Chapter 5: Sequences, Series and Progressions (Set-4)
In an AP, the 2nd term is 9 and the 8th term is 33; the common difference is
A 3
B 5
C 4
D 6
In an AP, a8−a2=(8−2)d=6da8−a2=(8−2)d=6d. Here 33−9=2433−9=24. So 6d=24⇒d=46d=24⇒d=4. The same difference repeats between every consecutive pair.
For the AP with first term 5 and common difference 3, the term 65 occurs at position
A 21
B 18
C 19
D 20
Use an=a+(n−1)dan=a+(n−1)d. So 65=5+3(n−1)⇒60=3(n−1)⇒n−1=20⇒n=2165=5+3(n−1)⇒60=3(n−1)⇒n−1=20⇒n=21. This counts how many equal steps are needed.
An AP has first term 12 and sum of first 12 terms is 342; the common difference is
A 2
B 4
C 3
D 5
Sn=n2[2a+(n−1)d]Sn=2n[2a+(n−1)d]. So 342=122[24+11d]=6(24+11d)=144+66d342=212[24+11d]=6(24+11d)=144+66d. Hence 66d=198⇒d=366d=198⇒d=3.
In an AP, the average of 4th and 10th terms is 38 and common difference is 4; the first term is
A 12
B 14
C 16
D 18
a4=a+12a4=a+12 and a10=a+36a10=a+36. Their average is (a+12)+(a+36)2=a+242(a+12)+(a+36)=a+24. Set a+24=38⇒a=14a+24=38⇒a=14.
A taxi charges ₹50 base plus ₹10 per km; the fare for 8 km is
A 130
B 120
C 140
D 150
Total fare = base + (rate × distance). So fare =50+10×8=50+80=130=50+10×8=50+80=130. This is a simple linear model and also forms an AP across distances.
Five arithmetic means are inserted between 6 and 36; the third inserted mean is
A 16
B 26
C 31
D 21
Total terms become 7, so common difference d=(36−6)/6=5d=(36−6)/6=5. Sequence: 6, 11, 16, 21, 26, 31, 36. The third inserted mean is 21.
In an AP, a5=20a5=20 and a15=50a15=50; the sum from 5th to 15th term is
A 330
B 420
C 385
D 450
From 5th to 15th gives 11 terms. First is 20 and last is 50. Sum =112(20+50)=112×70=385=211(20+50)=211×70=385. This uses AP block-sum idea.
An AP has a7=18a7=18 and a13=42a13=42; the 10th term is
A 30
B 28
C 32
D 34
In an AP, terms equally spaced around the middle average out. Since 10 is midway between 7 and 13, a10=a7+a132=18+422=30a10=2a7+a13=218+42=30.
If Sn=4n2−nSn=4n2−n is the sum of first n terms of an AP, then the common difference is
A 6
B 8
C 7
D 9
For an AP, Sn=d2n2+2a−d2nSn=2dn2+22a−dn. Compare coefficients with 4n2−n4n2−n. Thus d/2=4⇒d=8d/2=4⇒d=8. The nn-term coefficient fixes aa.
In the AP with a1=−5a1=−5 and d=2d=2, the first positive term is
A 3rd term
B 5th term
C 4th term
D 6th term
an=−5+(n−1)2=2n−7an=−5+(n−1)2=2n−7. First positive means 2n−7>0⇒n>3.52n−7>0⇒n>3.5. Smallest integer is 4. Then a4=1a4=1, positive.
In a GP, a3=12a3=12 and a6=96a6=96; the common ratio is
A 2
B 3
C 4
D 6
In GP, a6/a3=r6−3=r3a6/a3=r6−3=r3. Here 96/12=896/12=8. So r3=8⇒r=2r3=8⇒r=2. A fixed multiplier creates geometric growth.
A GP has first term 81 and fifth term 1 (positive ratio); the common ratio is
A 1/2
B 2/3
C 3
D 1/3
a5=ar4a5=ar4. So 1=81r4⇒r4=1/81=(1/3)41=81r4⇒r4=1/81=(1/3)4. With positive ratio, r=1/3r=1/3. Each term becomes one-third of the previous.
