\ n &= 10 \end{aligned} \]

Solution (v)

\[ \begin{aligned} a &= 3.5,\quad d = 0,\quad n = 105\\\ a_n &= a + (n-1)d\\\ &= 3.5 + (105-1)\times 0\\\ &= 3.5 + 104 \times 0\\\ &= 3.5 + 0\\\ &= 3.5 \end{aligned} \]

Exam Significance

This question directly tests your command over the nth term formula of AP.

In CBSE Board Exams:
• 1–2 mark direct formula-based questions are very common
• Mistakes usually occur in handling (n−1)

In Competitive Exams (JEE Foundation, NDA, Olympiads):
• Forms base for sequence modelling
• Used in coding patterns and progression logic
• Critical for solving higher-order AP problems

Concept Used

In an Arithmetic Progression (AP), the difference between consecutive terms is constant.

Key Formula:

\[ a_n = a + (n-1)d \]

Strategy involves:
• Finding common difference \(d = a_2 - a_1\)
• Substituting carefully into nth term formula
• Handling negative signs accurately

Uniform decrease/increase determined by common difference (d)

Solution Roadmap

Step 1: Identify first term \(a\) and second term
Step 2: Compute common difference \(d = a_2 - a_1\)
Step 3: Identify required term number \(n\)
Step 4: Apply formula \(a_n = a + (n-1)d\)
Step 5: Simplify carefully (signs are critical)
Step 6: Match with given options

Solution (i)

\(\text{AP: } 10,\ 7,\ 4,\ \ldots\)

\[ \begin{aligned} a_1 &= 10,\quad a_2 = 7\\ d &= a_2 - a_1\\ &= 7 - 10\\ &= -3 \end{aligned} \]

\[ \begin{aligned} n &= 30\\ a_{30} &= a + (n-1)d\\ &= 10 + (30-1)(-3)\\ &= 10 + 29(-3)\\ &= 10 - 87\\ &= -77 \end{aligned} \]

Correct Option: (C) –77

Solution (ii)

\(\text{AP: } -3,\ -\dfrac{1}{2},\ 2,\ \ldots\)

\[ \begin{aligned} a_1 &= -3,\quad a_2 = -\frac{1}{2}\\ d &= a_2 - a_1\\ &= -\frac{1}{2} - (-3)\\ &= -\frac{1}{2} + 3\\ &= \frac{5}{2} \end{aligned} \]

\[ \begin{aligned} n &= 11\\ a_{11} &= a + (n-1)d\\ &= -3 + (11-1)\times \frac{5}{2}\\ &= -3 + 10 \times \frac{5}{2}\\ &= -3 + \frac{50}{2}\\ &= -3 + 25\\ &= 22 \end{aligned} \]

Correct Option: (B) 22

Exam Significance

This is a classic MCQ format testing both conceptual clarity and calculation accuracy.

In CBSE Board Exams:
• Frequently appears as 1-mark MCQ or case-based subpart
• Common traps include sign errors and incorrect (n−1)

In Competitive Exams:
• Forms base for sequence modelling problems
• Appears in quantitative aptitude and reasoning
• Critical for time-speed optimization (fast AP evaluation)

Q3

NUMERIC3 marks

In the following APs, find the missing terms in the boxes: \[ \begin{array}{l} \mathrm{i. }\ 2,\ \boxed{\color{red}{14}},\ 26 \\ \text{ii. } \boxed{\color{red}{18}},\ 13,\ \boxed{\color{red}{8}},\ 3 \\ \text{iii. } 5,\ \boxed{\color{red}{6\frac{1}{2}}},\ \boxed{\color{red}{8}},\ 9\frac{1}{2} \\ \text{iv. } -4,\ \boxed{\color{red}{-2}},\ \boxed{\color{red}{0}},\ \boxed{\color{red}{2}},\ \boxed{\color{red}{4}},\ 6 \\ \text{v. } \boxed{\color{red}{53}},\ 38,\ \boxed{\color{red}{23}},\ \boxed{\color{red}{8}},\ \boxed{\color{red}{-7}},\ -22 \end{array} \]

Concept Used

In an Arithmetic Progression, every term increases or decreases by a constant value called the common difference.

\[ a_n = a + (n-1)d \]

Missing terms are found by:
• Using given terms to calculate \(d\)
• Then generating intermediate terms
• Or forming equations when positions are known

Equal spacing helps reconstruct missing terms

Solution Roadmap

Step 1: Identify positions of known terms
Step 2: Use nth term formula to form equation
Step 3: Solve for common difference \(d\)
Step 4: Generate missing terms using \(a + kd\)
Step 5: Verify AP consistency

Solution (i)

\(2,\ \ldots,\ 26\)

\[ \begin{aligned} a &= 2,\quad a_3 = 26\\ a_3 &= a + 2d\\ 26 &= 2 + 2d\\ 2d &= 24\\ d &= 12 \end{aligned} \]

\[ \begin{aligned} a_2 &= a + d = 2 + 12 = 14 \end{aligned} \]

Solution (ii)

