3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
An arithmetic progression has 3m terms a1,a2,…,a3m, where m≥1. Let A=a1+a4+a7+⋯+a3m−2 and C=a3+a6+a9+⋯+a3m. If A=160, C=190 and a3m=4a1, then the number of terms of the progression is ______. Exactly one option is correct.
Show answer and solution
Answer: B
Let d be the common difference. Each of A and C has m terms and a3j−a3j−2=2d, so C−A=2md=30, i.e. md=15.
From a3m=4a1: a1+(3m−1)d=4a1, so a1=3(3m−1)d.
Now A=∑j=1m(a1+3(j−1)d)=ma1+23m(m−1)d=md[33m−1+23(m−1)]=615(15m−11).
Setting this equal to 160 gives 15m−11=64, so m=5, d=3, a1=14.
Check: A=14+23+32+41+50=160, C=20+29+38+47+56=190, a15=56=4⋅14.
The progression has 3m=15 terms.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
A sequence (an)n≥1 of real numbers, with Sn=a1+a2+⋯+an, satisfies a1=1 and Sn2=an(Sn−21) for every integer n≥2. The value of a2+a3+⋯+a25 is ______. Exactly one option is correct.
A−2524
B−5150
C−4746
D−4948
Show answer and solution
Answer: D
For n≥2 write an=Sn−Sn−1: Sn2=(Sn−Sn−1)(Sn−21)=Sn2−21Sn−SnSn−1+21Sn−1.
Hence Sn−1−Sn=2SnSn−1.
If Sn=0 this forces Sn−1=0; since S1=1=0, induction gives Sn=0 for all n. Dividing by SnSn−1: Sn1−Sn−11=2.
So Sn1 is an AP with first term 1 and common difference 2: Sn=2n−11.
Mind the indexing: the requested sum starts at a2, so it equals S25−S1=491−1=−4948.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
For real x=−1 let t=x+12x−1, and let S(x)=n=0∑∞tn whenever this series converges. The set of all real x for which the series converges and S(x)<1 is ______. Exactly one option is correct.
A(−∞,21)∪(2,∞)
B(−1,21)
C(0,21)
D(0,2)
Show answer and solution
Answer: C
The geometric series converges if and only if ∣t∣<1, i.e. ∣2x−1∣<∣x+1∣. Squaring: (2x−1)2<(x+1)2⟺3x2−6x<0⟺0<x<2.
For such x, S(x)=1−t1=2−xx+1.
2−xx+1<1⟺2−x2x−1<0⟺x<21 or x>2.
Intersecting with (0,2) gives (0,21).
Trap: solving 2−xx+1<1 without the convergence condition gives option (A); the closed form is meaningless outside (0,2).
Difficulty: Very hard