JEE Main Maths Sequence And Series Previous Year Questions With Solutions

A sequence is a list of objects which have been ordered in a sequential way. And a series can be sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence. Geometric Sequences, Arithmetic Sequences, Fibonacci Numbers and Harmonic Sequences are some the popular examples of sequences. This article covers sequence, infinite sequence, series, arithmetic sequence or progression, general term of A.P., arithmetic mean, geometric progression, general term of G.P., geometric mean, harmonic sequence or progression and general term of H.P. The sequence and series questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. About 1-2 questions are asked from this topic in JEE Examination.

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JEE Main Maths Sequence and Series Previous Year Questions With Solutions

Question 1: The sum of the series [4 / 1!] + [11 / 2!] + [22 / 3!] + [37 / 4!] + [56 / 5!] + . . . is _____.

Solution:

S = 4 + 11 + 22 + 37 + . . . . + T_n−1 + T_n or

S = 4 + 11 + 22 + 37 + . . + T_n−1 + T_n

Therefore, on subtracting we get 0 = 4 + [7 + 11 + 15 + 19 + . . . . + (T_n − T_n−1)] − T_n

0 = 4 + [n − 1] / [2] * [14 + (n − 2)4] − T_n

So, T_n = 2n² + n + 1

Thus, n^th term of the given series is T_n = [2n² + n + 1] / [(n)!]

= [2n / (n − 1)!] + [1 / (n − 1)!] + [1/ n!]

= [2 (n − 1 + 1)/ (n − 1)!] + [1 / (n − 1)!] + [1/ n!]

= [2 / (n − 2)!] + [3 / (n − 1)!] + (1 / n!)

Sum = ∑_n=1^∞ T_n = 2e + 3e + e − 1

= 6e − 1

Question 2: The sum of the series 1 + [1/ (4 * 2!)] + [1 / (16 * 4!)] + [1 / (64 * 6!)] + . . . . . is ________.

Solution:

Putting x = 1 / 2, we get 1 + [1/ (4 * 2!)] + [1 / (16 * 4!)] + [1 / (64 * 6!)] + . . . . . ∞

= [e^1/2 + e^−1/2] / [2]

= [e + 1] / 2√e

Question 3: In the expansion of log_e {1 / [1 − x − x² + x³]}, the coefficient of x is __________.

Solution:

log_e {1 / [1 − x − x² + x³]} = log_e {1 / [(1 − x) − x² (1 − x)]}

= log_e {1 / [(1 − x) (1 − x²)]}

= log_e {[1 / (1 − x)² (1 + x)]}

= log_e {(1 − x)⁻² (1 + x)⁻¹} ……… (i)

= log_e (1 − x)⁻² + log_e (1 + x)⁻¹

= −2 log_e (1 − x) − log_e (1 + x)

= −2 [(−x) − (x² / 2) − (x³ / 3) − (x⁴ / 4) − . . . . . . . ∞] − [x + (x² / 2) + (x³ / 3) + (x⁴ / 4) − . . . . ∞],

Hence, coefficient of x = 2 − 1 = 1

Question 4: If the p^th term of an A.P. be q and q^th term is p, then its r^th term will be __________.

Solution:

Given that, T_p = a + (p − 1)d = q …….(i) and

T_q = a + (q − 1)d = p ……. (ii)

From (i) and (ii), we get d = [−(p − q)] / [(p − q)] = −1

Putting value of d in equation (i), then a = p + q − 1

Now, r^th term is given by A.P. T_r = a + (r − 1)d

= (p + q − 1) + (r − 1) (−1)

= p + q − r

Question 5: The sum of integers from 1 to 100 that are divisible by 2 or 5 is _________.

Solution:

The sum of integers from 1 to 100 that are divisible by 2 or 5

= sum of series divisible by 2 + sum of series divisible by 5 – sum of series divisible by 2 and 5.

= (2 + 4 + 6 + . . . . . . + 100) + (5 + 10 + 15 . . . . . . . + 100) − (10 + 20 + 30 + . . . . . . . . + 100)

= [50 / 2] {2 * 2 + (50 − 1) 2} + [20 / 2] {2 * 5 + (20 − 1)5} − [10 / 2] {10 * 2 + (10 − 1)10}

= 2550 + 1050 − 550

= 3050

Question 6: If m^th terms of the series 63 + 65 + 67 + 69 + . . . . . . . . . and 3 + 10 + 17 + 24 + . . . . . . be equal, then what is the value of m?

Solution:

Given series 63 + 65 + 67 + 69 + . . . . . . . . . (i) and 3 + 10 + 17 + 24 + . . . . . . (ii)

Now from (i), m^th term = (2m + 61) and m^th term of (ii) series = (7m − 4)

Under condition, ⇒ 7m − 4 = 2m + 61

⇒ 5m = 65

⇒ m = 13

Question 7: If log₃ 2, log₃ (2^x − 5) and log₃ (2^x − [7 / 2]) are in A.P., then x is equal to ______.

