Chapter 5: Sequences, Series and Progressions (Set-2)
An AP has a₁=12 and a₅=28; the common difference is
A 3
B 5
C 6
D 4
In an AP, a₅ = a₁ + 4d. So 28 = 12 + 4d ⇒ 4d = 16 ⇒ d = 4. The constant difference between consecutive terms is 4.
In an AP with a=9 and d=2, the 10th term is
A 27
B 25
C 29
D 31
Use aₙ = a + (n−1)d. Here a₁=9, d=2. So a₁₀ = 9 + 9×2 = 9 + 18 = 27. This matches repeated addition of 2.
If 7, x, 19 are in AP, x equals
A 11
B 13
C 12
D 14
For three-term AP, middle term is the average of extremes. So x = (7+19)/2 = 26/2 = 13. Also differences become 6 and 6, confirming AP.
If 4, x, 36 are in GP, x equals
A 10
B 14
C 16
D 12
In three-term GP, middle term satisfies x² = 4×36 = 144. So x = √144 = 12 (positive GM). Ratios are 3 and 3, confirming GP.
Number of terms in AP: a=3, d=2, last term 31
A 12
B 13
C 15
D 14
Use l = a + (n−1)d. So 31 = 3 + (n−1)2 ⇒ 28 = 2(n−1) ⇒ n−1=14 ⇒ n=15. Hence AP has 15 terms.
Sum of 15 terms of AP: a=3, d=2
A 240
B 270
C 285
D 255
First term a=3, n=15, last term l = 3 + 14×2 = 31. Sum S = n(a+l)/2 = 15(3+31)/2 = 15×34/2 = 255.
If an AP has a₃=11 and a₇=23, then d is
A 3
B 2
C 4
D 6
a₃=a+2d=11 and a₇=a+6d=23. Subtract: 4d=12 ⇒ d=3. This uses the constant gap in indices to find difference quickly.
If sum of first n terms of an AP is proportional to n², then d is
A Positive
B Zero
C Negative
D Not fixed
Sₙ = n/2[2a+(n−1)d] is quadratic in n when d≠0, but “proportional to n²” with no linear part requires a=0, not d=0. So statement is tricky; correct: d can be nonzero.
If Sₙ of an AP is exactly kn², then first term must be
A k
B 2k
C d
D 0
For AP, Sₙ = n/2[2a+(n−1)d] = (d/2)n² + (2a−d)/2 · n. For Sₙ = kn², linear coefficient must be 0 ⇒ 2a−d=0 ⇒ a=d/2.
If Sₙ = 5n² for an AP, then relation between a and d is
A 2a = d
B a = d
C a = 2d
D a = 0
Comparing Sₙ = (d/2)n² + (2a−d)/2·n with 5n², the linear term must vanish: 2a−d=0 ⇒ 2a=d. Then d/2=5 ⇒ d=10, a=5.
If a GP has a₁=2 and r=3, the 6th term is
A 162
B 729
C 486
D 486
aₙ = a·r^(n−1). So a₆ = 2·3^5 = 2·243 = 486. Each step multiplies by 3, so growth is fast compared to AP.
For GP with a=5 and r=2, sum of first 4 terms is
A 35
B 45
C 80
D 75
Terms: 5,10,20,40. Sum = 75. Or use S₄ = a(1−r⁴)/(1−r) = 5(1−16)/(1−2)=5(−15)/(−1)=75. Both match.
If |r|<1, and first term is 9, r=1/3, then S∞ is
A 12
B 13.5
C 15
D 18
S∞ = a/(1−r) = 9/(1−1/3)=9/(2/3)=9×3/2=13.5. Convergence happens because r is between −1 and 1.
Insert one AM between 8 and 20; the inserted number is
A 14
B 12
C 13
D 15
One AM between two numbers is their average: (8+20)/2=14. Then 8,14,20 forms a three-term AP with equal differences of 6.
Insert one GM between 2 and 18; the inserted number is
A 5
B 8
C 9
D 6
One GM between positive numbers is √(ab). Here √(2×18)=√36=6. Then 2,6,18 is a GP with common ratio 3.
If AM of x and 20 is 14, then x is
A 6
B 10
C 8
D 12
AM = (x+20)/2 = 14 ⇒ x+20 = 28 ⇒ x=8. Arithmetic mean is the midpoint on number line between the two numbers.
If GM of x and 36 is 12, then x is
A 4
B 3
C 6
D 9
GM means √(36x)=12 ⇒ 36x=144 ⇒ x=4. Geometric mean uses product, so it naturally fits GP middle-term condition.
For x,y>0, AM−GM is zero when
A x>y
B x<y
C xy=0
D x=y
AM ≥ GM for positive numbers, and equality happens only when both numbers are equal. Then AM and GM become the same value and their difference becomes zero.
If x+y is fixed (x,y>0), product xy is maximum when
A x=0
B x=y
C x=2y
D x=3y
By AM–GM, (x+y)/2 ≥ √(xy). For fixed x+y, √(xy) is maximized when equality holds, i.e., x=y. Thus xy is maximum at equal split.
If xy is fixed (x,y>0), sum x+y is minimum when
A x=y
B x=2y
C x=3y
D x=0
From AM–GM, (x+y)/2 ≥ √(xy). With fixed product xy, √(xy) is fixed, so x+y is minimized when equality holds, which requires x=y.
A sequence {aₙ} converges if aₙ approaches
A Infinity only
B Random values
C A real limit
D Zero always
A sequence converges when its terms get closer and closer to a single real number L as n grows. The limit can be any real number, not necessarily zero.
