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Concept 02 / 07Sequences and Series

Arithmetic Progressions: nth Term, Sum and Properties

The nth term a + (n - 1)d and the sum n/2 (2a + (n - 1)d) derived by pairing, recovering a term from sums (T_n = S_n - S_(n-1)), symmetric choice of terms, properties preserved under operations, and the arithmetic mean.

Figure 1
A sequence is a function on the natural numbers, n↦Tnn \mapsto T_{n}; a series is the sum of its terms. An arithmetic progression (AP) is a sequence with a constant difference between consecutive terms: Tn+1−Tn=dT_{n+1} - T_{n} = d for all nn. With first term aa, Tn=a+(n−1)dT_{n} = a + (n - 1)d. So the terms of an AP are values of a linear function of nn: plotted against nn they lie on a straight line with slope dd. Conversely, any sequence with Tn=pn+qT_{n} = pn + q is an AP with d=pd = p. Sum of the first nn terms. Write the sum forwards and backwards and add: each pair of vertically aligned terms sums to a+la + l, where l=a+(n−1)dl = a + (n - 1)d is the last term. There are nn such pairs, so 2Sn=n(a+l)2S_{n} = n(a + l). The figure shows this for 1+2+⋯+61 + 2 + \cdots + 6: two copies of the staircase fit together into a 6×76 \times 7 rectangle. Hence Sn=n2(a+l)=n2[2a+(n−1)d]S_{n} = \frac{n}{2}(a + l) = \frac{n}{2}[2a + (n - 1)d]. The sum is a quadratic in nn with no constant term: Sn=d2n2+(a−d2)nS_{n} = \frac{d}{2}n^{2} + (a - \frac{d}{2})n. Conversely, if Sn=pn2+qnS_{n} = pn^{2} + qn for all nn, the sequence is an AP with d=2pd = 2p. If there is a non-zero constant term, the sequence is an AP only from the second term on. Recovering terms from sums. For any sequence, Tn=Sn−Sn−1T_{n} = S_{n} - S_{n-1} for n≥2n \ge 2, and T1=S1T_{1} = S_{1}. This is the standard way to find the general term when the sum is given. Choosing terms symmetrically. When the problem gives the sum of three or four terms of an AP, choose them symmetrically so the sum eliminates dd: three terms as a−d,a,a+da - d, a, a + d (sum 3a3a); four terms as a−3d,a−d,a+d,a+3da - 3d, a - d, a + d, a + 3d (common difference 2d2d, sum 4a4a); five terms as a−2d,…,a+2da - 2d, \ldots, a + 2d. Properties. If each term is increased, decreased, multiplied or divided by the same non-zero constant, the result is still an AP. The sum of terms equidistant from the beginning and the end is constant: Tk+Tn−k+1=a+lT_{k} + T_{n-k+1} = a + l. a,b,ca, b, c are in AP if and only if 2b=a+c2b = a + c. If a1,a2,…a_{1}, a_{2}, \ldots is an AP, so are a1,a3,a5,…a_{1}, a_{3}, a_{5}, \ldots and ap,ap+q,ap+2q,…a_{p}, a_{p+q}, a_{p+2q}, \ldots. Term-by-term sums and differences of two APs are APs, but products and quotients generally are not. In an AP, Tn=Tn−k+Tn+k2T_{n} = \frac{T_{n-k} + T_{n+k}}{2}: each term is the average of terms equidistant from it. The arithmetic mean of aa and bb is A=a+b2A = \frac{a + b}{2}, the number that makes a,A,ba, A, b an AP. Inserting nn arithmetic means between