JEE‌ ‌Advanced‌ ‌Maths‌ ‌Sequences‌ ‌and‌ ‌Series‌ ‌ Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌

Students‌ ‌can‌ ‌find‌ ‌JEE‌ ‌Advanced‌ ‌previous‌ ‌year‌ ‌questions‌ ‌and‌ ‌solutions‌ ‌on‌ ‌sequences‌ ‌and‌ ‌series‌ ‌on‌ ‌this‌ ‌page.‌ ‌A‌ ‌sequence‌ ‌is‌ ‌a‌ ‌group‌ ‌of‌ ‌numbers‌ ‌in‌ ‌an‌ ‌ordered‌ ‌way‌ ‌following‌ ‌a‌ ‌pattern,‌ ‌whereas‌ ‌a‌ ‌series‌ ‌is‌ ‌the‌ ‌sum‌ ‌of‌ ‌all‌ ‌elements‌ ‌of‌ ‌a‌ ‌sequence.‌ ‌The‌ ‌important‌ ‌topics‌ ‌include‌ ‌arithmetic‌ ‌progression,‌ ‌geometric‌ ‌progression,‌ ‌harmonic‌ ‌progression,‌ ‌the‌ ‌relation‌ ‌between‌ ‌A.M,‌ ‌G.M,‌ ‌and‌ ‌H.M,‌ ‌etc.‌ ‌Students‌ ‌can‌ ‌expect‌ ‌questions‌ ‌from‌ ‌sequences‌ ‌and‌ ‌series‌ ‌for‌ ‌the‌ ‌JEE‌ ‌Advanced‌ ‌exam.‌ ‌Practising‌ ‌these‌ ‌solutions‌ ‌will‌ ‌definitely‌ ‌help‌ ‌students‌ ‌to‌ ‌improve‌ ‌their‌ ‌speed‌ ‌and‌ ‌accuracy.‌ ‌

Download ‌Sequences‌ ‌and‌ ‌Series‌ Previous Year Solved Questions PDF

Question 1: Let a₁, a₂, a₃,…be in harmonic progression with a₁ = 5 and a₂₀ = 25. The least positive integer n for which a_n < 0 is

(a) 22

(b) 23

(c) 24

(d) 25

Solution:

Given a₁, a₂, a₃,…are in HP.

Then 1/a₁, 1/a₂, 1/a₃ are in AP.

Take d as the common difference,

Given a₁ = 5 and a₂₀ = 25

1/a₂₀ = 1/a₁ + 19d

1/a₂₀ – 1/a₁ = 19d

1/25 – 1/5 = 19d

-4/25 = 19d

d = (-4/25) × 19

a + (n – 1)d < 0

⇒ (1/5) + (n. -1)(-4/25 × 19) < 0

4(n – 1)/19 × 5 > 1

n – 1 > 19 × 5/4

n > 19 × 5/4 + 1

n = 25

So the least positive value of n is 25.

Hence option d is the answer.

Question 2: Let a, b, and c be positive integers such that b/a is an integer. If a, b, c are in geometric progression and the arithmetic mean of a, b, c is b + 2, then the value of (a² + a – 14)/(a + 1) is

Solution:

Given that a, b, c are in geometric progression.

Let a, b, c be a, ar, ar² (where r is the common ratio)

Given (a + b + c)/3 = b + 2

⇒ (a + ar + ar²) = 3(ar) + 6 (since b = ar)

⇒ ar² – 2ar + a = 6

⇒ a(r² – 2r + 1) = 6

⇒ a(r – 1)² = 6

⇒ (r – 1)² = 6/a

6/a should be a perfect square.

So the possible value for a is 6.

(a² + a – 14)/(a+1) = (36 + 6 – 14)/7

= 28/7

= 4

Hence the value of (a² + a – 14)/(a + 1) is 4.