If aₙ = 1/n, then lim aₙ equals
A 1
B Infinity
C Does not exist
D 0
As n becomes very large, 1/n becomes very small. It approaches 0. This is a classic basic limit and often used to test convergence ideas in sequences.
If aₙ = (2n+1)/(n), then lim aₙ equals
A 2
B 1
C 3
D 0
(2n+1)/n = 2 + 1/n. As n→∞, 1/n→0, so limit becomes 2. Dividing numerator and denominator by n is the easy method.
If r=−1/2 in a GP, then |r| is
A −1/2
B 1/2
C 2
D 1
Absolute value removes sign. So |−1/2| = 1/2. This matters for checking convergence of infinite GP, where the condition is |r|<1.
Series 1 − 1/2 + 1/4 − 1/8 + … is
A AP series
B Harmonic series
C Square series
D GP series
Each term is multiplied by −1/2 to get the next. That constant ratio shows it is a GP series with a=1 and r=−1/2, and it converges since |r|<1.
Sum to infinity of 1 − 1/2 + 1/4 − 1/8 + … is
A 2/3
B 1/3
C 1/2
D 3/2
Infinite GP sum is a/(1−r). Here a=1 and r=−1/2. So S∞ = 1/(1−(−1/2)) = 1/(3/2) = 2/3.
If aₙ is bounded and increasing, then it must
A Diverge always
B Oscillate
C Converge
D Become zero
A basic result says a monotonic increasing sequence that is bounded above has a limit. So it converges. The bound prevents it from going to infinity.
If aₙ decreases and is bounded below, then it
A Must diverge
B Be periodic
C Be constant only
D Must converge
A monotonic decreasing sequence bounded below cannot decrease forever without limit. It approaches a greatest lower bound (infimum). Hence it converges to some real number.
The nth term test for series says: if aₙ does not go to 0, ∑aₙ
A Diverges
B Converges
C Equals aₙ
D Equals 0
If the terms of a series do not approach 0, partial sums cannot settle to a finite value. So the series must diverge. This is a simple necessary test.
If aₙ → 0, then ∑aₙ
A Must converge
B May diverge
C Must diverge
D Equals 0
aₙ → 0 is necessary but not sufficient for convergence. Example: harmonic series has aₙ=1/n → 0 but still diverges. More tests are needed.
Sum of first n terms of GP with r=1 equals
A a/n
B a²
C a−n
D an
If r=1, all terms are a. So sum of n terms is a+a+… (n times) = an. This is a simple special case of GP sum.
If aₙ = (−1)ⁿ, then sequence is
A Divergent
B Convergent
C Increasing
D Constant
(−1)ⁿ alternates between −1 and 1. It does not approach a single real limit. Hence the sequence diverges, though it remains bounded.
A weighted mean gives more importance to
A Smaller numbers
B Larger numbers
C Assigned weights
D Negative numbers
In weighted mean, each value is multiplied by its weight, and divided by total weight. So values with higher weights influence the mean more, regardless of their size.
Harmonic mean of two positive numbers x and y is
A (x+y)/2
B √(xy)
C (x−y)/2
D 2xy/(x+y)
Harmonic mean of two positives is defined as 2 divided by (1/x + 1/y), which simplifies to 2xy/(x+y). It is used in rates and average speed problems.
For positive numbers, which order is always true?
A AM ≤ GM ≤ HM
B AM ≥ GM ≥ HM
C GM ≥ AM ≥ HM
D HM ≥ GM ≥ AM
For positive numbers, arithmetic mean is at least geometric mean, and geometric mean is at least harmonic mean. Equality happens when all numbers are equal.
If 3 numbers are in AP, the middle equals
A Average of ends
B Sum of ends
C Product of ends
D Difference of ends
In three-term AP, the middle term is the arithmetic mean of the first and third. This ensures equal differences on both sides, making the progression symmetric.
If 3 numbers are in GP, the middle equals
A AM of ends
B HM of ends
C Sum of ends
D GM of ends
In three-term GP, the middle term is the geometric mean of the first and third terms. This makes ratios equal on both sides and gives the key condition b²=ac.
If an AP has 20 terms, then number of differences is
A 18
B 19
C 20
D 21
Differences are between consecutive terms. With 20 terms, there are 19 gaps: a₂−a₁ through a₂₀−a₁₉. Each gap equals the common difference d.
If an AP has odd number of terms, the middle term equals
A Average of extremes
B Sum average
C Product extremes
D Ratio extremes
In an AP with odd number of terms, the middle term lies exactly halfway. It equals the average of the first and last terms because terms are evenly spaced by d.
If a GP has odd number of terms, middle term equals
A (first+last)/2
B first−last
C first×last
D √(first×last)
In a GP with odd number of terms, the middle term is the geometric mean of first and last. This comes from equal ratios on both sides, giving middle² = first×last.
If a₁=2, a₂=5 in an AP, then d is
A 2
B 4
C 3
D 5
Common difference is a₂−a₁. Here d = 5−2 = 3. Then sequence continues 2, 5, 8, 11, … by adding 3 each time.
If a₁=3, a₂=12 in a GP, then r is
A 4
B 2
C 3
D 5
Common ratio r = a₂/a₁ = 12/3 = 4. Each term is obtained by multiplying the previous term by 4, giving 3, 12, 48, 192, …
For GP, if r=0, then terms after first are
A Same as first
B All ones
C All negative
D All zero
If r=0, then a₂=a₁·0=0, and all later terms remain 0. So the GP becomes a, 0, 0, 0, … This is a simple edge case.