Chapter 5: Sequences, Series and Progressions (Set-3)
An AP has first term 7 and sum of first 5 terms is 45; the common difference is
A 2
B 3
C 1
D 4
Use Sn=n2[2a+(n−1)d]Sn=2n[2a+(n−1)d]. Here 45=52[14+4d]45=25[14+4d]. So 90=5(14+4d)⇒18=14+4d⇒d=190=5(14+4d)⇒18=14+4d⇒d=1.
In an AP, the 8th term is 27 and the common difference is 3; the first term is
A 3
B 6
C 9
D 12
Use an=a+(n−1)dan=a+(n−1)d. Given 27=a+7⋅3=a+2127=a+7⋅3=a+21. So a=27−21=6a=27−21=6. The first term is found by reversing the equal-step increase.
A sequence is defined by an=4n−1an=4n−1; the common difference of this sequence is
A 4
B 2
C 3
D 5
For a linear term an=pn+qan=pn+q, consecutive difference is constant pp. Here an+1−an=[4(n+1)−1]−(4n−1)=4an+1−an=[4(n+1)−1]−(4n−1)=4. So it forms an AP with d=4d=4.
In an AP, the value of a3+a9a3+a9 is always equal to
A 2a52a5
B a6a6
C 2a62a6
D 3a63a6
Write a3=a+2da3=a+2d and a9=a+8da9=a+8d. Their sum is 2a+10d=2(a+5d)=2a62a+10d=2(a+5d)=2a6. Symmetric terms in an AP add to twice the middle term.
When three arithmetic means are inserted between 5 and 25, the second inserted mean is
A 10
B 15
C 20
D 12
With 3 means, total terms become 5: 5, x, y, z, 255,x,y,z,25. Common difference d=(25−5)/4=5d=(25−5)/4=5. Sequence is 5,10,15,20,255,10,15,20,25. Second inserted mean is 15.
A staircase has 12 steps; first step height is 15 cm and each next step is 1 cm higher; total height is
A 240 cm
B 252 cm
C 258 cm
D 246 cm
Step heights form an AP with a=15a=15, d=1d=1, n=12n=12. Last step l=15+11=26l=15+11=26. Total height S=122(15+26)=6⋅41=246S=212(15+26)=6⋅41=246 cm.
For an AP with first term 2 and common difference 5, the sum equals 154; the number of terms is
A 7
B 9
C 8
D 10
Sn=n2[2a+(n−1)d]=n2[4+5(n−1)]=n2(5n−1)Sn=2n[2a+(n−1)d]=2n[4+5(n−1)]=2n(5n−1). Set =154⇒n(5n−1)=308=154⇒n(5n−1)=308. Solving gives n=8n=8.
An AP has 20 terms, first term 3 and sum 430; the last term is
A 40
B 38
C 42
D 44
Use S=n2(a+l)S=2n(a+l). Here 430=202(3+l)=10(3+l)430=220(3+l)=10(3+l). So 3+l=43⇒l=403+l=43⇒l=40. This avoids finding dd directly.
In an AP, the first term is 2 and the fourth term is 14; the second term is
A 4
B 6
C 8
D 10
a4=a+3da4=a+3d. So 14=2+3d⇒d=414=2+3d⇒d=4. Then a2=a+d=2+4=6a2=a+d=2+4=6. Each term increases by the same constant difference.
If the sum of first n terms of an AP is Sn=3n2+nSn=3n2+n, the first term is
A 2
B 6
C 4
D 8
For AP, Sn=d2n2+2a−d2nSn=2dn2+22a−dn. Compare with 3n2+n3n2+n: d/2=3⇒d=6d/2=3⇒d=6. Also (2a−d)/2=1⇒2a−6=2⇒a=4(2a−d)/2=1⇒2a−6=2⇒a=4.
A GP has first term 1 and fifth term 81; the common ratio is
A 2
B 4
C 5
D 3
In GP, a5=ar4a5=ar4. Here 81=1⋅r4⇒r4=81⇒r=381=1⋅r4⇒r4=81⇒r=3 (positive ratio). Each term is three times the previous term.