Question 3: A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller to the numbers on the removed card is k, then k-20 equals

(a) 5

(b) 10

(c) 15

(d) 50

Solution:

Sum of n natural numbers = n(n + 1)/2

n(n + 1)/2 – (k + k + 1) = 1224

n(n + 1)/2 – 2k = 1225

n² + n – 2450 = 4k

⇒ k = (n + 50)(n – 49)/4

The smaller the numbers on the removed card is k.

So 1 ≤ k ≤ n-1

⇒ 1 ≤ (n + 50)(n – 49)/4 ≤ n – 1

49 < n < 51

⇒ n = 50

⇒ k = 100/4 = 25

k – 20 = 25 – 20 = 5

Hence option a is the answer.

Question 4: The sides of the right-angled triangle are in AP. If the triangle has an area 24, then what is the length of its smallest side?

Solution:

Let a – d, a, a + d be the sides of the right angled triangle.

Using Pythagoras theorem, we can write

(a + d)² = a² + (a – d)²

a² + 2ad + d² = a² + a²– 2ad + d²

a² – 4ad = 0

a(a – 4d) = 0

⇒ a = 4d or a = 0 (rejected)

⇒ a = 4d ….(i)

Area of triangle = ½ bh

24 = ½ (a – d)a

48 = 3d × 4d

48/12 = d²

⇒ d² = 4

⇒ d = 2

So a = 8

Length of smallest side = a – d

= 8 – 2

= 6

Hence the length of the smallest side is 6 units.

Question 5: If the sum of the first n terms of an AP is cn², then the sum of squares of these n terms is

(a) n(4n² – 1)c²/6

(b) n(4n² + 1)c²/3

(c) n(4n² – 1)c²/3

(d) n(4n² + 1)c²/6

Solution:

Let the sum of the first n terms of an AP be S_n.

Given S_n = cn²

S_n-1 = c (n – 1)²

T_n = S_n – S_n-1

= cn²– c(n – 1)²

= cn² – cn² + 2cn – c

= c(2n – 1)

Sum of squares = ∑Tn²

= c²[4 ∑n² +∑1 – 4∑n]

= [4c²(n(n + 1)(2n + 1)/6] + nc²– [2c²n(n + 1)]

= nc²(4n² + 6n + 2 + 3 – 6n – 6)/3

= nc²(4n² – 1)/3

= n(4n² – 1)c²/3

Hence option (c) is the answer.

Question 6: The third term of a geometric progression is 4. The product of the first five terms is

(a) 4³

(b) 4⁴

(c) 4⁵

(d) 4

Solution:

Let a be the first term and r be the common ratio.

Third term = ar² = 4

Product of first five terms = a(ar)(ar²)(ar³)(ar⁴)

= a⁵r¹⁰

= (ar²)⁵

= 4⁵

Hence option c is the answer.

Question 7: Let X be the set consisting of the first 2018 terms of an arithmetic progression, 1, 6, 11…., and Y be the set consisting of the first 2018 terms of arithmetic progression 9, 16, 23, …. Then, the number of elements in the set (X ⋃ Y) is

Solution:

X = {1, 6, 11, …}

Here a = 1, d = 5

We know t_n = a + (n – 1)d

2018^th term = 1 + 2017 × 5

= 10086

Y = { 9, 16, 23, ..}

Here a = 9, d = 7

2018^th term = 9 + 2017 × 7

= 14128

X⋂Y = {16, 51, 86, …}

10086 ≥ 16 + (n – 1)35

=> (n – 1)35 ≤ 10070

=> n ≤ 288.71

=> n = 288

The number of elements in the set X ⋃ Y is n(X ⋃ Y) = n(X) + n(Y) – n(X⋂Y)

= 2018 + 2018 – 288

= 3748

The number of elements in the set (X ⋃ Y) is 3748.

Question 8: The minimum value of the sum of real numbers a^-5, a^-4, 3a^-3, 1, a⁸ and a¹⁰ with a > 0 is

(a) 9

(b) 8

(c) 2

(d) 1

Solution:

Since a>0, a^-5, a^-4, 3a^-3, 1, a⁸ and a¹⁰ are positive.

