NDA Sequences and Series previous year questions with solutions

Answer: (a) 5R = 3S

Method. Q = 3P fixes 2a = (n + 1)d; then express R and S in n²d

S_k = k2[2a + (k − 1)d].
Q = 3P: n[2a + (2n − 1)d] = 3n2[2a + (n − 1)d], so 4a + (4n − 2)d = 6a + (3n − 3)d, giving 2a = (n + 1)d.
R = 3n2[(n + 1)d + (3n − 1)d] = 3n2 · 4nd = 6n²d.
S = 2n[(n + 1)d + (4n − 1)d] = 2n · 5nd = 10n²d.
5R = 30n²d = 3S.

Answer: (c) 11

Method. T_n = S_n - S_{n − 1}

T_n = (3n² + 5n) - [3(n − 1)² + 5(n − 1)] = 6n + 2.
6m + 2 = 68, so m = 11.

Answer: (c) 37

Method. common terms form an AP with difference lcm(2, 3) = 6

First AP: last term 19 + 109 × 2 = 237. Second AP: last term 19 + 74 × 3 = 241.
Common terms: 19, 25, 31, …, i.e. 19 + 6k, with difference lcm(2, 3) = 6.
They must not exceed 237: 19 + 6k ≤ 237 gives k ≤ 36.33, so k = 0, 1, …, 36.
Number of common terms = 37.

Answer: (d) n² = 2m(m + 2)

Method. sum of an AP versus sum of odd numbers

6 + 10 + 14 + … (m terms) = m2[12 + (m − 1)4] = m(2m + 4) = 2m(m + 2).
1 + 3 + 5 + … (n terms) = n².
So n² = 2m(m + 2).

Answer: (c) Two

Method. search n² = 2m(m + 2) with n < 25

n < 25 means 2m(m + 2) = n² ≤ 576, so m ≤ 16 (m = 17 gives 646).
n² is even, so n is even; put n = 2k: 2k² = m(m + 2), so m is even; put m = 2j: k² = 2j(j + 1).
j = 1: k² = 4, k = 2 (m = 2, n = 4). j = 2 to 7: 12, 24, 40, 60, 84, 112 are not squares. j = 8: k² = 144, k = 12 (m = 16, n = 24).
Two values of m: 2 and 16.

Answer: (b) The terms of S form an arithmetic progression with common difference 6.

Method. T_k = S_k - S_{k − 1}

T_k = (3k² + 5k) - [3(k − 1)² + 5(k − 1)] = 3(2k − 1) + 5 = 6k + 2.
T₁ = 8 = S₁, and T_{k + 1} - T_k = 6.
So the terms form an AP with common difference 6.

Shortcut. For S_k = Ak² + Bk, the series is an AP with common difference 2A = 6.

Answer: (a) 0

Method. write both terms as a + (n − 1)d and cancel (p − q)

p[a + (p − 1)d] = q[a + (q − 1)d].
a(p − q) + d[(p² - q²) − (p − q)] = 0.
Divide by p − q ≠ 0: a + (p + q − 1)d = 0.
But a + (p + q − 1)d is the (p + q)th term, so it equals 0.

Answer: (a) 4n − 1

Method. T_n = S_n - S_{n − 1}

S_n = 2n² + n.
T_n = (2n² + n) - [2(n − 1)² + (n − 1)] = 2(2n − 1) + 1 = 4n − 1.
Check: T₁ = S₁ = 3 = 4(1) − 1.

Answer: (c) The common difference is equal to twice the first term

Method. ratio of S_p and S_q from S_n = n2[2a + (n − 1)d]

S_pS_q = p[2a + (p − 1)d]q[2a + (q − 1)d] = p²q², so q[2a + (p − 1)d] = p[2a + (q − 1)d].
2aq + pqd − qd = 2ap + pqd − pd.
2a(q − p) = d(q − p); since p ≠ q, d = 2a.
The common difference is twice the first term.

Shortcut. S_n ∝ n² means S_n = kn², so the terms are k, 3k, 5k, …: a = k, d = 2k.

Answer: (c) 2920

Method. Split the linear general term into Σn and a constant

S = 17140Σn=1(2n + 5) = 17(2·140.1412 + 5.140).
= 17(19740 + 700) = 204407 = 2920.

Shortcut. AP with T₁ = 1, T₁₄₀ = 2857: S = 70(1 + 2857) = 70 + 2850 = 2920.

Answer: (b) 6

Method. General term of a decreasing AP; the first negative term is the largest negative term

a = 52, d = 2312 − 3012 = -712.
T_n = 52 − 7(n − 1)12 = 37 − 7n12.
T_n < 0 when n > 377 ≈ 5.29 (one decimal convention across the chapter; the q-packets currently mix 0.33, 5.19, 9.19 with 36.33)., so the first negative term is T₆ = -512.
The AP is decreasing, so later terms are smaller; T₆ is the largest negative term and n = 6.

