Arithmetic Progression • Topic 3 of 3

AP Applications

This is where the AP earns its place in CAT. Two ideas unlock most application questions. First, the neighbour-average property: in any AP each term is the exact average of the terms on either side, so a_n = (a_(n−1) + a_(n+1))/2 — three consecutive terms in AP means the middle one is their mean. Second, the symmetric-term trick: when a problem gives you the sum (and often the product) of an unknown set of terms in AP, do not write them as a, a + d, a + 2d. Instead centre them around the middle: take three terms as (a − d, a, a + d), four terms as (a − 3d, a − d, a + d, a + 3d) with common difference 2d, and five as (a − 2d, a − d, a, a + d, a + 2d). The plus and minus d-terms cancel when you add, so the sum collapses to (number of terms)×a and you read a off instantly, leaving just one equation for d. This converts ugly simultaneous systems into a single clean line — a genuine CAT time-saver.

✅ Solved examples

1. Three numbers in AP have sum 21 and product 231. Find the numbers.
Take them as (a − d), a, (a + d). Sum = 3a = 21 ⇒ a = 7. Product = 7(49 − d²) = 231 ⇒ 49 − d² = 33 ⇒ d² = 16 ⇒ d = 4. Numbers: 3, 7, 11.
2. If x + 1, 2x and 3x − 1 are three consecutive terms of an AP, find x.
The middle term is the average: 2(2x) = (x + 1) + (3x − 1) ⇒ 4x = 4x. This holds for all x, so check via equal differences: 2x − (x + 1) = (3x − 1) − 2x ⇒ x − 1 = x − 1, true for all x. They are in AP for every real x.
3. An auditorium has 20 rows; the first row has 18 seats and each next row has 2 more. Total seats?
AP with a = 18, d = 2, n = 20. Last row = 18 + 19×2 = 56. Total = 20/2 (18 + 56) = 10×74 = 740 seats.
4. A man repays a loan of ₹3,250 in monthly instalments, the first being ₹20 and each later one ₹15 more than the previous. In how many months is it cleared?
a = 20, d = 15, S_n = 3250. n/2 [40 + (n − 1)15] = 3250 ⇒ n(15n + 25) = 6500 ⇒ 3n² + 5n − 1300 = 0 ⇒ n = 20 months.

✏️ Practice — try these, take hints as needed

1. Three numbers in AP have sum 27 and product 504. Find the smallest.
Take (a − d), a, (a + d).
3a = 27 ⇒ a = 9.
9(81 − d²) = 504.
6 (numbers 6, 9, 12)
2. If k, 2k + 1 and 4k − 3 are in AP, find k.
Middle is the average.
2(2k + 1) = k + (4k − 3).
4k + 2 = 5k − 3.
5
3. A stack has 25 logs in the bottom row, 24 in the next, and so on up to 1. How many logs in all?
AP 25, 24, …, 1.
Sum of 1 to 25.
25×26/2.
325
4. Four numbers in AP have sum 32 and the product of the two middle terms is 60. Find the numbers.
Take (a − 3d), (a − d), (a + d), (a + 3d).
4a = 32 ⇒ a = 8.
(8 − d)(8 + d) = 60 ⇒ d = 2.
2, 6, 10, 14
5. A worker is paid ₹100 in week 1 and ₹20 more each following week. What is his total pay over 15 weeks?
a = 100, d = 20, n = 15.
S_n = n/2 [2a + (n − 1)d].
15/2 [200 + 14×20].
₹3,600

📝 Topic test — 8 questions

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