aa and bb creates an AP of n+2n + 2 terms, so d=b−an+1d = \frac{b - a}{n + 1}, and the sum of the nn means is n⋅a+b2n \cdot \frac{a + b}{2} (the sum of the whole AP minus the endpoints). Key Takeaways: - Tn=a+(n−1)dT_{n} = a + (n - 1)d; Sn=n2(a+l)=n2[2a+(n−1)d]S_{n} = \frac{n}{2}(a + l) = \frac{n}{2}[2a + (n - 1)d]. - TnT_{n} linear in nn   ⟺  \iff AP; SnS_{n} quadratic in nn with zero constant   ⟺  \iff AP. - Tn=Sn−Sn−1T_{n} = S_{n} - S_{n-1}. - Symmetric choices: a−d,a,a+da - d, a, a + d or a−3d,a−d,a+d,a+3da - 3d, a - d, a + d, a + 3d. - Tk+Tn−k+1=a+lT_{k} + T_{n-k+1} = a + l; a,b,ca, b, c in AP   ⟺  2b=a+c\iff 2b = a + c. - Sum of nn arithmetic means between aa and bb: n(a+b)2\frac{n(a + b)}{2}. - Pitfall: using Tn=Sn−Sn−1T_{n} = S_{n} - S_{n-1} for n=1n = 1, where S0S_{0} may not be 00 in the given formula. Brain-Teaser 1: The sum of the first pp terms of an AP is qq and the sum of the first qq terms is pp (p≠qp \neq q). Find the sum of the first p+qp + q terms. Solution: Write Sn=An2+BnS_{n} = An^{2} + Bn. Then Ap2+Bp=qAp^{2} + Bp = q and Aq2+Bq=pAq^{2} + Bq = p. Subtract: A(p2−q2)+B(p−q)=q−pA(p^{2} - q^{2}) + B(p - q) = q - p. Divide by p−q≠0p - q \neq 0: A(p+q)+B=−1A(p + q) + B = -1. So Sp+q=(p+q)[A(p+q)+B]=−(p+q)S_{p+q} = (p + q)[A(p + q) + B] = -(p + q). Using the quadratic form of SnS_{n} avoids solving for aa and dd separately, and the answer drops out after one subtraction. Brain-Teaser 2: If Sn=3n2+5nS_{n} = 3n^{2} + 5n, show that the terms form an AP and find the 2525th term. What changes if Sn=3n2+5n+2S_{n} = 3n^{2} + 5n + 2? Solution: Tn=Sn−Sn−1=3n2+5n−3(n−1)2−5(n−1)=6n+2T_{n} = S_{n} - S_{n-1} = 3n^{2} + 5n - 3(n - 1)^{2} - 5(n - 1) = 6n + 2 for n≥2n \ge 2, and T1=S1=8=6(1)+2T_{1} = S_{1} = 8 = 6(1) + 2. So Tn=6n+2T_{n} = 6n + 2 for all nn: an AP with d=6d = 6, and T25=152T_{25} = 152. If Sn=3n2+5n+2S_{n} = 3n^{2} + 5n + 2, then still Tn=6n+2T_{n} = 6n + 2 for n≥2n \ge 2, but T1=S1=10≠8T_{1} = S_{1} = 10 \neq 8. The sequence 10,14,20,26,…10, 14, 20, 26, \ldots is an AP only from the second term. The constant term in SnS_{n} is absorbed entirely by the first term. Brain-Teaser 3: Find the sum of all integers between 100100 and 10001000 that are divisible by 77 but not by 55. Solution: Multiples of 77 from 105105 to 994994: an AP with a=105a = 105, l=994l = 994, d=7d = 7, so n=994−1057+1=128n = \frac{994 - 105}{7} + 1 = 128, and the sum is 1282(105+994)=64×1099=70336\frac{128}{2}(105 + 994) = 64 \times 1099 = 70336. Multiples of 3535 in the same range: from 105105 to 980980, n=980−10535+1=26n = \frac{980 - 105}{35} + 1 = 26, sum =13×1085=14105= 13 \times 1085 = 14105. Subtracting, the answer is 70336−14105=5623170336 - 14105 = 56231. Each divisibility condition defines an AP, and 'but not' becomes a subtraction.

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Arithmetic Progressions: nth Term, Sum and Properties — DPP

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