Answer: (b) mx(m + n)1 − n

Method. Find d from S_n = 0, then S_{m + n} - S_n

S_n = n2(2x + (n − 1)d) = 0 gives d = -2xn − 1.
Sum of next m terms = S_{m + n} - S_n = S_{m + n} = m + n2(2x + (m + n − 1)d).
= (m + n)x(1 − m + n − 1n − 1) = (m + n)x·-mn − 1.
= mx(m + n)1 − n.

Answer: (d) 3/⁠5

Method. Ratio of sums at n = 1 gives the ratio of first terms

For n = 1, P and Q are the first terms of A and B.
Ratio = 5.1 + 49.1 + 6 = 915 = 35.

Answer: (c) 33/⁠59

Method. Ratio of m-th terms = ratio of sums at n = 2m − 1

PQ = n2(2a + (n − 1)d)n2(2A + (n − 1)D) = a + n − 12dA + n − 12D.
For the 10th terms a + 9d and A + 9D we need n − 12 = 9, i.e. n = 19.
Ratio = 5.19 + 49.19 + 6 = 99177 = 3359.

No correct option. This question is defective as printed in UPSC's paper.

Method. Match coefficients in the ratio-of-sums identity to get d: D

For every n: (2a + (n − 1)d)(9n + 6) = (5n + 4)(2A + (n − 1)D).
Compare the n² terms: 9d = 5D, so d = 59D; the constant terms (n = 1) give a = 35A (D > 0 because B is increasing).
So D > d (as 59D < D), and 12d = 203D < 7D, so 7D > 12d as well.
Example: A: 3, 193, … (d = 103); B: 5, 11, … (D = 6) gives P: Q = (5n + 4): (9n + 6), and both 6 > 103 and 42 > 40 hold.
Options (a) and (c) are both always correct.

Since d = 5D/⁠9 with D > 0, both D > d (option a) and 7D > 12d (option c) always hold, so the 'which one' question has two correct options; the setter evidently intended (c), the sharper inequality that implies (a).

Answer: (d) 0

Method. Equal partial sums force 2a + 19d = 0

S₉ = S₁₁: 92(2a + 8d) = 112(2a + 10d).
18a + 72d = 22a + 110d, so 4a + 38d = 0, i.e. 2a + 19d = 0.
S₂₀ = 202(2a + 19d) = 0.

Shortcut. S₉ = S₁₁ means T₁₀ + T₁₁ = 0, and S₂₀ = 10(T₁₀ + T₁₁) = 0.

Answer: (b) 25.5

Method. Find a and d from two terms, then S_n = n2(2a + (n − 1)d)

a + 4d = 110 and a + 9d = 15.
Subtract: 5d = 110, d = 150; a = 110 − 450 = 150.
S₅₀ = 25(250 + 4950) = 25·5150 = 512 = 25.5.

Shortcut. T_n = n50, so S₅₀ = 150·50.512 = 25.5.

Answer: (d) 102

Method. Mean of an AP = average of first and last terms

First term 4, d = 4, so the 50th term = 4 + 49.4 = 200.
Sum = 502(4 + 200) = 5100.
Arithmetic mean = 510050 = 102.

Answer: (c) 3

Method. Sum formula S_k = k2(2a + (k − 1)d)

S₃₀ = 15(2a + 29d).
S₂₀ - S₁₀ = 10(2a + 19d) − 5(2a + 9d) = 10a + 145d = 5(2a + 29d).
S₃₀S₂₀ - S₁₀ = 155 = 3.

Answer: (b) -250

Method. Sum formula S_n = n2(2a + (n − 1)d) with S₁₀ = 5S₅

Sum of next five terms = S₁₀ - S₅. Given S₅ = 14(S₁₀ - S₅), so S₁₀ = 5S₅.
With a = 2: S₅ = 52(4 + 4d) = 10 + 10d; S₁₀ = 5(4 + 9d) = 20 + 45d.
20 + 45d = 5(10 + 10d) = 50 + 50d gives d = -6.
S₁₀ = 20 + 45(-6) = -250.

Answer: (a) 1

Method. T_n = S_n - S_n-1

S₃ = 15, S₂ = 14.
T₃ = 15 − 14 = 1.

Answer: (b) 22

Method. count an AP of multiples of 4

Two-digit multiples of 4: 12, 16, …, 96.
Number = 96 − 124 + 1 = 22.

Answer: (c) 7

Method. Rewrite the sum in terms of the number of terms; for S_m = Am + Bm², d = 2B

Put m = 2n, so n = m2: S_m = 3m2 + 14·m²4 = 32m + 72m².
For an AP, S_m = m2(2a + (m − 1)d) = d2m² + (a − d2)m.
Compare the m² terms: d2 = 72, so d = 7.
Check: S₁ = 5, S₂ = 17, S₃ = 36, so the terms are 5, 12, 19, … with d = 7.