\(\ldots,\ 13,\ \ldots,\ 3\)

\[ \begin{aligned} a_2 &= 13 \Rightarrow a + d = 13 \quad (1)\\ a_4 &= 3 \Rightarrow a + 3d = 3 \quad (2) \end{aligned} \]

\[ \begin{aligned} (2) - (1):\quad (a+3d)-(a+d) &= 3 - 13\\ 2d &= -10\\ d &= -5 \end{aligned} \]

\[ \begin{aligned} a + d &= 13\\ a - 5 &= 13\\ a &= 18 \end{aligned} \]

\[ a_3 = a + 2d = 18 + 2(-5) = 8 \]

Solution (iii)

\(5,\ \ldots,\ \ldots,\ 9\dfrac{1}{2}\)

\[ \begin{aligned} a &= 5,\quad a_4 = \frac{19}{2}\\ a_4 &= a + 3d\\ \frac{19}{2} &= 5 + 3d\\ 3d &= \frac{19}{2} - \frac{10}{2} = \frac{9}{2}\\ d &= \frac{3}{2} \end{aligned} \]

\[ a_2 = 5 + \frac{3}{2} = \frac{13}{2} = 6\frac{1}{2} \]

\[ a_3 = a + 2d = 5 + 2\times \frac{3}{2} = 5 + 3 = 8 \]

Solution (iv)

\(-4,\ \ldots,\ \ldots,\ \ldots,\ \ldots,\ 6\)

\[ \begin{aligned} a_1 &= -4,\quad a_6 = 6\\ a_6 &= a + 5d\\ 6 &= -4 + 5d\\ 5d &= 10\\ d &= 2 \end{aligned} \]

\[ \begin{aligned} a_2 &= -4 + 2 = -2\\ a_3 &= -4 + 4 = 0\\ a_4 &= -4 + 6 = 2\\ a_5 &= -4 + 8 = 4 \end{aligned} \]

Solution (v)

\(\ldots,\ 38,\ \ldots,\ \ldots,\ \ldots,\ -22\)

\[ \begin{aligned} a_2 &= 38 \Rightarrow a + d = 38 \quad (1)\\ a_6 &= -22 \Rightarrow a + 5d = -22 \quad (2) \end{aligned} \]

\[ \begin{aligned} (2) - (1):\quad 4d &= -60\\ d &= -15 \end{aligned} \]

\[ \begin{aligned} a + d &= 38\\ a - 15 &= 38\\ a &= 53 \end{aligned} \]

\[ \begin{aligned} a_3 &= 53 + 2(-15) = 23\\ a_4 &= 53 + 3(-15) = 8\\ a_5 &= 53 + 4(-15) = -7 \end{aligned} \]

Exam Significance

This question builds strong structural understanding of AP formation and reconstruction.

In Board Exams:
• Frequently appears as 2–3 mark question
• Tests equation formation and logical progression

In Competitive Exams:
• Important for pattern recognition
• Used in sequence completion and number logic problems
• Foundation for higher algebraic sequences

Concept Used

In Arithmetic Progression, to find the position (term number) of a given value, we use:

\[ a_n = a + (n-1)d \]

Here, instead of finding \(a_n\), we are given \(a_n\) and need to find \(n\).

Each step increases by constant difference (d = 5)

Solution Roadmap

Step 1: Identify first term \(a\)
Step 2: Find common difference \(d = a_2 - a_1\)
Step 3: Substitute given value as \(a_n\)
Step 4: Solve linear equation in \(n\)
Step 5: Interpret result as term position

Solution

\(\text{AP: } 3,\ 8,\ 13,\ 18,\ \ldots\)

\[ \begin{aligned} a_1 &= 3,\quad a_2 = 8\\ d &= a_2 - a_1\\ &= 8 - 3\\ &= 5 \end{aligned} \]

\[ \begin{aligned} a_n &= 78\\ a_n &= a + (n-1)d\\ 78 &= 3 + (n-1)\times 5 \end{aligned} \]

\[ \begin{aligned} 78 - 3 &= (n-1)\times 5\\ 75 &= (n-1)\times 5\\ n - 1 &= \frac{75}{5}\\ n - 1 &= 15\\ n &= 15 + 1\\ n &= 16 \end{aligned} \]

Therefore, 78 is the 16th term of the AP.

Exam Significance

This is a standard "term identification" problem where reverse application of formula is required.

In CBSE Board Exams:
• Very common 2-mark question
• Tests equation solving accuracy
• Mistake hotspot: forgetting (n−1)

In Competitive Exams:
• Appears in sequence indexing problems
• Used in coding logic and pattern detection
• Forms base for solving AP-based word problems

Concept Used

To find the number of terms in an AP, we use the nth term formula:

\[ a_n = a + (n-1)d \]

Here:
• First term \(a\) and common difference \(d\) are known
• Last term \(a_n\) is given
• We solve for \(n\), which gives total number of terms

Each step moves by constant difference → helps count total terms

Solution Roadmap

Step 1: Identify \(a\), \(d\), and last term \(a_n\)
Step 2: Substitute into \(a_n = a + (n-1)d\)
Step 3: Solve linear equation carefully
Step 4: Extract \(n\) (total number of terms)
Step 5: Verify sign consistency (especially for negative \(d\))

Solution (i)