In the GP 5, 10, 20, … , 640, the number of terms is
A 7
B 9
C 8
D 10
an=5⋅2n−1=640⇒2n−1=128=27an=5⋅2n−1=640⇒2n−1=128=27. Thus n−1=7⇒n=8n−1=7⇒n=8. Counting powers of 2 makes it quick.
For GP with first term 4 and ratio 3, the sum of first 4 terms is
A 120
B 160
C 140
D 180
Terms are 4, 12, 36, 108. Sum =4+12+36+108=160=4+12+36+108=160. Also S4=ar4−1r−1=481−12=160S4=ar−1r4−1=4281−1=160. Both match.
The infinite GP has first term 7 and ratio −14−41; its sum to infinity is
A 28/5
B 35/4
C 21/5
D 7/5
Since ∣r∣<1∣r∣<1, sum exists: S∞=a1−r=71−(−1/4)=75/4=28/5S∞=1−ra=1−(−1/4)7=5/47=28/5. Negative ratio causes alternating signs but still converges.
Two geometric means are inserted between 9 and 72; the inserted terms are
A 12, 24
B 16, 32
C 18, 36
D 9, 27
Form 4-term GP: 9,9r,9r2,9r3=72⇒r3=8⇒r=29,9r,9r2,9r3=72⇒r3=8⇒r=2. Inserted terms are 9r=189r=18 and 9r2=369r2=36. Ratios stay constant.
In a GP, the product of the 1st and 5th terms equals the square of the
A 2nd term
B 4th term
C 5th term
D 3rd term
Let terms be a,ar,ar2,ar3,ar4a,ar,ar2,ar3,ar4. Product of 1st and 5th is a⋅ar4=a2r4a⋅ar4=a2r4. Square of 3rd is (ar2)2=a2r4(ar2)2=a2r4. Hence equal.
A culture triples every hour starting at 200 units; the amount after 4 hours is
A 16200
B 5400
C 8100
D 24300
Tripling each hour forms a GP with ratio 3. After 4 hours: 200×34=200×81=16200200×34=200×81=16200. Exponential growth means repeated multiplication, not addition.
In a GP with first term 3 and ratio −2, the term −24 occurs at
A 3rd term
B 5th term
C 4th term
D 6th term
an=3(−2)n−1an=3(−2)n−1. Solve 3(−2)n−1=−24⇒(−2)n−1=−8=(−2)33(−2)n−1=−24⇒(−2)n−1=−8=(−2)3. Thus n−1=3⇒n=4n−1=3⇒n=4. Sign alternates each step.
In a GP with first term 16 and ratio 1/2, the term 1 occurs at
A 4th term
B 5th term
C 6th term
D 7th term
16(1/2)n−1=1⇒(1/2)n−1=1/16=(1/2)416(1/2)n−1=1⇒(1/2)n−1=1/16=(1/2)4. So n−1=4⇒n=5n−1=4⇒n=5. Each term halves the previous one.
The arithmetic mean of two numbers is 13 and one number is 9; the other number is
A 15
B 16
C 17
D 18
AM condition: (x+9)/2=13⇒x+9=26⇒x=17(x+9)/2=13⇒x+9=26⇒x=17. Arithmetic mean is the midpoint, so the second number must be equally far from 13 as 9 is.
The geometric mean of two positive numbers is 15 and one number is 9; the other number is
A 25
B 20
C 30
D 35
GM means 9x=15⇒9x=225⇒x=259x=15⇒9x=225⇒x=25. Geometric mean links to product, so knowing GM and one factor fixes the other factor directly.
If two positive numbers have sum 20, their maximum possible product is
A 90
B 95
C 100
D 110
By AM–GM, x+y2≥xy2x+y≥xy. With x+y=20x+y=20, we get 10≥xy⇒xy≤10010≥xy⇒xy≤100. Maximum happens when x=y=10x=y=10.
For x>0x>0, the minimum value of 2x+18x2x+x18 is
A 10
B 12
C 14
D 16
Use AM–GM: 2x+18x≥22x⋅18x=236=122x+x18≥22x⋅x18=236=12. Equality when 2x=18x⇒x=32x=x18⇒x=3. So minimum is 12.