Shortcut. S_2n = 14n² + 3n means S_m has m²-coefficient 14/⁠4 = 7/⁠2; d is twice that, 7.

Answer: (b) 8

Method. T_n = S_n - S_n-1

S₄ = 4 × 5 = 20, S₃ = 3 × 4 = 12.
T₄ = 20 − 12 = 8.

Answer: (c) 1

Method. solve for a and d from the two given terms

T_m - T_n = (m − n)d = 1n − 1m = m − nmn ⇒ d = 1mn.
a = T_m - (m − 1)d = 1n − m − 1mn = 1mn.
T_mn = a + (mn − 1)d = 1mn + mn − 1mn = 1.

Answer: (c) 1635

Method. list the numbers as an AP and sum

Two-digit numbers of the form 3k + 2: 11, 14, …, 98, an AP with d = 3.
Number of terms = 98 − 113 + 1 = 30.
Sum = 302(11 + 98) = 15 × 109 = 1635.

Answer: (d) Q

Method. compare with S_n = n2[2a + (n − 1)d], or use T₂ - T₁

S_n = nP + n(n −⁠ 1)2Q has the form na + n(n −⁠ 1)2d with a = P, d = Q.
Check: T₁ = S₁ = P, S₂ = 2P + Q, so T₂ = P + Q and d = T₂ - T₁ = Q.

Answer: (b) 34 minutes

Method. sum of an AP set equal to the notes left, reject the root that gives negative terms

First 10 minutes: 10 × 150 = 1500 notes; 4500 − 1500 = 3000 remain.
From the 11^th minute the counts are 148, 146, 144, …; in k more minutes he counts k2[2(148) + (k − 1)(-2)] = k(149 − k).
k(149 − k) = 3000 ⇒ k² - 149k + 3000 = 0 ⇒ (k − 24)(k − 125) = 0.
k = 125 is rejected: a_10+k = 150 − 2k would be negative long before then (it reaches 0 at k = 75).
So k = 24, and the total time is 10 + 24 = 34 minutes.

Answer: (d) - (m + n)

Method. subtract the two sum equations of the AP

m2[2a + (m − 1)d] = n and n2[2a + (n − 1)d] = m, i.e. 2am + m(m − 1)d = 2n and 2an + n(n −⁠ 1)d = 2m.
Subtract: 2a(m − n) + (m² - n² - m + n)d = 2(n − m).
Divide by m − n(m ≠ n): 2a + (m + n − 1)d = -2.
S_m+n = m + n2[2a + (m + n − 1)d] = m + n2(-2) = -(m + n).

Answer: (b) 7

Method. T_n = S_n - S_n-1

S₅ = 25 − 10 = 15, S₄ = 16 − 8 = 8.
T₅ = S₅ - S₄ = 15 − 8 = 7.

Answer: (a) 2475

Method. sum of an AP: number of terms × average of first and last

Two-digit odd numbers: 11, 13, …, 99, an AP with d = 2.
Number of terms = 99 − 112 + 1 = 45.
Sum = 452(11 + 99) = 45 × 55 = 2475.

Answer: (d) 8q³

Method. the expression is (p + r)³

p³ + 3pr(p + r) + r³ = p³ + 3p²r + 3pr² + r³ = (p + r)³.
p, q, r in AP: p + r = 2q.
So the expression = (2q)³ = 8q³.

Answer: (c) 3 : 5 : 4

Method. equal AP gaps d, d with p make the GP condition d² = dp

Let q − p = r − q = d ≠ 0 (the numbers are unequal).
d, d, p in GP: d² = dp, so p = d.
Then p = d, q = 2d, r = 3d.
(p + q): (q + r): (r + p) = 3d: 5d: 4d = 3 : 5 : 4.

Answer: (b) 3k

Method. terms equidistant from T_p average to T_p

T_{p + q} = T_p + qd and T_{p − q} = T_p - qd.
T_p + T_{p + q} + T_{p − q} = 3T_p = 3k.

Answer: (a) -3

Method. GP condition (a + 6d)² = (a + 4d)(a + 12d)

T₅ = a + 4d, T₇ = a + 6d, T₁₃ = a + 12d.
(a + 6d)² = (a + 4d)(a + 12d).
a² + 12ad + 36d² = a² + 16ad + 48d², so -4ad = 12d².
With d ≠ 0, a = -3d, so a: d = -3.

Answer: (d) 6

Method. GP condition on terms 1, 2, 5 fixes d = 2a; then use the product

Let the terms be a, a + d, a + 2d, a + 3d, a + 4d.
First, second, fifth in GP: (a + d)² = a(a + 4d), so d² = 2ad, giving d = 2a (d ≠ 0).
The terms are a, 3a, 5a, 7a, 9a with product 945a⁵ = 229635.
a⁵ = 243, so a = 3 and d = 6.