So AM ≥ GM

⇒ (1/8) [a^-5+ a^-4+ a^-3 + a^-3 + a^-3 + 1 + a⁸ + a¹⁰] ≥ [a^-5 × a^-4 × a^-3 × a^-3 × a^-3 ×1 × a⁸ × a¹⁰]^1/8

⇒ (1/8) [a^-5+ a^-4+ a^-3 + a^-3 + a^-3 + 1 + a⁸ + a¹⁰] ≥ 1^1/8

The minimum value of the sum is 8.

Hence option b is the answer.

Question 9: Let S_n = ∑_k=1⁴ⁿ (-1)^k(k+1)/2 k². Then S_n can take values

(a) 1056

(b) 1088

(c) 1120

(d) 1332

Solution:

S_n = ∑_k=1⁴ⁿ (-1)^k(k+1)/2 k² = -1²– 2² + 3² + 4²– 5²– 6²+ 7² + 8² -….+(4n)²

= (-1² + 3²– 5² + 7² -…+ (4n-1)²)+ (-2² + 4² – 6²+ 8² -..+(4n²)

= 2(4 + 12 + 20+…(8n-4)) + 2(6 + 14 + 22+…(8n-2)) (Sum of n terms in A.P)

= n(8n) + n(8n+4) … (Use equation for sum of terms of A.P)

= n(16n+4)

Put n = 8, S_n = 1056.

Put n = 9, S_n = 1332

Hence option (a) and (d) are correct.

Question 10: If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M is equal to 2nd G.M, then m is equal to

Solution:

Let the AP be 3, a₁, a₂ …a_m, 243.

Common difference, d = (243 – 3)/(m + 1)

= 240/(m+1)

Let 3, G₁, G₂, G₃, and 243 be the GP.

Let r be the common ratio.

3r⁴ = 243

⇒ r⁴ = 243/3 = 81

⇒ r = 3

4th A.M is equal to 2nd G.M

G₂ = a₄

ar² = a + 4d

⇒ 27 = 3 + 4 × 240/(m + 1)

⇒ 24 = 960/(m + 1)

⇒ 960/24 = 1/(m + 1)

⇒ 1/40 = 1/(m + 1)

⇒ m + 1 = 40

So m = 39

Hence the value of m is 39.

Question 11: Let b_i > 1 for i = 1,2, …101. Suppose log_e b₁, log_e b₂, …log_e b₁₀ are in AP with the common difference log 2. Suppose a₁, a₂, ..a₁₀₁ are in AP such that a₁ = b₁ and a₅₁ = b₅₁. If t = b₁ + b₂ + …+ b₅₁ and s = a₁ + a₂ + …+ a₅₁ then

(a) s>t and a₁₀₁>b₁₀₁

(b) s>t and a₁₀₁<b₁₀₁

(c) s<t and a₁₀₁>b₁₀₁

(d) s<t and a₁₀₁<b₁₀₁

Solution:

Given log_e b₁, log_e b₂, …log_e b₁₀ are in AP.

log_e b₂ – log_e b₁ = log_e 2

log_e (b₂/b₁) = log_e 2

⇒ (b₂/b₁) = 2

Similarly b₃/b₂ = 2

b₁, b₂, …b₁₀₁ are in GP with common ratio r = 2.

b_n = 2^n-1b₁ …(i)

Given a₁, a₂, ..a₁₀₁ are in AP.

t = b₁ + b₂ + …+ b₅₁

s = a₁ + a₂ + …+ a₅₁

t = sum of 51 terms of G.P

= b₁(r⁵¹-1)/(r-1)

= b₁(2⁵¹-1)

s = sum of 51 terms of A.P

= (n/2)(first term+ last term)

= (51/2)(a₁ + a₅₁)

= (51/2)(b₁ + b₅₁)

= (51/2)(b₁ + 2⁵⁰b₁)

= b₁(51/2)(1 + 2⁵⁰)

It is clear that s> t.

