{"slug":"cbse-class-10-maths-arithmetic-progressions","title":"Arithmetic Progressions","description":"Recognising an AP, the nth term, the sum of the first n terms and the word problems built on them, weighted towards the production case study and sum-of-terms questions in the 2026 papers.","board":"CBSE","grade":"Class 10","subject":"Maths","chapter":"Arithmetic Progressions","url":"https://www.flipnlearn.app/deck/cbse-class-10-maths-arithmetic-progressions","studyUrl":"https://www.flipnlearn.app/deck/cbse-class-10-maths-arithmetic-progressions/study","cards":[{"id":"arithmetic-progression-ap","kind":"term-definition","front":"Arithmetic progression (AP)","back":"A list of numbers in which every term after the first is got by adding the same fixed number to the term before it. Examples are 1, 2, 3, 4, ... and 100, 70, 40, 10, ..."},{"id":"common-difference","kind":"term-definition","front":"Common difference","back":"The fixed number d added to each term of an AP to get the next, found as a(k+1) − a(k) for any consecutive pair. It may be positive, negative or zero, and you always subtract the earlier term from the later one, even when the later term is smaller."},{"id":"general-form-of-an-ap","kind":"term-definition","front":"General form of an AP","back":"a, a + d, a + 2d, a + 3d, ..., with first term a and common difference d. Knowing both a and d is enough to write out the whole AP; either one alone is not."},{"id":"finite-ap","kind":"term-definition","front":"Finite AP","back":"An AP with only a finite number of terms, so it has a last term, such as the monthly balances 950, 900, 850, ..., 50. An AP that goes on without end has no last term and is called infinite."},{"id":"how-to-check-whether-a-list-is-an-ap","kind":"worked-step","front":"How do you check whether a list of numbers such as −2, 2, −2, 2, ... forms an AP?","back":"Find the differences between consecutive terms and see whether they are all equal. Here 2 − (−2) = 4 but −2 − 2 = −4, so the differences change and the list is not an AP. For 4, 10, 16, 22 every difference is 6, so it is."},{"id":"general-term-of-an-ap","kind":"formula","front":"nth term (general term) of an AP","back":"aₙ = a + (n − 1)d, with first term a and common difference d. In an AP with m terms, the last term aₘ is often written l."},{"id":"why-the-difference-of-two-terms-depends-only-on-d","kind":"concept-question","front":"Why does the difference between two terms of an AP, such as a₁₇ − a₁₀, depend only on the common difference?","back":"Writing both with aₙ = a + (n − 1)d gives (a + 16d) − (a + 9d). The first term cancels, leaving 7d, so in general aₘ − aₙ = (m − n)d whatever the first term is."},{"id":"how-to-find-which-term-has-a-given-value","kind":"worked-step","front":"How do you find which term of the AP 21, 18, 15, ... equals −81?","back":"Put a = 21, d = −3 and aₙ = −81 into aₙ = a + (n − 1)d, giving −81 = 21 + (n − 1)(−3). This simplifies to −105 = −3n, so n = 35 and −81 is the 35th term."},{"id":"why-a-non-integer-n-means-not-a-term","kind":"concept-question","front":"Why does getting n = 151/3 show that 301 is not a term of 5, 11, 17, 23, ...?","back":"n counts the position of a term, so it must be a positive integer. Solving 301 = 5 + (n − 1) × 6 gives a fraction, which means no position in the list holds 301."},{"id":"how-to-find-an-ap-from-two-of-its-terms","kind":"worked-step","front":"How do you find an AP when you know its 3rd term is 5 and its 7th term is 9?","back":"Write each given term using aₙ = a + (n − 1)d: a + 2d = 5 and a + 6d = 9. Solve this pair of linear equations to get a = 3 and d = 1, so the AP is 3, 4, 5, 6, ..."},{"id":"how-to-count-multiples-in-a-range","kind":"worked-step","front":"How do you count the two-digit numbers divisible by 3 using an AP?","back":"List them as an AP from the first to the last such number: 12, 15, ..., 99, so a = 12, d = 3 and the last term is 99. Then 99 = 12 + (n − 1) × 3 gives n − 1 = 29, so there are 30 of them."},{"id":"how-to-find-a-term-from-the-end","kind":"worked-step","front":"How do you find the 11th term from the last term of the AP 10, 7, 4, ..., −62?","back":"Find the number of terms first: −62 = 10 + (n − 1)(−3) gives n = 25. Counting back 11 from the 25th reaches the 15th term, 10 + 14 × (−3) = −32. Alternatively, reverse the AP so a = −62 and d = 3, and find its 11th term."},{"id":"sum-of-first-n-terms","kind":"formula","front":"Sum of first n terms of an AP","back":"Sₙ = (n/2)[2a + (n − 1)d], with first term a and common difference d. It links four quantities, Sₙ, a, d and n, so any three of them determine the fourth."},{"id":"sum-using-first-and-last-term","kind":"formula","front":"Sum using first and last term of an AP","back":"S = (n/2)(a + l), where l is the last (nth) term. It is the handy form when the first and last terms are known but the common difference is not."},{"id":"relation-between-term-and-sum","kind":"formula","front":"Relation between term and sum in an AP","back":"aₙ = Sₙ − Sₙ₋₁: the nth term is the sum of the first n terms minus the sum of the first (n − 1) terms. It recovers a term when only a formula for the sum is given."},{"id":"sum-of-first-n-positive-integers","kind":"formula","front":"Sum of the first n positive integers","back":"Sₙ = n(n + 1)/2, from S = (n/2)(a + l) with a = 1 and l = n. For instance, 1 + 2 + ... + 1000 = 500500."},{"id":"why-write-the-sum-twice","kind":"concept-question","front":"Why does writing an AP's sum forwards and then backwards help find it?","back":"Adding the two versions term by term pairs the first with the last, the second with the second-last, and so on, and every pair has the same total. So twice the sum is n equal pairs, as when Gauss got 2S = 100 × 101 for 1 to 100."},{"id":"why-two-values-of-n-give-the-same-sum","kind":"concept-question","front":"In the AP 24, 21, 18, ..., why do both 4 terms and 13 terms give a sum of 78?","back":"The first term is positive and the common difference negative, so later terms turn negative. The 5th to 13th terms add up to zero, leaving the sum unchanged from 4 terms to 13; that is why the quadratic in n has two admissible roots."},{"id":"how-to-find-d-from-a-given-sum","kind":"worked-step","front":"If the first 14 terms of an AP add up to 1050 and the first term is 10, how do you find the 20th term?","back":"Substitute into Sₙ = (n/2)[2a + (n − 1)d]: 1050 = 7(20 + 13d) = 140 + 91d, so d = 10. Then use aₙ = a + (n − 1)d to get a₂₀ = 10 + 19 × 10 = 200."},{"id":"how-to-solve-an-ap-production-problem","kind":"worked-step","front":"A factory makes 600 TV sets in year 3 and 700 in year 7, rising uniformly. How do you find the first year's output and the 7-year total?","back":"Treat yearly output as an AP: a + 2d = 600 and a + 6d = 700 give d = 25 and a = 550. Then S₇ = (7/2)[2 × 550 + 6 × 25] = 4375 sets over the first seven years."},{"id":"arithmetic-mean","kind":"formula","front":"Arithmetic mean of a and c","back":"If a, b and c are in AP, then b = (a + c)/2; b is then the arithmetic mean of a and c, lying midway between them."}]}