For x>0x>0, the minimum value of x2+16x2x2+x216 is
A 8
B 6
C 10
D 12
AM–GM gives x2+16×2≥2×2⋅16×2=216=8×2+x216≥2×2⋅x216=216=8. Equality when x2=16×2⇒x=2×2=x216⇒x=2. Hence minimum is 8.
If AM equals GM for positive numbers with sum 18, then their product is
A 72
B 90
C 96
D 81
AM = GM occurs only when numbers are equal. With sum 18, both are 9. Product =9×9=81=9×9=81. This is the equality condition behind AM–GM.
Three positive numbers have sum 12; the maximum possible product is
A 48
B 60
C 64
D 72
By AM–GM for three numbers, x+y+z3≥xyz33x+y+z≥3xyz. With sum 12, average is 4, so xyz≤43=64xyz≤43=64. Maximum occurs at x=y=z=4x=y=z=4.
For positive x and y, the minimum value of xy+yxyx+xy is
A 2
B 1
C 3
D 4
Let t=xy>0t=yx>0. Then expression is t+1t≥2t+t1≥2 by AM–GM. Equality occurs when t=1⇒x=yt=1⇒x=y. So the minimum is 2.
The harmonic mean of 4 and 12 is
A 5
B 6
C 7
D 8
HM for two positives is 2xyx+yx+y2xy. So HM =2⋅4⋅124+12=9616=6=4+122⋅4⋅12=1696=6. Harmonic mean is often used in rate-type averages.
Weighted mean of 40 (weight 3) and 70 (weight 1) equals
A 45
B 50
C 47.5
D 52.5
Weighted mean =40⋅3+70⋅13+1=120+704=1904=47.5=3+140⋅3+70⋅1=4120+70=4190=47.5. The value 40 dominates because it has higher weight.
The limit of the sequence n2+1n2+3n2+3n2+1 as n→∞n→∞ is
A 0
B 2
C No limit
D 1
Divide numerator and denominator by n2n2: 1+1/n21+3/n21+3/n21+1/n2. As n→∞n→∞, both 1/n21/n2 terms go to 0, leaving 1/1=11/1=1. So it converges to 1.
The sequence an=(−32)nan=(−23)n is
A Divergent
B Convergent to 0
C Convergent to 1
D Constant
Here ∣−3/2∣>1∣−3/2∣>1, so the magnitude grows without bound while signs alternate. Because terms do not approach a single fixed real value, the sequence has no limit and diverges.
The infinite series ∑n=1∞(13)n∑n=1∞(31)n equals
A 1/3
B 1/2
C 2/3
D 3/2
This is an infinite GP with first term a=1/3a=1/3 and ratio r=1/3r=1/3. Sum S∞=a1−r=1/31−1/3=1/32/3=1/2S∞=1−ra=1−1/31/3=2/31/3=1/2.
The infinite series 52+54+58+⋯25+45+85+⋯ has sum
A 4
B 6
C 5
D 7
This is an infinite GP with a=5/2a=5/2 and r=1/2r=1/2. Since ∣r∣<1∣r∣<1, sum is S∞=a1−r=5/21/2=5S∞=1−ra=1/25/2=5. The terms keep halving.
The value of 1+2+3+⋯+201+2+3+⋯+20 is
A 210
B 200
C 220
D 230
Sum of first n natural numbers is n(n+1)/2n(n+1)/2. For n=20: 20⋅21/2=10⋅21=21020⋅21/2=10⋅21=210. Pairing first and last terms gives the same result quickly.
The value of 12+22+⋯+15212+22+⋯+152 is
A 1200
B 1300
C 1240
D 1350
Use ∑k=1nk2=n(n+1)(2n+1)6∑k=1nk2=6n(n+1)(2n+1). For n=15: 15⋅16⋅316=74406=1240615⋅16⋅31=67440=1240. This formula is standard and stable.
The value of 13+23+⋯+10313+23+⋯+103 is
A 2500
B 3520
C 4000
D 3025
∑k=1nk3=[n(n+1)2]2∑k=1nk3=[2n(n+1)]2. For n=10: [10⋅112]2=552=3025[210⋅11]2=552=3025. It equals the square of the triangular sum.
The value of ∑n=19(1n−1n+1)∑n=19(n1−n+11) is
A 9/10
B 1/10
C 10/9
D 1
This telescopes: most terms cancel in partial sums. Writing out gives 1−12+12−13+⋯+19−1101−21+21−31+⋯+91−101. Everything cancels except 1−110=9101−101=109.