Answer: (c) 75

Method. terms are a, 3a, 5a, 7a, 9a with a = 3

As in the previous item, d = 2a and 945a⁵ = 229635, so a⁵ = 243 and a = 3.
The terms are 3, 9, 15, 21, 27.
Sum = 25a = 75.

Answer: (a) 0

Method. in an AP, a_i + a_j is constant when i + j is constant

The index pairs 1 + 34, 5 + 30, 10 + 25, 15 + 20 all sum to 35, so a₁ + a₃₄ = a₅ + a₃₀ = a₁₀ + a₂₅ = a₁₅ + a₂₀.
The given sum is 4(a₁ + a₃₄) = 300, so a₁ + a₃₄ = 75.
Required = (a₁ + a₃₄) + (a₅ + a₃₀) − (a₁₀ + a₂₅) − (a₁₅ + a₂₀) = 75 + 75 − 75 − 75 = 0.

Answer: (d) 1275

Method. S_n = n2(first term + last term)

From the stem, a₁ + a₃₄ = a₅ + a₃₀ = a₁₀ + a₂₅ = a₁₅ + a₂₀, so 4(a₁ + a₃₄) = 300 and a₁ + a₃₄ = 75.
34Σn=1 a_n = 342(a₁ + a₃₄) = 17 × 75 = 1275.

Answer: (a) 6

Method. Symmetric terms of an AP have equal sums: p + s = q + r

In an AP, p + s = q + r, so q + r = 8 and qr = 15.
q, r are roots of t² - 8t + 15 = 0: 3 and 5, so the common difference is ±2.
The numbers are 1, 3, 5, 7 (in some order of direction).
Largest − smallest = 7 − 1 = 6.

Answer: (c) 98

Method. Express pq as a quadratic in d and minimise

Let the AP have first term p and common difference d.
Fifth term: p + 4d = 3, so p = 3 − 4d; third term q = p + 2d = 3 − 2d.
pq = (3 − 4d)(3 − 2d) = 8d² - 18d + 9.
This upward parabola is least at d = 182.8 = 98.

Answer: (c) x ∈ {-2, 4}

Method. Middle term of an AP is the mean of its neighbours

x², x, -8 in AP: 2x = x² + (-8).
x² - 2x − 8 = 0, i.e. (x − 4)(x + 2) = 0.
x = 4 or x = -2. Check: 16, 4, -8 (d = -12) and 4, -2, -8 (d = -6) are both APs.
So x ∈ {-2, 4}.

Answer: (c) 32

Method. sum of the inserted means = (number inserted) × (m + n)/⁠2

x² - 16x + 39 = (x − 3)(x − 13), so m = 3, n = 13.
3, p, q, r, s, 13 is an AP with 5d = 10, d = 2: p, q, r, s = 5, 7, 9, 11.
p + q + r + s = 32.

Shortcut. The four inserted terms average 3 + 132 = 8, so their sum is 4 × 8 = 32.

Answer: (d) 1, 2 and 3

Method. test each statement with 2b = a + c

1. 2(kb) = k(a + c) = ka + kc: true.
2. 2(k − b) = 2k − (a + c) = (k − a) + (k − c): true.
3. 2·bk = a + ck = ak + ck (k ≠ 0): true.
All three are correct.

Answer: (c) infinite number of APs

Method. common difference d = 4/⁠p works for every positive integer p

Take d = 4/⁠p for any positive integer p: then 5 = 1 + p·d and 25 = 1 + 6p·d are both terms, so every p gives a different AP containing 1, 5, 25 (p = 1: d = 4; p = 2: d = 2; p = 4: d = 1; …). Hence infinitely many APs.
So q = 5p; for each p = 1, 2, 3, … take d = 4p: e.g. d = 4 (1, 5, 9, …, 25), d = 2, d = 1, ….
Hence infinitely many APs. (They also lie in infinitely many GPs, ratio 5^1/⁠p, so (d) is false.)

Answer: (c) Twice the p^th term

Method. write both terms with a and d

T_p+q + T_p-q = [a + (p + q − 1)d] + [a + (p − q − 1)d] = 2a + 2(p − 1)d.
= 2[a + (p − 1)d] = 2T_p.

Answer: (b) 18⁠√3⁠

Method. T₅² = T₃T₇ fixes x; then r² = T₅T₃

(x² + 2)² = x(x³ + 10): x⁴ + 4x² + 4 = x⁴ + 10x, so 2x² - 5x + 2 = 0.
x = 2 or 12; x > 1 gives x = 2.
T₃ = 2, T₅ = 6, T₇ = 18, so r² = 3 and r = ⁠√3⁠ (r = -⁠√3⁠ gives -18⁠√3,⁠ not offered).
T₈ = T₇ · r = 18⁠√3.⁠

Answer: (b) 2m

Method. GP relations g₁² = pg₂ and g₂² = g₁q

In the GP p, g₁, g₂, q: g₁² = pg₂ and g₂² = g₁q.
So g₁²g₂ = p and g₂²g₁ = q.
Sum = p + q = 2m.