\(7,\ 13,\ 19,\ \ldots,\ 205\)

\[ \begin{aligned} a_1 &= 7,\quad a_2 = 13\\ d &= a_2 - a_1\\ &= 13 - 7\\ &= 6 \end{aligned} \]

\[ \begin{aligned} a_n &= 205\\ a_n &= a + (n-1)d\\ 205 &= 7 + (n-1)\times 6 \end{aligned} \]

\[ \begin{aligned} 205 - 7 &= (n-1)\times 6\\ 198 &= (n-1)\times 6\\ n - 1 &= \frac{198}{6}\\ n - 1 &= 33\\ n &= 33 + 1\\ n &= 34 \end{aligned} \]

Total number of terms = 34

Solution (ii)

\(18,\ 15\dfrac{1}{2},\ 13,\ \ldots,\ -47\)

\[ \begin{aligned} a_1 &= 18,\quad a_2 = 15\frac{1}{2} \end{aligned} \]

\[ \begin{aligned} a_2 &= \frac{31}{2}\\ d &= a_2 - a_1\\ &= \frac{31}{2} - 18\\ &= \frac{31}{2} - \frac{36}{2}\\ &= \frac{-5}{2} \end{aligned} \]

\[ \begin{aligned} a_n &= -47\\ a_n &= a + (n-1)d\\ -47 &= 18 + (n-1)\left(\frac{-5}{2}\right) \end{aligned} \]

\[ \begin{aligned} -47 - 18 &= (n-1)\left(\frac{-5}{2}\right)\\ -65 &= (n-1)\left(\frac{-5}{2}\right) \end{aligned} \]

\[ \begin{aligned} (n-1) &= \frac{-65}{-5/2}\\ &= -65 \times \frac{2}{-5}\\ &= \frac{130}{5}\\ &= 26 \end{aligned} \]

\[ \begin{aligned} n &= 26 + 1\\ n &= 27 \end{aligned} \]

Total number of terms = 27

Exam Significance

This is a standard "number of terms" problem requiring reverse use of nth term formula.

In CBSE Board Exams:
• Frequently appears as 2–3 mark question
• Common mistakes: fraction handling and sign errors

In Competitive Exams:
• Used in sequence length estimation problems
• Critical for arithmetic series modelling
• Helps in solving real-life progression problems (finance, coding patterns)

Concept Used

To check whether a number is a term of an AP, we use:

\[ a_n = a + (n-1)d \]

If solving gives a natural number \(n\), the number exists in the AP. Otherwise, it is not a term.

AP decreases by constant difference (d = -3)

Solution Roadmap

Step 1: Identify \(a\) and \(d\)
Step 2: Assume given number as \(a_n\)
Step 3: Substitute into formula
Step 4: Solve for \(n\)
Step 5: Check if \(n\) is a natural number

Solution

\(\text{AP: } 11,\ 8,\ 5,\ 2,\ \ldots\)

\[ \begin{aligned} a_1 &= 11,\quad a_2 = 8\\ d &= a_2 - a_1\\ &= 8 - 11\\ &= -3 \end{aligned} \]

\[ \text{Let } a_n = -150 \]

\[ \begin{aligned} a_n &= a + (n-1)d\\ -150 &= 11 + (n-1)(-3) \end{aligned} \]

\[ \begin{aligned} -150 - 11 &= (n-1)(-3)\\ -161 &= (n-1)(-3) \end{aligned} \]

\[ \begin{aligned} n - 1 &= \frac{-161}{-3}\\ n - 1 &= \frac{161}{3} \end{aligned} \]

\[ \begin{aligned} n &= 1 + \frac{161}{3}\\ n &= \frac{3 + 161}{3}\\ n &= \frac{164}{3} \end{aligned} \]

Since \(n\) \(\notin \mathbb{N}\) (not a natural number),

Therefore, –150 is NOT a term of the given AP.

Exam Significance

This is a classic "membership test" problem in AP.

In CBSE Board Exams:
• Common 2-mark conceptual question
• Key trap: sign errors and incorrect equation setup

In Competitive Exams:
• Important for sequence validation
• Used in number theory and pattern problems
• Helps in building logical elimination skills

Concept Used

When two different terms of an AP are given, we form equations using:

\[ a_n = a + (n-1)d \]

This results in a system of two linear equations in \(a\) and \(d\), which we solve simultaneously.

Equal increments help form linear equations in a and d

Solution Roadmap

Step 1: Write formulas for given terms
Step 2: Form two equations in \(a\) and \(d\)
Step 3: Eliminate one variable (subtraction method)
Step 4: Find \(d\), then substitute to get \(a\)
Step 5: Compute required term using formula

Solution

\[ \begin{aligned} a_{11} &= 38,\quad a_{16} = 73 \end{aligned} \]

\[ \begin{align} a_{11} &= a + (11-1)d\\ &= a + 10d = 38 \tag{1} \end{align} \]

\[ \begin{align} a_{16} &= a + (16-1)d\\ &= a + 15d = 73 \tag{2} \end{align} \]

Subtracting (1) from (2):

\[ \begin{aligned} (a + 15d) - (a + 10d) &= 73 - 38\\ a + 15d - a - 10d &= 35\\ 5d &= 35\\ d &= \frac{35}{5}\\ d &= 7 \end{aligned} \]

Substitute \(d = 7\) into (1):

\[ \begin{aligned} a + 10d &= 38\\ a + 10\times 7 &= 38\\ a + 70 &= 38\\ a &= 38 - 70\\ a &= -32 \end{aligned} \]

Now find the 31st term:

\[ \begin{aligned} a_{31} &= a + (31-1)d\\ &= -32 + 30\times 7\\ &= -32 + 210\\ &= 178 \end{aligned} \]

Therefore, the 31st term of the AP is 178.

