A sequence is an ordered list of numbers; formally, a finite sequence on n terms is a function f:{1,2,…,n}→R with f(k)=ak, while an infinite sequence is a function on all of N. Terms may repeat (unlike the elements of a set).
Arithmetic progression (AP):a,a+d,a+2d,…, where d is the common difference. The nth term is
Tn=a+(n−1)d.
Every term (after the first) is the arithmetic mean of its neighbours: ak=2ak−1+ak+1 (Result 5.1) — equivalently, Tm+n+Tm−n=2Tm for any valid m,n.
Geometric progression (GP):a,ar,ar2,… (a=0,r=0), where r is the common ratio. The nth term is
Tn=arn−1.
Every term is the geometric mean of its neighbours: ak=ak−1ak+1 (Result 5.2), so tn−k,tn,tn+k is again a GP for any k. Taking logs of a GP with r>0 turns it into an AP with common difference logr.
Arithmetico-geometric progression (AGP): term-by-term product of an AP and a GP, a,(a+d)r,(a+2d)r2,…, with nth term Tn=(a+(n−1)d)rn−1. Setting r=1 recovers an AP; setting d=0 recovers a GP — AP and GP are special cases of AGP.
Harmonic progression (HP): a sequence whose reciprocals form an AP: h1,h2,… is an HP exactly when h11,h21,… is an AP, giving the general HP form a1,a+d1,a+2d1,… (provided no denominator vanishes). If a,b,c are in HP then b=a+c2ac.
Means. For n numbers a1,…,an: the arithmetic mean is na1+⋯+an; for nnon-negative numbers the geometric mean is na1a2⋯an; the harmonic mean of npositive numbers is a11+⋯+an1n. For two numbers a,b: AM=2a+b, GM=ab, HM=a+b2ab, and always AM≥GM≥HM (equality throughout iff a=b). A striking consequence: AM×HM=GM2, so AM,GM,HM of two positive numbers are themselves in GP.
Classifying a sequence. Given a formula for an, list several terms and test: is the difference of consecutive terms constant (AP)? Is the ratio constant (GP)? Do the reciprocals form an AP (HP)? Does it factor as an AP times a GP term-by-term (AGP)? A constant nonzero sequence is simultaneously an AP (d=0) and a GP (r=1); a sequence that fails every test is classified as "none of them".
Spot the pattern in each listed sequence and write a single closed formula (piecewise where needed).
✓Final answer
an={n+1nn oddn even
an=n+1n
an=2n2n−1
an={7−nn+8n oddn even
For each listed sequence, look for the underlying pattern in the index — a repeated-pair rule, a numerator/denominator pattern, or two interleaved arithmetic progressions on alternating positions.
Step 1. (i) 2,2,4,4,6,6,…. This is exactly the sequence from Exercise 5.2 Q2(i): each even number is repeated twice. So an=n+1 when n is odd (giving the even number n+1), and an=n when n is even (already even). an={n+1nn oddn even.
Step 2. (ii) 21,32,43,54,65,…. Numerator =n, denominator =n+1: an=n+1n.
Step 3. (iii) 21,43,65,87,109,…. Numerator is the odd number 2n−1, denominator is the even number 2n: an=2n2n−1.
Step 4. (iv) 6,10,4,12,2,14,0,16,−2,…. The odd-position terms (1st,3rd,5th,…) are 6,4,2,0,−2,… — an AP with first term 6, common difference −2; for n=2k−1, value =6−2(k−1)=8−2k=7−n (checking n=1→6, n=3→4, etc.). The even-position terms (2nd,4th,…) are 10,12,14,16,… — an AP with first term 10, common difference 2; for n=2k, value =10+2(k−1)=8+2k=n+8 (checking n=2→10, n=4→12, etc.).
an={7−nn+8n oddn even
✓Final answer
an={n+1nn oddn even;
an=n+1n;
an=2n2n−1;
an={7−nn+8n oddn even
Pattern-spotting; splitting interleaved sequences into two APs on alternating positions
Trying to fit sequence (iv) with a single non-piecewise formula, when it is genuinely two interleaved APs
Off-by-one when converting the sub-AP's own index k back to the original index n