Answer: (c) Three

Method. S₈ = S₄(1 + r⁴); keep the case S₄ = 0

The 5th to 8th terms are r⁴ times the first four, so S₈ = S₄ + r⁴S₄ = S₄(1 + r⁴).
S₄(1 + r⁴) = 5S₄ gives S₄(r⁴ - 4) = 0.
r⁴ = 4 gives r = ⁠√2⁠ or r = -⁠√2.⁠
S₄ = a(1 + r)(1 + r²) = 0 (a ≠ 0) gives r = -1: then S₄ = S₈ = 0, and 0 = 5 × 0 holds.
So r ∈ {⁠√2,⁠ -⁠√2,⁠ -1}: three real values.

Not flawed, but a trap: dividing by S₄ (or using r⁸ - 1r⁴ - 1 = r⁴ + 1) loses r = -1, which leaves only ±⁠√2⁠ and option (b); the complete count is three, option (c).

Answer: (c) in HP

Method. logs of a GP form an AP; reciprocals of an AP form an HP

x, y, z in GP: y² = xz, so 2 ln y = ln x + ln z; ln x, ln y, ln z are in AP.
Adding 1 to each: 1 + ln x, 1 + ln y, 1 + ln z are in AP (all positive since x, y, z > 1).
Reciprocals of an AP of non-zero terms are in HP, so the three fractions are in HP.

Answer: (c) One

Method. recover f, then use f(2x)² = f(x)f(4x)

f(2x) = (2x)² + 1, so f(t) = t² + 1. Then f(x) = x² + 1 and f(4x) = 16x² + 1.
GM condition: (4x² + 1)² = (x² + 1)(16x² + 1).
16x⁴ + 8x² + 1 = 16x⁴ + 17x² + 1, so 9x² = 0.
Only x = 0 (where f(2x) = 1 = ⁠√(1⁠ · 1)): one value.

Answer: (c) t₁₁

Method. product of terms of a GP equidistant from the middle term

Let t_n = Ar^{n − 1}. The terms t₁, t₃, …, t₂₁ are the 11 odd-numbered terms.
Product = A¹¹r^{0 + 2 + 4 + … + 20} = A¹¹r¹¹⁰.
(A¹¹r¹¹⁰)^1/⁠11 = Ar¹⁰ = t₁₁.

Shortcut. The 11 terms are a GP with middle term t₁₁, so their geometric mean is t₁₁.

Answer: (c) 10

Method. f(x) = 2^x from the functional equation; GP sum

f(x + y) = f(x)f(y) with f(1) = 2 gives f(2) = f(1)² = 4 and, by induction, f(x) = 2^x for x ∈ N.
nΣx=2 2^x = 2² + … + 2ⁿ = 2^{n + 1} - 4.
2^{n + 1} - 4 = 2044 gives 2^{n + 1} = 2048 = 2¹¹, so n = 10.

Answer: (b) 682

Method. f(x) = 2^x; sum of odd powers of 2 as a GP

f(2x − 1) = 2^{2x − 1}, so the sum is 2 + 2³ + 2⁵ + 2⁷ + 2⁹.
GP with first term 2, ratio 4, 5 terms: 2(4⁵ - 1)4 − 1 = 2.10233 = 682.
Check: 2 + 8 + 32 + 128 + 512 = 682.

Answer: (d) 5460

Method. f(x) = 2^x, so 2^xf(x) = 4^x; GP sum

2^xf(x) = 2^x·2^x = 4^x.
6Σx=1 4^x = 4(4⁶ - 1)4 − 1 = 4.40953 = 5460.

Answer: (d) 1, 2 and 3

Method. Powers of a positive GP form a GP

a, b, c > 0 in GP: b² = ac.
1. (b²)² = (ac)² = a²c², so a², b², c² are in GP.
2. (1b)² = 1ac = 1a·1c, so 1a, 1b, 1c are in GP.
3. (⁠√b)²⁠ = b = ⁠√(ac)⁠ = ⁠√a·√c⁠ (all positive), so ⁠√a,⁠ ⁠√b,⁠ ⁠√c⁠ are in GP.
All three are correct.

Answer: (a) 1 only

Method. Logarithms of a GP form an AP; adding a constant to an AP gives an AP, not an HP

x, y, z in GP (positive, as ln is taken): y² = xz, so 2 ln y = ln x + ln z.
ln(3x) = ln 3 + ln x, etc.; adding the constant ln 3 to an AP gives an AP. Statement 1 is correct.
xyz + ln x, xyz + ln y, xyz + ln z is the AP ln x, ln y, ln z shifted by the constant xyz, so it is an AP.
An AP with distinct terms is never an HP (e.g. x = 1, y = e, z = e²: e³, e³ + 1, e³ + 2 are not in HP). Statement 2 is not correct.