In a GP, the second term is 6 and the common ratio is 2; the first term is
A 2
B 3
C 4
D 6
a2=ara2=ar. Given 6=a⋅2⇒a=36=a⋅2⇒a=3. A GP grows by multiplication, so dividing by the ratio gives the previous term.
A GP has first term 2 and sum of first 3 terms equals 26; the (positive) ratio is
A 2
B 4
C 3
D 5
First three terms are 2,2r,2r22,2r,2r2. Sum 2(1+r+r2)=26⇒1+r+r2=13⇒r2+r−12=02(1+r+r2)=26⇒1+r+r2=13⇒r2+r−12=0. Positive root is r=3r=3.
In the GP 3, 6, 12, … , 384, the number of terms is
A 7
B 8
C 9
D 10
Here a=3a=3, r=2r=2. nth term =3⋅2n−1=384⇒2n−1=128=27=3⋅2n−1=384⇒2n−1=128=27. So n−1=7⇒n=8n−1=7⇒n=8.
For a GP with a=1 and r=3, the sum of first 4 terms is
A 40
B 39
C 41
D 42
Terms are 1,3,9,271,3,9,27. Sum =1+3+9+27=40=1+3+9+27=40. You can also use S4=a(1−r4)/(1−r)=1(1−81)/(1−3)=40S4=a(1−r4)/(1−r)=1(1−81)/(1−3)=40.
The infinite series 6 + 3 + 1.5 + … has sum
A 9
B 10
C 12
D 14
This is an infinite GP with a=6a=6, r=1/2r=1/2 and ∣r∣<1∣r∣<1. Sum to infinity is S∞=a/(1−r)=6/(1−1/2)=12S∞=a/(1−r)=6/(1−1/2)=12.
A GP has first term 8 and ratio −12−21; the third term is
A -2
B 2
C 4
D -4
Terms are 8, 8(−1/2)=−4, −4(−1/2)=28,8(−1/2)=−4,−4(−1/2)=2. Negative ratio alternates signs. Multiplying by the same ratio each step gives the correct third term.
Two geometric means are inserted between 4 and 108; the inserted terms are
A 9 and 27
B 18 and 54
C 6 and 18
D 12 and 36
Form 4-term GP: 4,4r,4r2,4r3=1084,4r,4r2,4r3=108. So r3=27⇒r=3r3=27⇒r=3. Inserted terms are 4r=124r=12 and 4r2=364r2=36.
A quantity doubles every year; the sequence of yearly amounts forms a GP with ratio
A 1/2
B 3
C 2
D 4
“Doubles” means each new value is the previous value multiplied by 2. That constant multiplier is the common ratio. So the amounts follow a GP with r=2r=2.
If AM of two positive numbers is 15 and GM is 12, their product equals
A 144
B 120
C 180
D 225
For two numbers x,y>0x,y>0, GM=xyGM=xy. Given GM=12⇒xy=12⇒xy=144GM=12⇒xy=12⇒xy=144. AM information is extra here but still consistent.
For positive x, the minimum value of x+16xx+x16 is
A 6
B 8
C 10
D 12
By AM–GM, x+16x≥2x⋅16x=216=8x+x16≥2x⋅x16=216=8. Equality occurs when x=16x⇒x=4x=x16⇒x=4.
For positive x, the minimum value of x+9xx+x9 is
A 5
B 7
C 6
D 8
AM–GM gives x+9x≥29=6x+x9≥29=6. Equality when x=9x⇒x=3x=x9⇒x=3. This is a common basic optimization pattern using AM–GM.
If x+y = 18 with x,y positive, the maximum value of xy is
A 72
B 81
C 80
D 90
AM–GM: x+y2≥xy2x+y≥xy. With x+y=18x+y=18, we get 9≥xy⇒xy≤819≥xy⇒xy≤81. Maximum occurs at equality x=y=9x=y=9.
If xy = 49 with x,y positive, the minimum value of x+y is
A 14
B 12
C 16
D 18
AM–GM: x+y2≥xy=49=72x+y≥xy=49=7. So x+y≥14x+y≥14. Equality occurs when x=y=7x=y=7. That gives the smallest possible sum.