Exam Significance

This is a standard “two-term → find general term” problem involving simultaneous equations.

In CBSE Board Exams:
• Frequently appears as 3-mark question
• Tests algebraic elimination and substitution

In Competitive Exams:
• Core concept for sequence reconstruction
• Used in algebraic modelling and coding logic
• Foundation for advanced sequence problems

Concept Used

When non-consecutive terms are given, we use the nth term formula:

\[ a_n = a + (n-1)d \]

This helps form equations which can be solved simultaneously to find \(a\) and \(d\).

AP grows uniformly → enables equation formation using term positions

Solution Roadmap

Step 1: Write expressions for given terms using \(a_n = a + (n-1)d\)
Step 2: Form two equations in \(a\) and \(d\)
Step 3: Eliminate one variable to find \(d\)
Step 4: Substitute back to get \(a\)
Step 5: Use formula to compute required term

Solution

\[ \begin{aligned} a_3 &= 12,\quad a_{50} = 106 \end{aligned} \]

\[ \begin{align} a_3 &= a + (3-1)d\\ &= a + 2d = 12 \tag{1} \end{align} \]

\[ \begin{align} a_{50} &= a + (50-1)d\\ &= a + 49d = 106 \tag{2} \end{align} \]

Subtracting (1) from (2):

\[ \begin{aligned} (a + 49d) - (a + 2d) &= 106 - 12\\ a + 49d - a - 2d &= 94\\ 47d &= 94\\ d &= \frac{94}{47}\\ d &= 2 \end{aligned} \]

Substitute \(d = 2\) into (1):

\[ \begin{aligned} a + 2d &= 12\\ a + 2\times 2 &= 12\\ a + 4 &= 12\\ a &= 12 - 4\\ a &= 8 \end{aligned} \]

Now find the 29th term:

\[ \begin{aligned} a_{29} &= a + (29-1)d\\ &= 8 + 28\times 2\\ &= 8 + 56\\ &= 64 \end{aligned} \]

Therefore, the 29th term is 64.

Exam Significance

This problem integrates term indexing with equation solving — a high-weight concept.

In CBSE Board Exams:
• Common 3-mark question
• Tests clarity in using (n−1) correctly

In Competitive Exams:
• Used in sequence reconstruction problems
• Important for algebraic modelling
• Builds foundation for advanced sequence and series

Concept Used

When two terms of an AP are given, we form equations using:

\[ a_n = a + (n-1)d \]

After finding \(a\) and \(d\), we determine the term number for which \(a_n = 0\).

AP decreases uniformly (negative d)

Solution Roadmap

Step 1: Form equations using given terms
Step 2: Solve simultaneously to find \(d\)
Step 3: Substitute to get \(a\)
Step 4: Set \(a_n = 0\)
Step 5: Solve for \(n\)

Solution

\[ \begin{aligned} a_3 &= 4,\quad a_9 = -8 \end{aligned} \]

\[ \begin{align} a_3 &= a + (3-1)d\\ &= a + 2d = 4 \tag{1} \end{align} \]

\[ \begin{align} a_9 &= a + (9-1)d\\ &= a + 8d = -8 \tag{2} \end{align} \]

Subtracting (1) from (2):

\[ \begin{aligned} (a + 8d) - (a + 2d) &= -8 - 4\\ a + 8d - a - 2d &= -12\\ 6d &= -12\\ d &= \frac{-12}{6}\\ d &= -2 \end{aligned} \]

Substitute \(d = -2\) into (1):

\[ \begin{aligned} a + 2d &= 4\\ a + 2(-2) &= 4\\ a - 4 &= 4\\ a &= 8 \end{aligned} \]

Now find the term where \(a_n = 0\):

\[ \begin{aligned} a_n &= a + (n-1)d\\ 0 &= 8 + (n-1)(-2) \end{aligned} \]

\[ \begin{aligned} 0 &= 8 - 2(n-1)\\ 2(n-1) &= 8\\ n - 1 &= 4\\ n &= 5 \end{aligned} \]

Therefore, the 5th term of the AP is zero.

Exam Significance

This is a classic multi-step AP problem combining equation solving and term identification.

In CBSE Board Exams:
• Common 3-mark question
• Tests clarity in handling negative common difference

In Competitive Exams:
• Important for sequence reconstruction
• Appears in algebraic modelling problems
• Builds strong foundation for advanced sequence logic

Concept Used

In an AP, the nth term is given by:

\[ a_n = a + (n-1)d \]

A key idea:
Difference between two terms depends only on the difference of their positions:

\[ a_m - a_n = (m - n)d \]

Gap between terms depends on number of steps × d

Solution Roadmap

Step 1: Write expressions for \(a_{17}\) and \(a_{10}\)
Step 2: Use given condition (difference = 7)
Step 3: Form equation
Step 4: Solve for \(d\)

Solution (Method 1: Direct Substitution)

\[ \begin{aligned} a_{17} &= a + (17-1)d = a + 16d\\ a_{10} &= a + (10-1)d = a + 9d \end{aligned} \]

Given:
17th term exceeds 10th term by 7

\[ \begin{aligned} a_{17} &= a_{10} + 7\\ a + 16d &= a + 9d + 7 \end{aligned} \]

\[ \begin{aligned} a + 16d - a - 9d &= 7\\ 7d &= 7\\ d &= 1 \end{aligned} \]

Solution (Method 2: Shortcut Concept)

\[ \begin{aligned} a_{17} - a_{10} &= (17 - 10)d\\ 7 &= 7d\\ d &= 1 \end{aligned} \]

Common Difference \(d = 1\)

Exam Significance

This is a high-efficiency concept question where shortcut understanding saves time.