Answer: (c) Both 1 and 2

Method. Scaling a GP keeps the common ratio

Let the GP be a, ar, ar², … and k ≠ 0.
Multiplying: ka, kar, kar², … has each term r times the previous, so it is a GP (ratio r). Statement 1 is correct.
Dividing by k is multiplying by 1k ≠ 0, so the same argument applies. Statement 2 is correct.

Answer: (b) 243

Method. Product of terms symmetric about the middle term of a GP

Let the first five terms be ar², ar, a, ar, ar², with middle (third) term a = 3.
Product = a⁵ = 3⁵ = 243.
The ratio cancels, so the data are sufficient.

Answer: (c) Both 1 and 2

Method. GP condition b² = ac, then take square roots and logarithms

p², q², r² in GP: q⁴ = p²r².
All positive, so take the positive square root: q² = pr, hence p, q, r are in GP. Statement 1 is correct.
Take ln of q² = pr: 2 ln q = ln p + ln r, hence ln p, ln q, ln r are in AP. Statement 2 is correct.

Answer: (b) n/⁠m

Method. each even term is r times the odd term before it

Odd terms: a, ar², …, ar¹⁹⁸ (100 terms), sum m.
Even terms: ar, ar³, …, ar¹⁹⁹ = r(a + ar² + … + ar¹⁹⁸), so n = rm.
r = nm.

Answer: (a) q² = pr

Method. Terms equally spaced in a GP form a GP

Let the GP be A, AR, AR², … so T_k = AR^{k − 1}.
p = AR², q = AR⁷, r = AR¹².
pr = A²R¹⁴ = (AR⁷)² = q².
So q² = pr.

Answer: (d) (-1)^n-1 5ⁿ⁺¹

Method. GP with first term 25 and ratio -5

T_n = 25(-5)^n-1 = 5² × (-1)^n-15^n-1 = (-1)^n-15ⁿ⁺¹.
Check: n = 1 gives 25, n = 2 gives -125.

Answer: (c) 0 < x < 10

Method. x = 5(1 − r) with -1 < r < 1, r ≠ 0

x1 − r = 5 ⇒ x = 5(1 − r), with |r| < 1 for the infinite sum to exist.
-1 < r < 1 ⇒ 0 < 1 − r < 2 ⇒ 0 < x < 10.

Answer: (c) 94

Method. sum of an infinite GP

First term 3, common ratio -13.
Sum = 31 + 13 = 343 = 94.

Answer: (c) 4, 2, 1, 12, 12², …

Method. ar = 2 and a/⁠(1 − r) = 8

a = 8(1 − r), so 8r(1 − r) = 2 ⇒ 4r² - 4r + 1 = 0 ⇒ (2r − 1)² = 0 ⇒ r = 12.
a = 8 × 12 = 4: the GP is 4, 2, 1, 12, 12², …

Answer: (a) x = 11 + y

Method. sum of an infinite GP with ratio -y

First term 1, common ratio -y, |-y| < 1.
x = 11 − (-y) = 11 + y.

Answer: (c) 243

Method. terms equidistant from the middle of a GP multiply to the square of the middle term

Write the first five terms as ar², ar, a, ar, ar², with the third term a = 3.
Product = a⁵ = 3⁵ = 243.

Answer: (a) 6

Method. add the exponents: an infinite GP

Product = 612 + 14 + 18 + ….
Exponent = 121 − 12 = 1.
Product = 6¹ = 6.

Answer: (a) x = y1 + y

Method. sum of an infinite GP, then solve for x

y = x1 − x (first term x, ratio x, |x| < 1).
y − xy = x ⇒ y = x(1 + y) ⇒ x = y1 + y.

Answer: (c) HP

Method. clear denominators and reduce to b = 2aca + c

(b − c) + (b − a)(b − a)(b − c) = 2b.
b[2b − (a + c)] = 2[b² - b(a + c) + ac].
2b² - b(a + c) = 2b² - 2b(a + c) + 2ac, so b(a + c) = 2ac.
b = 2aca + c: a, b, c are in HP.

No correct option. This question is defective as printed in UPSC's paper.

Method. AP and GP conditions give p + q = 2, pq = 4

p, 1, q in AP: p + q = 2. p, 2, q in GP: pq = 4.
Then p and q are roots of t² - 2t + 4 = 0, whose discriminant is 4 − 16 < 0: no real p, q exist.
Formally, 2pqp + q = 82 = 4 and 1p + 1q = p + qpq = 12 = 2 · 14, so both I and II would hold.

Defective as printed: p + q = 2 and pq = 4 have no real solution (discriminant −⁠12), so no real p, q satisfy both conditions. Taken formally (or with complex p, q), both statements I and II hold, which is option (c).

