Exercise-Based Problem Solving and Applications

Theory: Many practical cases—like fare calculation, loan repayment, or tree planting—can be modeled using AP, especially when values change regularly.

Example:

Question: A well digging starts at ₹150 for the first meter and increases by ₹50 for every additional meter. Write the AP.

Solution: AP: 150, 200, 250, 300, …

Theory: A sequence is an AP if the difference between all consecutive terms is constant.

Example:

Question: Is the sequence 5, 8, 12, 17 an AP?

Solution: 8 – 5 = 3, 12 – 8 = 4 → Not equal, hence Not an AP

Theory: When certain conditions like “3rd term = 5, 7th term = 9” are given, create equations using: \[a_n = a + (n – 1)d\]

Example:

Question: Determine the AP if 3rd term = 5 and 7th term = 9.

Solution: \[a + 2d = 5\]\[a + 6d = 9 ⇒ d = 1, a = 3\]\[AP: 3, 4, 5, 6, …\]

Theory: If some terms are missing, use known terms to determine a and d, then fill the blanks.

Example:

Question: Fill the blanks in the AP: 2, __, __, __, 26

Solution: \(a_5 = a + 4d = 26 ⇒ d = 6\)
Sequence: 2, 8, 14, 20, 26

Theory: Many application-based questions require modeling the situation into an AP to find total cost, term value, or number of steps.

Example:

Question: Ramkali saves ₹5 in first week and increases by ₹1.75 weekly. In which week will her saving be ₹20.75?

Solution: a = 5, d = 1.75 \[20.75 = 5 + (n – 1)×1.75 ⇒ n = 10\]

Theory: From interest calculation to counting multiples, AP appears in finance, scheduling, and planning scenarios.

Example:

Question: Find how many 3-digit numbers are divisible by 7.

Solution: First: 105, Last: 994, d = 7 \[a_n = a + (n – 1)d ⇒ 994 = 105 + (n – 1)×7 ⇒ n = 128\]

Theory: When term position or total sum is known, you can reverse-engineer the AP formula.

Example:

Question: Find the common difference if 17th term exceeds 10th term by 7.

Solution: \[a + 16d – (a + 9d) = 7 ⇒ d = 1\]

Theory: If the nth term is given as a formula, plug in values to form the AP.

Example:

Question: If \(a_n = 3 + 2n\), find the first 4 terms and their sum.

Solution: \[a_1 = 5, a_2 = 7, a_3 = 9, a_4 = 11\] \[Sum: 5 + 7 + 9 + 11 = 32\]

Theory: Patterns like logs in a pyramid, or steps in a terrace, often follow AP for dimensions or counts.

Example:

Question: A terrace has 15 steps each 0.25 m high and 0.5 m wide. Find total volume if each is 50 m long.

Solution: \[V = 50 × ∑(0.25 × 0.5n) = AP sum with a = 0.125\]

Theory: Olympiad-style or NEP-aligned questions combine multiple AP concepts like position, sum, comparison.

Example:

Question: Which term of AP: 3, 15, 27, … is 132 more than the 54th term?

Solution: \(d = 12 ⇒ Let term = a_n\) \[a_n – a_54 = 132 ⇒ (n – 54)×12 = 132 ⇒ n = 65\]