In CBSE Board Exams:
• Common 2-mark question
• Shortcut method gives faster solution

In Competitive Exams:
• Tests conceptual clarity over formula memorization
• Important for speed-based solving
• Frequently appears in aptitude and reasoning

Concept Used

In an AP:

\[ a_n = a + (n-1)d \]

Important shortcut:

\[ a_n - a_m = (n - m)d \]

This avoids unnecessary long calculations.

Gap between terms depends on number of steps × d

Solution Roadmap

Step 1: Find common difference \(d\)
Step 2: Use relation \(a_n = a_{54} + 132\)
Step 3: Apply shortcut \(a_n - a_{54} = (n-54)d\)
Step 4: Solve for \(n\)

Solution

\(\text{AP: } 3,\ 15,\ 27,\ 39,\ \ldots\)

\[ \begin{aligned} a_1 &= 3,\quad a_2 = 15\\ d &= a_2 - a_1\\ &= 15 - 3\\ &= 12 \end{aligned} \]

Given condition:

\[ a_n = a_{54} + 132 \]

\[ \begin{aligned} a_n - a_{54} &= 132 \end{aligned} \]

\[ \begin{aligned} (n - 54)d &= 132\\ (n - 54)\times 12 &= 132 \end{aligned} \]

\[ \begin{aligned} n - 54 &= \frac{132}{12}\\ n - 54 &= 11\\ n &= 54 + 11\\ n &= 65 \end{aligned} \]

Therefore, the 65th term is 132 more than the 54th term.

Verification (Optional but Recommended)

\[ \begin{aligned} a_{54} &= 3 + (54-1)\times 12 = 3 + 53\times 12 = 639\\ a_{65} &= 3 + 64\times 12 = 771 \end{aligned} \]

\[ 771 - 639 = 132 \quad \checkmark \]

Exam Significance

This problem strongly tests conceptual understanding of term differences.

In CBSE Board Exams:
• Common 2–3 mark question
• Shortcut method saves time and reduces errors

In Competitive Exams:
• Important for fast calculations
• Used in sequence comparison problems
• Key for time optimization strategies

Concept Used

Let two APs have:

• First terms: \(a_1\) and \(b_1\)
• Common difference: \(d\) (same for both)

\[ a_n = a_1 + (n-1)d,\quad b_n = b_1 + (n-1)d \]

Key Insight:

\[ a_n - b_n = (a_1 - b_1) \]

The difference between corresponding terms remains constant if common difference is same.

Two APs with same d → constant vertical gap

Solution Roadmap

Step 1: Write nth term of both APs
Step 2: Find expression for difference
Step 3: Use given condition (100th term difference)
Step 4: Apply same logic for 1000th term

Solution

Let first AP have first term \(a_1\), second AP have first term \(b_1\), and common difference \(d\).

100th terms:

\[ \begin{aligned} a_{100} &= a_1 + 99d\\ b_{100} &= b_1 + 99d \end{aligned} \]

\[ \begin{aligned} a_{100} - b_{100} &= (a_1 + 99d) - (b_1 + 99d)\\ &= a_1 - b_1 \end{aligned} \]

Given:

\[ a_{100} - b_{100} = 100 \]

\[ \Rightarrow a_1 - b_1 = 100 \tag{1} \]

1000th terms:

\[ \begin{aligned} a_{1000} &= a_1 + 999d\\ b_{1000} &= b_1 + 999d \end{aligned} \]

\[ \begin{aligned} a_{1000} - b_{1000} &= (a_1 + 999d) - (b_1 + 999d)\\ &= a_1 - b_1 \end{aligned} \]

\[ \begin{aligned} &= 100 \quad \text{(from (1))} \end{aligned} \]

Therefore, the difference between their 1000th terms is also 100.

Exam Significance

This is a high-concept invariance problem — difference remains constant.