The harmonic series 1+12+13+⋯1+21+31+⋯ is
A Convergent
B Geometric series
C Divergent
D Telescoping series
Even though 1/n→01/n→0, the partial sums grow without bound. The harmonic series diverges, which shows that “terms go to 0” alone does not guarantee a finite sum.
The series ∑n=1∞1n3/2∑n=1∞n3/21 is
A Divergent
B Convergent
C Alternating only
D Not defined
This is a p-series ∑1/np∑1/np with p=3/2>1p=3/2>1. Any p-series with exponent greater than 1 converges to a finite value because terms decrease fast enough.
If a series ∑an∑an converges, then anan must
A Equal 1
B Grow large
C Approach 0
D Alternate always
A necessary condition for convergence is an→0an→0. If terms do not go to 0, partial sums cannot settle to a finite limit. This is the basic nth-term test idea.
The series ∑n=0∞3nn!∑n=0∞n!3n is
A Convergent
B Divergent
C Telescoping
D Harmonic type
Using ratio test: an+1an=3n+1/(n+1)!3n/n!=3n+1→0anan+1=3n/n!3n+1/(n+1)!=n+13→0. Since the ratio limit is less than 1, the series converges.
The infinite GP 1−1+1−1+⋯1−1+1−1+⋯ is
A Convergent
B Sum equals 0
C Sum equals 1
D Divergent
Partial sums are 1, 0, 1, 0, … which do not approach a single value. Since the sequence of partial sums has no limit, the series does not converge in the usual sense.
A recursion is given by a1=2a1=2 and an=2an−1+1an=2an−1+1; the value of a3a3 is
A 9
B 10
C 11
D 12
Compute stepwise: a2=2⋅2+1=5a2=2⋅2+1=5. Then a3=2⋅5+1=11a3=2⋅5+1=11. Recursive rules use earlier terms to build the next, so careful step calculation is enough.
A sequence satisfies a1=1,a2=3,an=an−1+an−2a1=1,a2=3,an=an−1+an−2; the value of a5a5 is
A 9
B 11
C 10
D 12
Build terms: a3=1+3=4a3=1+3=4, a4=3+4=7a4=3+4=7, a5=4+7=11a5=4+7=11. This Fibonacci-type recursion grows by adding the previous two terms.
The limit of an=2+(−1)nnan=2+n(−1)n as n→∞n→∞ is
A 2
B 0
C 1
D No limit
The oscillating part (−1)nnn(−1)n goes to 0 because 1/n→01/n→0. So anan approaches 2+0=22+0=2. The alternation becomes negligible for large n.
The sum 1+2+4+8+⋯1+2+4+8+⋯ (first n terms) equals 1023; the value of n is
A 9
B 11
C 10
D 12
This is a GP with a=1,r=2a=1,r=2. Sum Sn=2n−12−1=2n−1Sn=2−12n−1=2n−1. Set 2n−1=1023⇒2n=1024=2102n−1=1023⇒2n=1024=210. So n=10n=10.
The sum of the AP 1,4,7,…1,4,7,… is 145; the number of terms is
A 8
B 9
C 11
D 10
Here a=1,d=3a=1,d=3. Sn=n2[2+(n−1)3]=n2(3n−1)Sn=2n[2+(n−1)3]=2n(3n−1). Set =145⇒n(3n−1)=290=145⇒n(3n−1)=290. Solving gives n=10n=10.
An infinite GP has sum 9 and first term 6; the common ratio is
A 1/4
B 1/3
C 1/2
D 2/3
For ∣r∣<1∣r∣<1, S∞=a1−rS∞=1−ra. So 9=61−r⇒1−r=69=23⇒r=139=1−r6⇒1−r=96=32⇒r=31. This ensures convergence.
The series ∑n=1∞1n2+n∑n=1∞n2+n1 is
A Convergent
B Divergent
C Alternating only
D Geometric only
1n2+n=1n(n+1)n2+n1=n(n+1)1, which behaves like 1/n21/n2 for large n. Since ∑1/n2∑1/n2 converges, this series also converges by comparison (and it telescopes too).