Answer: (c) Both 1 and 2

Method. chain the AP, GP and HP conditions to get c² = ae

a, b, c in AP: b = a + c2. c, d, e in HP: d = 2cec + e.
b, c, d in GP: c² = bd = a + c2 · 2cec + e = (a + c)cec + e.
So c(c + e) = (a + c)e, i.e. c² + ce = ae + ce, giving c² = ae: a, c, e are in GP (1 correct).
Taking reciprocals, (1c)² = 1a · 1e, so 1a, 1c, 1e are in GP (2 correct).

Answer: (b) 1 and 2 only

Method. substitute b = 2aca + c for an HP

b − a = 2ac − a² - aca + c = a(c − a)a + c, and b − c = c(a − c)a + c.
1b − a + 1b − c = a + cc − a(1a − 1c) = a + cc − a · c − aac = 1a + 1c.
For an HP, 1a + 1c = 2b, so 1 and 2 are both correct.
3: 12(1a + 1b + 1c) = 12(2b + 1b) = 32b ≠ 2b, so 3 is wrong.

Answer: (c) a, b and c are in GP

Method. harmonic mean of the outer terms equals the middle term of an HP

Since (a + b), 2b, (b + c) are in HP, 2b = 2(a + b)(b + c)(a + b) + (b + c).
So b(a + 2b + c) = (a + b)(b + c).
ab + 2b² + bc = ab + ac + b² + bc.
b² = ac, so a, b, c are in GP.

Answer: (a) a, b, c are in AP

Method. Exponents of a GP of powers of 2 form an AP

2^1/⁠c, 2^b/⁠ac, 2^1/⁠a in GP: (2^b/⁠ac)² = 2^1/⁠c·2^1/⁠a.
Equate exponents: 2bac = 1c + 1a = a + cac.
So 2b = a + c: a, b, c are in AP.

Answer: (b) a, b, c are in GP

Method. HP condition: middle term = 2xyx + y

With x = a + b2, y = b + c2: b = 2xyx + y.
2xy = (a + b)(b + c)2, x + y = a + 2b + c2, so b = (a + b)(b + c)a + 2b + c.
ab + 2b² + bc = ab + ac + b² + bc.
b² = ac, so a, b, c are in GP.

Answer: (a) 130

Method. AM ≥ GM, with equality when x = y

x + y ≥ 2⁠√(xy)⁠ = 2⁠√4225⁠ = 2.65 = 130.
Equality needs x = y = 65, which are natural numbers (65² = 4225).
Minimum of x + y is 130.

Answer: (a) 1 only

Method. Reciprocals of an HP are in AP

1b + c, 1c + a, 1a + b in HP means b + c, c + a, a + b are in AP.
2(c + a) = (b + c) + (a + b) gives a + c = 2b, so a, b, c are in AP. Statement 1 is correct.
Statement 2 needs (c + a)⁴ = (b + c)²(a + b)². Write the AP as m − k, m, m + k: this needs m² = |m² - k²|, which fails in general.
Counter-example: a = 1, b = 2, c = 3 gives b + c = 5, c + a = 4, a + b = 3; 25, 16, 9 are not in GP (16² ≠ 225). Statement 2 is not correct.

Answer: (b) 2.4v km/⁠hr

Method. Equal distances at two speeds: average speed is the harmonic mean

Time for first 60 km = 603v = 20v h; time for next 60 km = 602v = 30v h.
Total time = 50v h for 120 km.
Average speed = 120 ÷ 50v = 2.4v km/⁠hr.

Shortcut. Equal distances: average speed = 2.3v·2v3v + 2v = 12v5 = 2.4v.

Answer: (c) 1 or ab or ac

Method. evaluate the ratio separately for AP, GP and HP

AP: a − b = b − c, so the ratio is 1.
GP: b = ar, c = ar²: a(1 − r)ar(1 − r) = 1r = ab.
HP: 1b − 1a = 1c − 1b ⇒ a − bab = b − cbc ⇒ a − bb − c = abbc = ac.
So the ratio is 1 or ab or ac.

Answer: (c) 9 : 1

Method. set a/⁠b = t² and solve the quadratic in t

a + b2⁠√(ab)⁠ = 53. Divide top and bottom by b and put a/⁠b = t² (t > 0): t² + 12t = 53.
3t² - 10t + 3 = 0 ⇒ (3t − 1)(t − 3) = 0 ⇒ t = 3 or 13.
a: b = t² = 9 : 1 (or 1 : 9); among the options only 9 : 1 appears.

No correct option. This question is defective as printed in UPSC's paper.

