Chapter 5: Sequences, Series and Progressions (Set-5)
An AP has sum of first 10 terms 210 and sum of first 20 terms 820; the first term is
A 5
B 6
C 7
D 8
Use Sn=n2[2a+(n−1)d]Sn=2n[2a+(n−1)d]. From S10=210⇒5(2a+9d)=210⇒2a+9d=42S10=210⇒5(2a+9d)=210⇒2a+9d=42. From S20=820⇒10(2a+19d)=820⇒2a+19d=82S20=820⇒10(2a+19d)=820⇒2a+19d=82. Subtract: 10d=40⇒d=410d=40⇒d=4. Then 2a+36=42⇒a=62a+36=42⇒a=6.
In an AP, a4+a14=60a4+a14=60 and a9=28a9=28; the common difference is
A 2
B 3
C 5
D 4
In an AP, symmetric terms add to twice the middle term. Here 4 and 14 are symmetric around 9, so a4+a14=2a9a4+a14=2a9. That gives 60=2⋅2860=2⋅28, which is false unless data forces d. Compute: a4=a+3da4=a+3d, a14=a+13da14=a+13d. Sum 2a+16d=60⇒a+8d=302a+16d=60⇒a+8d=30. Also a9=a+8d=28a9=a+8d=28. Hence a+8da+8d equals 28, so 30 vs 28 mismatch means d must adjust? Actually both give same expression, so impossible data. Correct option: No solution.
The correct conclusion for a4+a14=60a4+a14=60 and a9=28a9=28 in an AP is
A d equals 4
B a equals 2
C No such AP
D Infinite solutions
In any AP, a4+a14=2a9a4+a14=2a9 because 4 and 14 are equally spaced around 9. So left side must be 2×28=562×28=56, not 60. Hence no AP satisfies both conditions.
An AP has positive terms with a1=2a1=2 and an=50an=50. If sum of all terms is 520, the number of terms is
A 16
B 20
C 18
D 22
Use S=n2(a1+an)S=2n(a1+an). Here 520=n2(2+50)=n2⋅52=26n520=2n(2+50)=2n⋅52=26n. So n=520/26=20n=520/26=20. This uses sum with first and last directly.
An AP has a3=7a3=7 and a12=34a12=34. The sum of terms from 3rd to 12th equals
A 205
B 185
C 215
D 225
From 3rd to 12th there are 10 terms. First in this block is 7 and last is 34. Block sum =102(7+34)=5⋅41=205=210(7+34)=5⋅41=205. AP block sums use count and endpoints.
If SnSn of an AP is Sn=5n2+3nSn=5n2+3n, the common difference is
A 8
B 9
C 12
D 10
For an AP, Sn=d2n2+2a−d2nSn=2dn2+22a−dn. Compare with 5n2+3n5n2+3n. Then d/2=5⇒d=10d/2=5⇒d=10. The linear coefficient sets aa, but asked only dd.
In an AP, the product of 5th and 9th terms is 441 and the 7th term is 21. The common difference is
A 0
B 3
C 2
D 4
Let a7=21a7=21. Then a5=21−2da5=21−2d and a9=21+2da9=21+2d. Product =(21−2d)(21+2d)=441−4d2=(21−2d)(21+2d)=441−4d2. Given product is 441, so 441−4d2=441⇒d=0441−4d2=441⇒d=0. Hence d=0, not 3.
Correct value of common difference for the AP in previous question is
A 1
B 2
C 0
D 3
With middle term 21, symmetric terms are 21±2d21±2d. Their product equals 212−4d2212−4d2. If the product equals 212212, then 4d2=0⇒d=04d2=0⇒d=0. So the AP is constant.
A GP has a2=6a2=6 and a5=162a5=162. The common ratio is
A 2
B 6
C 4
D 3
In GP, a5/a2=r3a5/a2=r3. Here 162/6=27162/6=27. So r3=27⇒r=3r3=27⇒r=3. Once ratio is known, the entire GP is fixed.
A GP has positive terms and S4=80S4=80 with first term 5. The common ratio is
A 2
B 3
C 1
D 1/2
S4=5(1+r+r2+r3)=80⇒1+r+r2+r3=16S4=5(1+r+r2+r3)=80⇒1+r+r2+r3=16. Try r=2r=2: 1+2+4+8=151+2+4+8=15 not 16. Try r=1r=1: sum 4. So none? If ratio 1/2: sum 1+0.5+0.25+0.125=1.8751+0.5+0.25+0.125=1.875. No match. Hence no real r.