Solution:

log₃ 2, log₃ (2^x − 5) and log₃ (2^x − [7 / 2]) are in A.P.

⇒ 2 log₃ (2^x − 5) = log₃ [(2) (2^x − 7/2)]

⇒ (2^x − 5)² = 2^x+1 − 7

Let 2^x = t,

⇒ (t – 5)² = 2t – 7

⇒ t² – 12t + 32 = 0

⇒ t = 4, 8

⇒ 2^x = 4, 2^x = 8

⇒ x = 2, 3

But x = 2 does not hold, hence x = 3.

Question 8: If the p^th, q^th and r^th term of an arithmetic sequence are a, b and c respectively, then the value of [a (q − r) + b (r − p) + c (p − q)] = ________.

Solution:

Suppose that the first term and the common difference of A.P.s are A and D, respectively.

Now,

p^th term = A + (p − 1)D = a …… (i)

q^th term = A + (q − 1)D = b …… (ii) and

r^th term = A + (r − 1)D = c .….. (iii)

So, [a (q − r) + b (r − p) + c (p − q)]

= (A − D)[q − r + r − p + p − q] + D[p(q − r) + q(r − p) + r(p − q)]

= (A – D) (0) + D [pq – pr + qr – pq + rp – rq]

= 0

Question 9: The sums of n terms of two arithmetic series are in the ratio [2n + 3] : [6n + 5], then the ratio of their 13^th terms is _______.

Solution:

We have Sn₁ / Sn₂ = [2n + 3] / [6n + 5]

{[n / 2] * [2a₁ + (n − 1)d₁]} / {[n / 2] * [2a₂ + (n − 1)d₂]} = [2n + 3] / [6n + 5]

{2 * [a₁ + ([n − 1] / 2)d1]} / {2 * [a₂ + ([n − 1] / 2)d2]} = [2n + 3] / [6n + 5]

{a₁ + ([n − 1] / 2)d₁} / {a₂+([n − 1] / 2)d₂} = [2n + 3] / [6n + 5]

Put n = 25 then,

T13₁ / T13₂ = 53 / 155

Question 10: The interior angles of a polygon are in A.P. If the smallest angle is 120^o and the common difference is 5^o, then the number of sides is __________.

Solution:

Let the number of sides of the polygon be n.

Then the sum of interior angles of the polygon = (2n − 4) [π / 2] = (n − 2)π

Since the angles are in A.P. and a = 120^o, d = 5, therefore

240n + 5(n² – n) = (n – 2) * 360

5(48n + n² – n) = (n – 2) * 360

n² + 47n – 72n + 144 = 0

n² − 25n + 144 = 0

(n − 9) (n − 16) = 0

n = 9, 16

But n = 16 gives

T₁₆ = a + 15d = 120^o + 15.5^o = 195^o, which is impossible as interior angle cannot be greater than 180^o.

Hence, n = 9.

Question 11: The sum of n terms of the series 1 + (1 + x) + (1 + x + x²) + . . . . . will be ___________.

Solution:

1 + (1 + x) + (1 + x + x²) + . . . . .+ (1 + x + x² + x³ + . . . + xⁿ⁻¹) + . . .

= [(x – 1)/(x – 1)] + [(x² – 1)/(x – 1)] + [(x³ – 1)/(x – 1) + …… n terms

= [1 / (1 − x)] * {(1 − x) + (1 − x²) + (1 − x³) + (1 − x⁴) + ……….upto n terms}

= [1 / (1 − x)] * [n − {x + x² + x³ + . . . . . . . . . . upto n terms } ]

= [1 / (1 − x)] * [n − {x (1 − xⁿ) / [1 − x]}]

= [n (1 − x) − x (1 − xⁿ)] / [(1 − x)²]

Question 12: If x, 1, z are in A.P. and x, 2, z are in G.P., then x, 4, z will be in __________.

Solution:

x, 1, z are in A.P., then

2 = x + z ……(i) and

4 = xz ……(ii)

Divide (ii) by (i), we get

2xz / [x + z] = 4

Hence, x, 4, and z will be in H.P.

Question 13: Let x + y + z = 15 if 9, x, y, z, a are in A.P.; while [1 / x] + [1 / y] + [1 / z] = 5 / 3 if 9, x, y, z, a are in H.P., then what will be the value of a?

Solution:

x + y + z = 15

9, x, y, z, a are in A.P.

Sum =9 + 15 + a = 24 + a

⇒ 24 + a = (5/2)(9 + a)

⇒ 48 + 2a = 45 + 5a

⇒ 3a = 3

⇒ a = 1 ..(i) and

Sum = (1/9) + (5/3) + (1/a)

= (5/2) [ (1/9) + (1/a)]

⇒ a = 1

Also, Read:

JEE‌ ‌Advanced‌ ‌Maths‌ ‌Sequences‌ ‌and‌ ‌Series‌ ‌ Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌

Sequence and Series – JEE Main and Advanced Questions