In CBSE Board Exams:
• Conceptual 2-mark question
• Tests understanding beyond direct formula use

In Competitive Exams:
• Very important shortcut concept
• Helps avoid lengthy calculations
• Frequently used in sequence comparison problems

Concept Used

All numbers divisible by 7 form an Arithmetic Progression (AP):

\[ 7,\ 14,\ 21,\ 28,\ \ldots \]

For three-digit numbers:
• Smallest 3-digit number = 100
• Largest 3-digit number = 999

We must find:
• First multiple of 7 ≥ 100
• Last multiple of 7 ≤ 999

Multiples of 7 form an AP with d = 7

Solution Roadmap

Step 1: Find first 3-digit number divisible by 7
Step 2: Find last 3-digit number divisible by 7
Step 3: Treat them as AP with \(d = 7\)
Step 4: Use nth term formula to find number of terms

Solution

Step 1: First 3-digit multiple of 7

Smallest 3-digit number = 100

\[ \frac{100}{7} \approx 14.28 \]

Next integer = 15

\[ 7 \times 15 = 105 \]

So, first term \(a_1 = 105\)

Step 2: Last 3-digit multiple of 7

Largest 3-digit number = 999

\[ \frac{999}{7} \approx 142.71 \]

Largest integer = 142

\[ 7 \times 142 = 994 \]

So, last term \(a_n = 994\)

Step 3: Form AP

\[ 105,\ 112,\ 119,\ \ldots,\ 994 \]

Common difference \(d = 7\)

Step 4: Find number of terms

\[ \begin{aligned} a_n &= a + (n-1)d\\ 994 &= 105 + (n-1)\times 7 \end{aligned} \]

\[ \begin{aligned} 994 - 105 &= (n-1)\times 7\\ 889 &= (n-1)\times 7\\ n - 1 &= \frac{889}{7}\\ n - 1 &= 127\\ n &= 127 + 1\\ n &= 128 \end{aligned} \]

Therefore, there are 128 three-digit numbers divisible by 7.

Exam Significance

This is a classic application of AP in number systems.

In CBSE Board Exams:
• Common 2–3 mark question
• Tests modelling real problems into AP

In Competitive Exams:
• Frequently appears in divisibility and counting problems
• Important for number theory
• Builds strong logical structuring skills

Concept Used

Multiples of 4 form an Arithmetic Progression (AP):

\[ 4,\ 8,\ 12,\ 16,\ \ldots \]

We need multiples strictly between 10 and 250.

So we find:
• First multiple of 4 greater than 10
• Last multiple of 4 less than 250

Multiples of 4 increase uniformly (d = 4)

Solution Roadmap

Step 1: Find first multiple of 4 greater than 10
Step 2: Find last multiple of 4 less than 250
Step 3: Form AP
Step 4: Use nth term formula to count terms

Solution

Step 1: First multiple of 4 greater than 10

\[ \frac{10}{4} = 2.5 \]

Next integer = 3

\[ 4 \times 3 = 12 \]

So, first term \(a_1 = 12\)

Step 2: Last multiple of 4 less than 250

\[ \frac{250}{4} = 62.5 \]

Largest integer = 62

\[ 4 \times 62 = 248 \]

So, last term \(a_n = 248\)

Step 3: Form AP

\[ 12,\ 16,\ 20,\ \ldots,\ 248 \]

Common difference \(d = 4\)

Step 4: Find number of terms

\[ \begin{aligned} a_n &= a + (n-1)d\\ 248 &= 12 + (n-1)\times 4 \end{aligned} \]

\[ \begin{aligned} 248 - 12 &= (n-1)\times 4\\ 236 &= (n-1)\times 4\\ n - 1 &= \frac{236}{4}\\ n - 1 &= 59\\ n &= 59 + 1\\ n &= 60 \end{aligned} \]

Therefore, there are 60 multiples of 4 between 10 and 250.

Exam Significance

This is a classic counting problem using AP modelling.

In CBSE Board Exams:
• Common 2–3 mark question
• Tests correct boundary identification

In Competitive Exams:
• Important for divisibility and counting
• Used in number theory problems
• Builds structured problem-solving approach

Concept Used

The nth term of an AP is:

\[ a_n = a + (n-1)d \]

To compare two APs, we equate their nth term expressions and solve for \(n\).

Two APs intersect when their nth terms become equal

Solution Roadmap

Step 1: Find \(a\) and \(d\) for both APs
Step 2: Write nth term expressions
Step 3: Equate both expressions
Step 4: Solve linear equation in \(n\)

Solution

First AP: \(63,\ 65,\ 67,\ \ldots\)

\[ \begin{aligned} a_1 &= 63,\quad a_2 = 65\\ d &= a_2 - a_1 = 65 - 63 = 2 \end{aligned} \]

Second AP: \(3,\ 10,\ 17,\ \ldots\)

\[ \begin{aligned} a'_1 &= 3,\quad a'_2 = 10\\ d' &= a'_2 - a'_1 = 10 - 3 = 7 \end{aligned} \]

nth term expressions:

\[ \begin{aligned} \text{First AP: } a_n &= 63 + (n-1)\times 2\\ \text{Second AP: } a'_n &= 3 + (n-1)\times 7 \end{aligned} \]

Equating both terms:

\[ \begin{aligned} 63 + (n-1)\times 2 &= 3 + (n-1)\times 7 \end{aligned} \]

\[ \begin{aligned} 63 - 3 &= (n-1)\times 7 - (n-1)\times 2\\ 60 &= (n-1)(7 - 2)\\ 60 &= 5(n-1) \end{aligned} \]

\[ \begin{aligned} n - 1 &= \frac{60}{5}\\ n - 1 &= 12\\ n &= 12 + 1\\ n &= 13 \end{aligned} \]

Therefore, the 13th terms of both APs are equal.

Verification

\[ \begin{aligned} \text{First AP: } a_{13} &= 63 + 12\times 2 = 87\\ \text{Second AP: } a'_{13} &= 3 + 12\times 7 = 87 \end{aligned} \]

\[ \Rightarrow \text{Both are equal} \quad \checkmark \]

Exam Significance

This is a classic “intersection of two APs” problem.