Correct conclusion for GP with first term 5 and S4=80S4=80 is
A r equals 2
B r equals 3
C No real ratio
D r equals 1/2
S4=5(1+r+r2+r3)=80⇒1+r+r2+r3=16S4=5(1+r+r2+r3)=80⇒1+r+r2+r3=16. The cubic r3+r2+r−15=0r3+r2+r−15=0 has r=2r=2 giving −1−1, r=3r=3 giving +18+18, so no integer root. It has one real root near 2.07, but options don’t include it, so no correct option among given ratios.
In a GP, if a1=9a1=9 and a4=1a4=1, the common ratio is
A 1/2
B 1/3
C 1/4
D 1/6
a4=a1r3⇒1=9r3⇒r3=1/9a4=a1r3⇒1=9r3⇒r3=1/9. Since 1/9=(1/3)21/9=(1/3)2, we get r=(1/9)1/3=1/93r=(1/9)1/3=1/39, not 1/31/3. So none fits.
Correct common ratio for a1=9a1=9 and a4=1a4=1 in a GP is
A 1/3
B 1/9
C 1/27
D 1/931/39
In a GP, a4=9r3=1a4=9r3=1. So r3=1/9r3=1/9. Taking cube root gives r=(1/9)1/3=1/93r=(1/9)1/3=1/39. This keeps the GP consistent.
The sum to infinity of an infinite GP is 12 and first term is 3. The common ratio is
A 1/2
B 3/4
C 1/3
D 2/3
S∞=a1−rS∞=1−ra. So 12=31−r⇒1−r=1/4⇒r=3/412=1−r3⇒1−r=1/4⇒r=3/4. Since ∣r∣<1∣r∣<1, the infinite sum is valid.
A ball travels 16 m first bounce and each bounce is 3/4 of previous. Total distance (infinite) is
A 64 m
B 48 m
C 80 m
D 96 m
Distances form infinite GP: 16+16⋅34+16(34)2+⋯16+16⋅43+16(43)2+⋯. Sum =161−3/4=64=1−3/416=64. This models repeated shrinking distances.
Two numbers have AM 25 and GM 20. The difference between numbers is
A 15
B 20
C 30
D 10
Let numbers be x,yx,y. AM gives x+y=50x+y=50. GM gives xy=400xy=400. Then (x−y)2=(x+y)2−4xy=2500−1600=900(x−y)2=(x+y)2−4xy=2500−1600=900. So ∣x−y∣=30∣x−y∣=30. Correct is 30.
Correct difference between numbers when AM 25 and GM 20 is
A 20
B 10
C 30
D 5
Using x+y=50x+y=50 and xy=400xy=400, compute (x−y)2=502−4⋅400=2500−1600=900(x−y)2=502−4⋅400=2500−1600=900. Thus ∣x−y∣=30∣x−y∣=30. Both numbers are positive and distinct.
For x>0x>0, the minimum value of x+25x+6x+x25+6 is
A 17
B 16
C 18
D 19
By AM–GM, x+25x≥225=10x+x25≥225=10. Adding 6 gives minimum 10+6=1610+6=16. Equality occurs at x=5x=5, making the bound achievable.
For positive x,yx,y with xy=36xy=36, the minimum value of x+yx+y is
A 10
B 11
C 14
D 12
AM–GM: x+y2≥xy=36=62x+y≥xy=36=6. So x+y≥12x+y≥12. Equality when x=y=6x=y=6. This gives the smallest possible sum.
For positive x,y,zx,y,z with x+y+z=18x+y+z=18, maximum xyzxyz equals
A 196
B 243
C 216
D 324
AM–GM for three numbers: x+y+z3≥xyz33x+y+z≥3xyz. With sum 18, average is 6, so xyz≤63=216xyz≤63=216. Maximum occurs at x=y=z=6x=y=z=6.
The sequence an=3+5n−2n2an=3+n5−n22 converges to
A 1
B 3
C 5
D 8
As n→∞n→∞, 5n→0n5→0 and 2n2→0n22→0. So an→3+0−0=3an→3+0−0=3. Small rational parts vanish for large n.
The sequence an=(1+2n)nan=(1+n2)n approaches
A e
B 2e
C 4e
D e²
A standard limit is (1+kn)n→ek(1+nk)n→ek. Here k=2k=2, so the limit is e2e2. This comes from exponential growth approximation.
The series ∑n=1∞n2n∑n=1∞2nn equals
A 2
B 1
C 3
D 4
Known result for ∣r∣<1∣r∣<1: ∑n=1∞nrn=r(1−r)2∑n=1∞nrn=(1−r)2r. With r=1/2r=1/2, sum =1/2(1/2)2=2=(1/2)21/2=2. This converges fast.
The series ∑n=1∞n3n∑n=1∞3nn equals
A 1/2
B 1
C 3/4
D 3/2
Use ∑nrn=r(1−r)2∑nrn=(1−r)2r for ∣r∣<1∣r∣<1. Here r=1/3r=1/3. Sum =1/3(2/3)2=1/34/9=34=(2/3)21/3=4/91/3=43.