Given a₅₁ = b₅₁amd a₁ = b₁.

a₅₁ = a₁ +50d

b₅₁ = b₁ + 50d

⇒ d = (b₅₁ – b₁)/50

a₁₀₁ = a₁+100d

= b₁+ 100 (b₅₁ – b₁)/50

= b₁+ 2(b₅₁ – b₁)

= 2b₅₁ – b₁

= 2 × 2⁵⁰b₁ – b₁ (from (i))

= b₁(2⁵¹ -1)

b₁₀₁ = 2¹⁰⁰b₁

It is clear that 2¹⁰⁰b₁ > b₁(2⁵¹ -1)

⇒ b₁₀₁ > a₁₀₁

Hence option b is the answer.

Question 12: The harmonic mean and geometric mean of two positive numbers be the ratio 4:5. Then the two numbers are in the ratio

(a) 1 : 4

(b) 4 : 1

(c) 3 : 4

(d) 4 : 3

Solution:

Let a and b be the two positive numbers.

HM = 2ab/(a + b)

GM = √(ab)

HM/GM = 4/5

2ab/(a + b) ÷ √(ab) = 4/5

⇒ 2√(ab)/(a + b) = 4/5

√ab/(a + b) = 2/5

Cross multiply

5√ab = 2(a + b)

Squaring both sides

25ab = 4(a + b)²

25ab = 4(a² + 2ab + b²)

4a² – 17ab + 4b² = 0

(a – 4b) (4a – b) = 0

⇒ (a – 4b) = 0 or (4a – b) = 0

=> a/b = 4/1 or a/b = ¼

Hence option a and option b are correct.

Question 13: If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + …. is (102)m, then m is equal to

(a) 20

(b) 5

(c) 10

(d) 25

Solution:

Given 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + …. = 102m

⇒ (3 + 4) + (8 + 9) + (13 + 14) + (18 + 19) + …. = 102m

⇒ 7 + 17 + 27 + …20 terms = 102m

Here a = 7, d = 10, n = 20

Sum of n terms of AP = (n/2)[2a + (n – 1)d]

= 10(14 + 190)

= 2040

= 102 m

So m = 2040/102

= 20

Hence option a is the answer.

Question 14: The number of terms in an AP is even; the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by

(a) 4

(b) 8

(c) 12

(d) 16

Solution:

Let 2n be the number of terms in the AP.

Sum of odd terms = (a₁ + a₃ + …a_2n-1) = 24 …(i)

Sum of even terms = (a₂ + a₄ + …a_2n) = 30…(ii)

a_2n = a₁ + (2n-1)d

a_2n – a₁ = 2nd – d

21/2 = 2nd – d… (iii)

(ii) – (i)

a₂ – a₁ + a₄ – a₃ + …a_2n – a_2n-1 = 30 – 24

d + d + …. n times = 6

nd = 6 …(iv)

Substitute (iv) in (iii)

21/2 = 2 × 6 – d

12 – d = 21/2

⇒ d = (24 – 21)/2

= 3/2

Put d in (iv)

3n/2 = 6

⇒ n = 12/3 = 4

So 2n = 8

Hence option b is the answer.

Question 15: The value of 1² + 3² + 5² + ….+ 25² is

(a) 2925

(b) 1469

(c) 1728

(d) 1456

Solution:

1² + 3² + 5² + ….+ 25² = (1² + 2² + 3² + ….+ 25²) – (2² + 4² + ….+ 24²)

= (1² + 2² + 3² + ….+ 25²) – 2²(1² + 2² + ….+ 12²)

= (25 × 26 × 51/6) – (4 × 12 × 13 × 25/6)

= 5525 – 2600

= 2925

Hence option a is the answer.

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