Method. AM and GM fix x + z and xz; the HM is 2xz/⁠(x + z)

AP: x + z = 2 × 32 = 3. GP: xz = 3² = 9.
Then x and z are the roots of t² - 3t + 9 = 0, whose discriminant is 9 − 36 = -27 < 0: no real x, z exist.
Formally, x, h, z are in HP when h = 2xzx + z = 2 × 93 = 6, which is option (a).

Defective as printed: x + z = 3 and xz = 9 have no real solution (discriminant −⁠27), so no real x, z make x, 32, z an AP and x, 3, z a GP. Read formally, the harmonic mean 2xz/⁠(x + z) = 6 gives x, 6, z in HP, option (a).

Answer: (c) |⁠√x₁⁠ - ⁠√x₂|⁠ > ⁠√2⁠

Method. AM − GM = 12(⁠√x₁⁠ - ⁠√x₂)²⁠

AM − GM = x₁ + x₂2 - ⁠√(x₁x₂)⁠ = (⁠√x₁⁠ - ⁠√x₂)²2.⁠
AM − GM > 1 ⇔ (⁠√x₁⁠ - ⁠√x₂)²⁠ > 2 ⇔ |⁠√x₁⁠ - ⁠√x₂|⁠ > ⁠√2.⁠

Answer: (d) 111111111

Method. sum of the first n odd numbers is n²

1 + 3 + 5 + … (n terms) = n².
n² = 12345678987654321 = 111111111² (the palindrome 1 2 3 … 9 … 3 2 1 is the square of nine 1s).
n = 111111111.

Answer: (a) 1 only

Method. AP sum for statement 1; a counter-example for statement 2

1. 2 + 4 + … + 2n = 2(1 + 2 + … + n) = 2·n(n + 1)2 = n² + n. Correct.
2. n = 40: 40² + 40 + 41 = 1681 = 41², not prime. Not correct.

Answer: (a) 10ⁿ p

Method. Repunit as a GP sum: p = 10ⁿ - 19

p = 1 + 10 + 10² + … + 10^{n − 1} = 10ⁿ - 19, so 9p + 1 = 10ⁿ.
9p² + p = p(9p + 1) = 10ⁿp.
Check n = 1: p = 1, 9 + 1 = 10 = 10¹·1.

Answer: (d) 11! - 1

Method. Telescoping: n(n!) = (n + 1)! - n!

n(n!) = (n + 1 − 1)n! = (n + 1)! - n!.
Sum for n = 1 to 10: (2! - 1!) + (3! - 2!) + … + (11! - 10!).
All middle terms cancel, leaving 11! - 1! = 11! - 1.

Answer: (a) 51

Method. pair the terms

(1 − 2) + (3 − 4) + … + (99 − 100) + 101: 50 pairs, each -1.
Value = -50 + 101 = 51.

Answer: (b) ⁠√2⁠ + 1

Method. the continued fraction repeats itself: x = 2 + 1/⁠x

Let x = 2 + 12 + …; then x = 2 + 1x, with x > 2.
x² - 2x − 1 = 0 ⇒ x = 1 ± ⁠√2.⁠
Take the positive root: x = ⁠√2⁠ + 1.

Answer: (d) n + 1, 3

Method. arithmetico-geometric series: S − 3S

S = 1.3 + 2.3² + … + n·3ⁿ; 3S = 1.3² + … + (n − 1)3ⁿ + n·3ⁿ⁺¹.
S − 3S = 3 + 3² + … + 3ⁿ - n·3ⁿ⁺¹ = 3(3ⁿ - 1)2 − n·3ⁿ⁺¹.
-2S = 3ⁿ⁺¹ - 3 − 2n·3ⁿ⁺¹2 ⇒ S = (2n − 1)3ⁿ⁺¹ + 34.
So a = n + 1 and b = 3.

Shortcut. Put n = 1: the left side is 3, so (3^a + b)/⁠4 = 3; with a = 2, b = 3 this gives 12/⁠4 = 3 and only (d) fits.

Answer: (a) 13[n − 19(1 − 110ⁿ)]

Method. write each term as 13(1 − 10^-k) and sum a GP

0.3 = 13(0.9), 0.33 = 13(0.99), …, so the k^th term is 13(1 − 110^k).
Sum = 13[n − (110 + 110² + … + 110ⁿ)].
The GP sums to 110(1 − 110ⁿ)1 − 110 = 19(1 − 110ⁿ).
Sum = 13[n − 19(1 − 110ⁿ)], option (a).

Shortcut. Put n = 1: the sum must be 0.3 = 13(1 − 110); only (a) gives this.

Answer: (c) 2^-n + n − 1

Method. write each term as 1 − 2^-k and sum a GP

The k^th term is 2^k - 12^k = 1 − 12^k.
Sum = n − (12 + 14 + … + 12ⁿ) = n − (1 − 12ⁿ).
= 2^-n + n − 1, option (c).

Shortcut. Put n = 1: sum = 12; only (c) gives 2^-1 + 1 − 1 = 12.