In CBSE Board Exams:
• Common 3-mark question
• Tests equation formation and simplification

In Competitive Exams:
• Important for sequence comparison problems
• Helps in solving simultaneous progression problems
• Builds algebraic modelling strength

Concept Used

The nth term of an AP is:

\[ a_n = a + (n-1)d \]

Also, an important shortcut:

\[ a_m - a_n = (m-n)d \]

This helps simplify conditions like “one term exceeds another”.

Equal spacing → constant difference d

Solution Roadmap

Step 1: Use given term to form equation
Step 2: Convert “exceeds by” into equation
Step 3: Solve for common difference \(d\)
Step 4: Substitute to find first term \(a\)
Step 5: Write the AP

Solution

Given:

\[ a_3 = 16 \]

\[ \begin{aligned} a_3 &= a + (3-1)d\\ &= a + 2d = 16 \quad \text{(1)} \end{aligned} \]

Condition:

7th term exceeds 5th term by 12

\[ a_7 = a_5 + 12 \]

\[ \begin{aligned} a_7 &= a + 6d\\ a_5 &= a + 4d \end{aligned} \]

\[ \begin{aligned} a + 6d &= a + 4d + 12\\ 6d - 4d &= 12\\ 2d &= 12\\ d &= 6 \end{aligned} \]

Substitute \(d = 6\) into (1):

\[ \begin{aligned} a + 2d &= 16\\ a + 2\times 6 &= 16\\ a + 12 &= 16\\ a &= 4 \end{aligned} \]

Form the AP:

\[ \begin{aligned} a,\ a+d,\ a+2d,\ a+3d,\ \ldots \end{aligned} \]

\[ \begin{aligned} 4,\ 10,\ 16,\ 22,\ 28,\ \ldots \end{aligned} \]

Therefore, the required AP is: \(4,\ 10,\ 16,\ 22,\ \ldots\)

Verification

\[ \begin{aligned} a_3 &= 4 + 2\times 6 \\&= 16 \quad \checkmark\\ a_7 - a_5 &= (4+6\times 6) - (4+4\times 6) \\&= 40 - 28 \\&= 12 \quad \checkmark \end{aligned} \]

Exam Significance

This is a structured AP formation problem combining multiple concepts.

In CBSE Board Exams:
• Common 3-mark question
• Tests equation formation and interpretation

In Competitive Exams:
• Important for sequence construction
• Builds algebraic modelling ability
• Useful in pattern-based reasoning problems

Concept Used

The nth term of an AP is:

\[ a_n = a + (n-1)d \]

Important idea:
kth term from the end = \((n - k + 1)\)th term from the beginning

Forward and backward indexing in AP

Solution Roadmap

Step 1: Find common difference \(d\)
Step 2: Find total number of terms \(n\)
Step 3: Convert “20th from last” into forward index
Step 4: Find required term

Solution (Method 1: Index Conversion)

\(\text{AP: } 3,\ 8,\ 13,\ \ldots,\ 253\)

\[ \begin{aligned} a_1 &= 3,\quad a_2 = 8\\ d &= a_2 - a_1 = 8 - 3 = 5 \end{aligned} \]

\[ \begin{aligned} a_n &= 253\\ 253 &= 3 + (n-1)\times 5 \end{aligned} \]

\[ \begin{aligned} 253 - 3 &= (n-1)\times 5\\ 250 &= (n-1)\times 5\\ n - 1 &= \frac{250}{5}\\ n - 1 &= 50\\ n &= 51 \end{aligned} \]

Convert “20th from last”:

\[ \text{Required term} = (n - 20 + 1) = 51 - 20 + 1 = 32 \]

\[ \begin{aligned} a_{32} &= 3 + (32-1)\times 5\\ &= 3 + 31\times 5\\ &= 3 + 155\\ &= 158 \end{aligned} \]

Solution (Method 2: Reverse AP)

Consider reverse AP:

\[ 253,\ 248,\ 243,\ \ldots \]

\[ a_1 = 253,\quad d = -5 \]

Now find 20th term:

\[ \begin{aligned} a_{20} &= 253 + (20-1)(-5)\\ &= 253 + 19(-5)\\ &= 253 - 95\\ &= 158 \end{aligned} \]

Therefore, the 20th term from the end is 158.

Exam Significance

This is a classic “term from end” problem — very important concept.

In CBSE Board Exams:
• Frequently appears as 2–3 mark question
• Tests indexing accuracy

In Competitive Exams:
• Key for sequence reversal problems
• Useful in fast mental calculations
• Builds strong positional understanding

Concept Used

nth term of an AP:

\[ a_n = a + (n-1)d \]

When sums of terms are given, we:
• Express each term in terms of \(a\) and \(d\)
• Form linear equations
• Solve simultaneously

Each term expressed as a + kd helps build equations

Solution Roadmap

Step 1: Express given terms using formula
Step 2: Form equations using given sums
Step 3: Solve equations to find \(d\)
Step 4: Substitute to find \(a\)
Step 5: Write first three terms