Evaluate the telescoping sum ∑n=1∞1(n+1)(n+2)∑n=1∞(n+1)(n+2)1
A 1/3
B 1/2
C 2/3
D 1
1(n+1)(n+2)=1n+1−1n+2(n+1)(n+2)1=n+11−n+21. Summing from 1 to N gives 12−1N+221−N+21. As N→∞N→∞, it approaches 1/21/2. So correct is 1/2.
Correct value of ∑n=1∞1(n+1)(n+2)∑n=1∞(n+1)(n+2)1 is
A 1/3
B 2/3
C 1/4
D 1/2
Write as partial fractions: 1(n+1)(n+2)=1n+1−1n+2(n+1)(n+2)1=n+11−n+21. All middle terms cancel in partial sums, leaving 1221 as the limit.
Use 1n(n+3)=13(1n−1n+3)n(n+3)1=31(n1−n+31). In partial sums, most terms cancel, leaving first three harmonic pieces. Final sum is 11/1811/18.
The p-series ∑n=1∞1n0.9∑n=1∞n0.91 is
A Divergent
B Convergent
C Alternating
D Telescoping
A p-series ∑1/np∑1/np converges only if p>1p>1. Here p=0.9<1p=0.9<1, so partial sums grow without bound. Hence the series diverges.
The series ∑n=2∞1nlnn∑n=2∞nlnn1 is
A Convergent
B Geometric type
C Telescoping type
D Divergent
This is a known comparison with integral test: ∫1xlnxdx=ln(lnx)∫xlnx1dx=ln(lnx), which grows without bound. So ∑1nlnn∑nlnn1 diverges.
The series ∑n=1∞2nn!∑n=1∞n!2n is
A Divergent
B Harmonic type
C Convergent
D Telescoping type
Apply ratio test: an+1an=2n+1/(n+1)!2n/n!=2n+1→0anan+1=2n/n!2n+1/(n+1)!=n+12→0. Since ratio limit is less than 1, the series converges.
The series ∑n=1∞(3n5n+2)n∑n=1∞(5n+23n)n is
A Divergent
B Convergent
C Not defined
D Finite terms
Use root test: ann=3n5n+2→35<1nan=5n+23n→53<1. If the root limit is less than 1, the series converges absolutely. So it converges.
The series ∑n=1∞n23n∑n=1∞3nn2 is
A Divergent
B Alternating only
C Not a series
D Convergent
Exponential 3n3n grows much faster than polynomial n2n2. By ratio test or comparison with a convergent geometric series, the terms shrink rapidly. Hence the series converges.
The sum ∑k=0n(nk)∑k=0n(kn) equals
A 2n2n
B n!n!
C n2n2
D 2n2n
By the binomial theorem, (1+1)n=∑k=0n(nk)1n−k1k=∑k=0n(nk)(1+1)n=∑k=0n(kn)1n−k1k=∑k=0n(kn). So the sum equals 2n2n.
The sequence defined by a1=1a1=1, an+1=2an+3an+1=2an+3 has closed form
A 2n−32n−3
B 2n+32n+3
C 2n+1−32n+1−3
D 3⋅2n−33⋅2n−3
Compute a few terms: 1, 5, 13, 29. Pattern fits an=2n+1−3an=2n+1−3. Check: an+1=2(2n+1−3)+3=2n+2−3an+1=2(2n+1−3)+3=2n+2−3, correct.
Correct closed form for a1=1a1=1, an+1=2an+3an+1=2an+3 is
A 2n−32n−3
B 2n+1−32n+1−3
C 3⋅2n3⋅2n
D 2n+32n+3
Assume an=2n+1−3an=2n+1−3. It matches a1=4−3=1a1=4−3=1. Also recursion gives an+1=2an+3=2(2n+1−3)+3=2n+2−3an+1=2an+3=2(2n+1−3)+3=2n+2−3. So it holds.
For the Fibonacci-type sequence a1=2,a2=3,an=an−1+an−2a1=2,a2=3,an=an−1+an−2, the value of a7a7 is
A 21
B 29
C 55
D 34
Build terms: a3=5a3=5, a4=8a4=8, a5=13a5=13, a6=21a6=21, a7=34a7=34. Each term is sum of previous two, so careful stepwise addition gives the result.
The limit of an=(−2)n3nan=3n(−2)n is
A 1
B -1
C 0
D No limit
an=(−23)nan=(3−2)n. Since ∣−2/3∣<1∣−2/3∣<1, powers shrink to 0. Even though signs alternate, magnitude approaches 0, so the limit is 0.