For positive x and y, which expression is always non-negative
A GM − AM
B AM + GM
C AM − GM
D AM × GM
AM–GM states AM≥GMAM≥GM for positive numbers. So AM−GM≥0AM−GM≥0 always. The difference is zero only when all numbers are equal, otherwise it is positive.
Weighted mean of 10 (weight 2) and 20 (weight 1) equals
A 30/3
B 50/3
C 60/3
D 40/3
Weighted mean =10⋅2+20⋅12+1=20+203=403=2+110⋅2+20⋅1=320+20=340. The value 10 gets double importance because its weight is higher.
Harmonic mean of 6 and 12 equals
A 6
B 8
C 9
D 10
Harmonic mean of two positives is 2xyx+yx+y2xy. Here 2⋅6⋅126+12=14418=86+122⋅6⋅12=18144=8. HM is useful in average rate problems.
If two positive numbers are equal, then
A Only AM equal
B Only GM equal
C All three equal
D Only HM equal
When the numbers are equal, AM, GM, and HM all become that same common value. This is exactly the equality condition for AM–GM and also for the chain AM≥GM≥HMAM≥GM≥HM.
The limit of the sequence an=nn+1an=n+1n is
A 0
B 2
C No limit
D 1
nn+1=11+1/nn+1n=1+1/n1. As n→∞n→∞, 1/n→01/n→0, so the expression approaches 1. The terms get closer to 1 from below.
The sum of the infinite series ∑n=1∞1n(n+2)∑n=1∞n(n+2)1 equals
A 1/2
B 3/4
C 1
D 5/4
1n(n+2)=12(1n−1n+2)n(n+2)1=21(n1−n+21). Partial sums cancel in pairs. Sum becomes 12(1+12)=3421(1+21)=43 as the tail terms vanish.
The series 1+14+19+116+⋯1+41+91+161+⋯ is
A Convergent
B Divergent
C Alternating
D Not a series
This is ∑1n2∑n21, a standard p-series with p=2>1p=2>1. Such series converges to a finite value. Terms decrease fast enough for convergence.
For an infinite GP with first term 5 and ratio 1, the series is
A Convergent
B Sum equals 5
C Divergent
D Sum equals 0
If r=1r=1, every term stays 5, so partial sums are 5,10,15,…5,10,15,… which grow without bound. Since ∣r∣<1∣r∣<1 is required, the infinite sum does not exist.
Evaluate ∑k=152k∑k=152k
A 20
B 30
C 40
D 50
∑k=152k=2(1+2+3+4+5)=2⋅15=30∑k=152k=2(1+2+3+4+5)=2⋅15=30. This uses basic summation and shows how sigma notation represents repeated addition of terms.
The sum of first 5 terms of GP 2, 6, 18, … is
A 242
B 240
C 244
D 246
It is a GP with a=2a=2, r=3r=3. Sum S5=ar5−1r−1=2⋅243−12=242S5=ar−1r5−1=2⋅2243−1=242. Direct addition also gives 2+6+18+54+162.
Find the sum of AP 11, 14, 17, … , 41
A 275
B 297
C 286
D 308
Here a=11a=11, d=3d=3, last l=41l=41. Number of terms n=41−113+1=11n=341−11+1=11. Sum S=112(11+41)=112⋅52=286S=211(11+41)=211⋅52=286.
In the AP 5, 9, 13, … , is 35 a term
A Yes
B Always yes
C Cannot decide
D No
nth term is an=5+(n−1)4an=5+(n−1)4. Set 5+4(n−1)=35⇒4(n−1)=30⇒n−1=7.55+4(n−1)=35⇒4(n−1)=30⇒n−1=7.5, not an integer. So 35 cannot appear as an AP term.
In the GP 2, 6, 18, … , 54 is the
A 3rd term
B 4th term
C 5th term
D 6th term
General term an=2⋅3n−1an=2⋅3n−1. Set 2⋅3n−1=54⇒3n−1=27=332⋅3n−1=54⇒3n−1=27=33. So n−1=3⇒n=4n−1=3⇒n=4.