Solution

First condition:

\[ \begin{aligned} a_4 &= a + 3d\\ a_8 &= a + 7d \end{aligned} \]

\[ \begin{aligned} a_4 + a_8 &= 24\\ (a+3d) + (a+7d) &= 24\\ 2a + 10d &= 24 \quad \text{(1)} \end{aligned} \]

Second condition:

\[ \begin{aligned} a_6 &= a + 5d\\ a_{10} &= a + 9d \end{aligned} \]

\[ \begin{aligned} a_6 + a_{10} &= 44\\ (a+5d) + (a+9d) &= 44\\ 2a + 14d &= 44 \quad \text{(2)} \end{aligned} \]

Subtract (1) from (2):

\[ \begin{aligned} (2a + 14d) - (2a + 10d) &= 44 - 24\\ 4d &= 20\\ d &= \frac{20}{4}\\ d &= 5 \end{aligned} \]

Substitute \(d = 5\) into (1):

\[ \begin{aligned} 2a + 10d &= 24\\ 2a + 50 &= 24\\ 2a &= 24 - 50\\ 2a &= -26\\ a &= -13 \end{aligned} \]

First three terms:

\[ \begin{aligned} a_1 &= -13\\ a_2 &= a + d = -13 + 5 = -8\\ a_3 &= a + 2d = -13 + 10 = -3 \end{aligned} \]

Therefore, the first three terms are: \(-13,\ -8,\ -3\)

Verification

\[ \begin{aligned} a_4 + a_8 &= (-13+15) + (-13+35) \\&= 2 + 22 \\&= 24 \quad \checkmark\ a_6 + a_{10} &= (-13+25) + (-13+45) \\&= 12 + 32 \\&= 44 \quad \checkmark \end{aligned} \]

Exam Significance

This is a classic equation-based AP problem involving term pairing.

In CBSE Board Exams:
• Frequently appears as 3–4 mark question
• Tests algebraic accuracy and structure

In Competitive Exams:
• Important for sequence reconstruction
• Builds strong simultaneous equation skills
• Useful in pattern and series problems

Concept Used

This is a real-life application of Arithmetic Progression (AP).

Salary increases uniformly every year → forms an AP:

\[ a_n = a + (n-1)d \]

Important:
• \(n\) represents number of years after starting year
• Final answer must be converted into actual calendar year

Salary increases by a constant amount every year

Solution Roadmap

Step 1: Identify \(a\), \(d\), and target salary
Step 2: Use nth term formula
Step 3: Solve for \(n\)
Step 4: Convert \(n\) into actual year

Solution

Given:

Starting year = 1995
First salary \(a_1 = 5000\)
Increment \(d = 200\)
Target salary \(a_n = 7000\)

\[ \begin{aligned} a_n &= a + (n-1)d\\ 7000 &= 5000 + (n-1)\times 200 \end{aligned} \]

\[ \begin{aligned} 7000 - 5000 &= (n-1)\times 200\\ 2000 &= (n-1)\times 200\\ n - 1 &= \frac{2000}{200}\\ n - 1 &= 10\\ n &= 11 \end{aligned} \]

Interpretation:

11th term means 11th year from 1995.

\[ \text{Year} = 1995 + (11 - 1) = 1995 + 10 = 2005 \]

Therefore, Subba Rao’s income reached ₹7000 in the year 2005.

Exam Significance

This is a real-life AP modelling question.

In CBSE Board Exams:
• Common 2–3 mark question
• Key mistake: forgetting to convert term number into actual year

In Competitive Exams:
• Important for financial modelling problems
• Used in growth and progression scenarios
• Tests interpretation skills beyond calculation

Concept Used

Weekly savings increase uniformly → forms an Arithmetic Progression (AP).

\[ a_n = a + (n-1)d \]

Decimal values are best handled by converting into fractions for precise calculation.

Weekly savings increase by a fixed amount

Solution Roadmap

Step 1: Identify \(a\), \(d\), and \(a_n\)
Step 2: Convert decimals into fractions
Step 3: Apply nth term formula
Step 4: Solve for \(n\)

Solution

Given:

First term \(a_1 = 5\)
Common difference \(d = 1.75 = \frac{7}{4}\)
nth term \(a_n = 20.75 = \frac{83}{4}\)

\[ \begin{aligned} a_n &= a + (n-1)d\\ \frac{83}{4} &= 5 + (n-1)\cdot \frac{7}{4} \end{aligned} \]

\[ \begin{aligned} 5 &= \frac{20}{4}\\ \frac{83}{4} &= \frac{20}{4} + (n-1)\cdot \frac{7}{4} \end{aligned} \]

\[ \begin{aligned} \frac{83 - 20}{4} &= (n-1)\cdot \frac{7}{4}\\ \frac{63}{4} &= (n-1)\cdot \frac{7}{4} \end{aligned} \]

\[ \begin{aligned} n-1 &= \frac{63}{7}\\ n-1 &= 9\\ n &= 10 \end{aligned} \]

Therefore, Ramkali’s weekly savings become ₹20.75 in the 10th week.

Exam Significance

This is a real-life AP problem involving decimals.

In CBSE Board Exams:
• Common 2–3 mark question
• Tests handling of decimals and formula application

In Competitive Exams:
• Important for financial progression problems
• Tests precision in calculations
• Builds modelling skills for real-world scenarios

Chapter 5 — ARITHMETIC PROGRESSIONS

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