Q.Write the first 6 terms of the sequences whose nth term an is given below.
Concept understanding — Finite Sequences
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 n non-negative numbers the geometric mean is na1a2⋯an; the harmonic mean of n positive 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".
Apply each piecewise/recursive rule term by term for n=1,…,6.
- 2,2,4,4,6,6
- 1,2,3,5,8,13
- 1,2,3,6,11,20
Substitute directly for the piecewise rule (i), and build up term-by-term for the two recursively-defined sequences (ii),(iii).
Step 1. (i) an=n+1 (odd n), an=n (even n).
n=1(odd):2; n=2(even):2; n=3(odd):4; n=4(even):4; n=5(odd):6; n=6(even):6.
Sequence: 2,2,4,4,6,6.
Step 2. (ii) a1=1,a2=2,an=an−1+an−2 for n>2.
a3=a2+a1=2+1=3; a4=a3+a2=3+2=5; a5=a4+a3=5+3=8; a6=a5+a4=8+5=13.
Sequence: 1,2,3,5,8,13.
Step 3. (iii) an=n for n=1,2,3; an=an−1+an−2+an−3 for n>3.
a1=1,a2=2,a3=3. a4=a3+a2+a1=3+2+1=6; a5=a4+a3+a2=6+3+2=11; a6=a5+a4+a3=11+6+3=20.
Sequence: 1,2,3,6,11,20.
(i) 2,2,4,4,6,6; (ii) 1,2,3,5,8,13; (iii) 1,2,3,6,11,20.
- Off-by-one error applying the odd/even rule in (i)
- Forgetting a term when accumulating the running three-term sum in (iii) (e.g. using only two previous terms instead of three)
- CBSE 2023Set ANNUAL1 markMCQQ.The sequence 31,3+21,3+221,… forms an:(a) Harmonic Progression(b) Arithmetic Progression(c) Arithmetico-Geometric Progression(d) Geometric Progression
›Reveal solutionSolution
A sequence is a Harmonic Progression exactly when the reciprocals of its terms form an Arithmetic Progression — that's the case here.
Look at the reciprocals of the given terms: 3, 3+2, 3+22,…
Consecutive differences: (3+2)−3=2, and (3+22)−(3+2)=2.
Since the reciprocals differ by the same constant 2 each time, the reciprocals form an Arithmetic Progression with common difference 2.
By definition, a sequence whose reciprocals form an AP is a Harmonic Progression.
✓Final answerHarmonic Progression.
- CBSE 2022Set ANNUAL1 markMCQQ.The nth term of the sequence 21,43,87,1615,..... is:(a) 2−n+n−1(b) 2n−n−1(c) 2n−1(d) 1−2−n
›Reveal solutionSolution
Each term 2n2n−1 can be written as 1−2−n, matching the sequence 21,43,87,1615,…
Write each given term with denominator a power of 2: 21=2121−1, 43=2222−1, 87=2323−1, 1615=2424−1.
So the nth term is an=2n2n−1=1−2n1=1−2−n.
✓Final answerThe correct option is (d) 1−2−n.
- CBSE 2019Set ANNUAL1 markMCQQ.The nth term of the sequence 2,7,14,23,… is:(a) n2+2n+1(b) n2+2n−1(c) n2−2n−1(d) n2−2n+1
›Reveal solutionSolution
Since the sequence's differences grow arithmetically, the nth term is quadratic; checking n=1,2,3,4 against n2+2n−1 confirms it fits all four given terms.
The sequence is 2,7,14,23,…. First differences: 7−2=5, 14−7=7, 23−14=9. These differences themselves increase by 2 each time, which is the signature of a quadratic nth term an2+bn+c.
Rather than solving the full system, test the candidate options directly:
- n=1: 12+2(1)−1=1+2−1=2 ✓
- n=2: 4+4−1=7 ✓
- n=3: 9+6−1=14 ✓
- n=4: 16+8−1=23 ✓
All four match, so the formula is confirmed.
✓Final answerThe correct option is (b) n2+2n−1.
- CBSE 2018Set ANNUAL1 markMCQQ.The number of bacteria in a certain culture doubles every hour. If there were 40 bacteria present in the culture originally, the number of bacteria present at the end of 2nd hour and 4th hour are respectively:(a) 40, 160(b) 40, 80(c) 80, 640(d) 160, 640
›Reveal solutionSolution
With population doubling each hour, count after n hours =40×2n; at n=2 this is 160, and at n=4 it is 640.
Starting count =40. Since the population doubles every hour, after n hours the count is 40×2n (a geometric progression with first term 40 and common ratio 2).
At the end of the 2nd hour (n=2): 40×22=40×4=160.
At the end of the 4th hour (n=4): 40×24=40×16=640.
✓Final answerThe correct option is (d) 160, 640.
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