Four arithmetic means are inserted between 10 and 35; the third inserted mean is
A 20
B 30
C 25
D 15
With 4 means, total terms are 6, so d=35−105=5d=535−10=5. The sequence is 10,15,20,25,30,3510,15,20,25,30,35. Inserted means are 15,20,25,30; third is 25.
One geometric mean is inserted between 3 and 48; the mean is
A 9
B 12
C 10
D 16
One GM between positive numbers is abab. Here 3⋅48=144=123⋅48=144=12. Then 3, 12, 48 forms a GP with common ratio 4.
If three numbers are both in AP and GP, then the numbers must be
A All equal
B First is zero
C Ratio is zero
D Difference is 1
For three terms a,b,ca,b,c, AP gives 2b=a+c2b=a+c and GP gives b2=acb2=ac. Solving together forces a=b=ca=b=c. So only equal numbers can satisfy both conditions.
In an AP, if ap=aqap=aq for distinct p and q, then the common difference must be
A 1
B -1
C 0
D Not fixed
In AP, ap=a+(p−1)dap=a+(p−1)d and aq=a+(q−1)daq=a+(q−1)d. If p≠qp=q and values are equal, subtraction gives (p−q)d=0⇒d=0(p−q)d=0⇒d=0. So the sequence is constant.
A sequence is defined by a1=3a1=3 and an=an−1+4an=an−1+4; the 6th term is
A 19
B 27
C 31
D 23
This is an AP with a=3a=3, d=4d=4. So a6=3+(6−1)4=3+20=23a6=3+(6−1)4=3+20=23. Adding 4 repeatedly gives 3,7,11,15,19,23.
For an AP with 9 terms, if the 5th term is 20, then the sum of 9 terms is
A 160
B 180
C 170
D 190
In an AP with odd number of terms, the middle term equals the average, and the sum equals nn times the middle term. So S9=9⋅a5=9⋅20=180S9=9⋅a5=9⋅20=180.
The sequence an=(−1)nnan=n(−1)n has limit
A 1
B -1
C 0
D No limit
Although (−1)n(−1)n alternates sign, the magnitude is 1/n1/n, which goes to 0. So terms approach 0 from alternating sides. Hence the sequence converges to 0.
For the alternating series 1−12+13−14+⋯1−21+31−41+⋯, the terms anan satisfy
A an→0an→0
B an→1an→1
C an→−1an→−1
D anan constant
The nth term in magnitude is 1/n1/n, which decreases to 0 as nn increases. This is a necessary condition for possible convergence of a series, even though it alone is not sufficient.
If a series has partial sums SnSn approaching 10, then the series is
A Divergent
B Convergent
C Always zero
D Not defined
A series converges when its partial sums SnSn approach a finite real number. If Sn→10Sn→10, then the series has sum 10. This is the definition of convergence for series.
In a GP, which equation is always true for consecutive terms a2,a3,a4a2,a3,a4
A 2a3=a2+a42a3=a2+a4
B a3=a2−a4a3=a2−a4
C a32=a2a4a32=a2a4
D a3=a2+a4a3=a2+a4
In a GP, a3=a2ra3=a2r and a4=a3ra4=a3r. Then a2a4=a2(a3r)=a2(a2r2)=a32a2a4=a2(a3r)=a2(a2r2)=a32. This is the geometric-mean property.
The sequence an=5+1nan=5+n1 is
A Increasing, bounded
B Increasing, unbounded
C Decreasing, unbounded
D Decreasing, bounded
As nn increases, 1/n1/n decreases, so 5+1/n5+1/n decreases toward 5. It is bounded below by 5 and above by 6, so it is decreasing and bounded.
For a GP with first term 9 and ratio 1/31/3, the 4th term is
A 1/3
B 1
C 1/9
D 1/27
a4=ar3=9(13)3=9⋅127=13a4=ar3=9(31)3=9⋅271=31? Wait carefully: 9/27=1/39/27=1/3. So the 4th term is 1/31/3.
Correct 4th term for GP with first term 9 and ratio 1/31/3 is
A 1/3
B 1
C 1/9
D 3
Use an=arn−1an=arn−1. Here a4=9(13)3=9⋅127=13a4=9(31)3=9⋅271=31. Each term is one-third of the previous: 9, 3, 1, 1/3.