If an AP has positive terms and a1a3=48a1a3=48 with a2=6a2=6, then a1a1 is
A 5
B 4
C 6
D 8
In three-term AP: a1=6−da1=6−d, a3=6+da3=6+d. Product (6−d)(6+d)=36−d2=48(6−d)(6+d)=36−d2=48 gives d2=−12d2=−12, impossible. So no such AP exists with real terms.
Correct conclusion for a2=6a2=6 and a1a3=48a1a3=48 in an AP is
A a1=4a1=4
B a1=8a1=8
C d=2d=2
D No real AP
For an AP with middle term 6, extremes are 6−d6−d and 6+d6+d. Their product is 36−d236−d2, which cannot exceed 36 for real d. Since 48>36, no real AP fits.
In a GP, if a1a3=64a1a3=64 and a2=8a2=8, then the common ratio can be
A Both ±2
B 2
C -2
D 1/2
With a2=8a2=8, let ratio be r. Then a1=8/ra1=8/r and a3=8ra3=8r. Product a1a3=(8/r)(8r)=64a1a3=(8/r)(8r)=64, always true for any nonzero r. But GP terms depend on r; common choices giving integer neighbors are r=2r=2 or r=−2r=−2.
The sum of first n terms of AP is zero, and first term is 7. If n is even, the last term is
A -6
B -7
C -5
D -4
For even n, sum zero means average of first and last is zero: a1+an2=0⇒a1+an=02a1+an=0⇒a1+an=0. With a1=7a1=7, we get an=−7an=−7. This uses symmetry in sums.
If a,b,ca,b,c are in GP and a+b+c=21a+b+c=21 with b=7b=7, then acac equals
A 36
B 42
C 56
D 49
In three-term GP, b2=acb2=ac. Given b=7b=7, ac=49ac=49. The extra condition a+b+c=21a+b+c=21 only confirms that aa and cc are symmetric around 7 in log scale.
For x>0x>0, the maximum value of 1x+14−xx1+4−x1 on 0
A 1
B 2
C No maximum
D 4
As x→0+x→0+, 1/x→∞1/x→∞. As x→4−x→4−, 1/(4−x)→∞1/(4−x)→∞. The expression becomes arbitrarily large near endpoints, so no finite maximum exists.
The minimum value of 1x+14−xx1+4−x1 on 0
A 2
B 1
C 4
D 8
Use AM–GM: 1x+14−x≥4x(4−x)x1+4−x1≥x(4−x)4 is not direct. Better: by symmetry, minimum occurs at x=2x=2. Value becomes 1/2+1/2=11/2+1/2=1. So minimum is 1.
Evaluate ∑n=1∞(12n−12n+1)∑n=1∞(2n1−2n+11)
A 1/2
B 1
C 3/2
D 2
This telescopes: term is 1/2n−1/2n+11/2n−1/2n+1. Partial sum up to N equals 1/2−1/2N+11/2−1/2N+1. As N→∞N→∞, last part goes to 0, leaving 1/21/2.
A series has nth term an=n+1n⋅12nan=nn+1⋅2n1. The series ∑an∑an is
A Divergent
B Oscillatory
C Not defined
D Convergent
n+1n=1+1nnn+1=1+n1 stays near 1, so anan behaves like 1/2n1/2n. Since ∑1/2n∑1/2n converges, this series also converges by comparison.
The limit of sequence an=1+2+3+⋯+nn2an=n21+2+3+⋯+n is
A 0
B 1
C 1/2
D 2
1+2+⋯+n=n(n+1)21+2+⋯+n=2n(n+1). Divide by n2n2: n(n+1)2n2=12(1+1n)→122n2n(n+1)=21(1+n1)→21. So limit is 1/2.
If r≠1r=1, the sum of first n terms of a GP can be written using last term ll as
A l−ar−1r−1l−a
B lr−ar−1r−1lr−a
C a−lr−1r−1a−l
D l−a1−r1−rl−a
For GP, Sn=a1−rn1−rSn=a1−r1−rn. Since l=arn−1l=arn−1, also arn=lrarn=lr. Then Sn=a−arn1−r=a−lr1−r=lr−ar−1Sn=1−ra−arn=1−ra−lr=r−1lr−a, equivalent forms.
In an AP, if the ratio of 7th term to 3rd term is 3:13:1 and a3=4a3=4, then a1a1 is
A 1
B 2
C 3
D 0
Let a1=aa1=a, common difference dd. Then a3=a+2d=4a3=a+2d=4. Also a7=a+6da7=a+6d. Given (a+6d)/(a+2d)=3(a+6d)/(a+2d)=3. So a+6d=3a+6d⇒2a=0⇒a=0a+6d=3a+6d⇒2a=0⇒a=0. Hence first term is 0.