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.

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 + . . . . + Tn−1 + Tn or

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

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

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

So, Tn = 2n2 + n + 1

Thus, nth term of the given series is Tn = [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=1Tn = 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:

[ex + e−x] / [2] = 1 + [x2 / 2!] + [x4 / 4!] + [x6 / 6!] + . . . . ∞

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

= [e1/2 + e−1/2] / [2]

= [e + 1] / 2√e

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

Solution:

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

= loge {1 / [(1 − x) (1 − x2)]}

= loge {[1 / (1 − x)2 (1 + x)]}

= loge {(1 − x)−2 (1 + x)−1} ……… (i)

= loge (1 − x)−2 + loge (1 + x)−1

= −2 loge (1 − x) − loge (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 pth term of an A.P. be q and qth term is p, then its rth term will be __________.

Solution:

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

Tq = 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, rth term is given by A.P. Tr = 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 mth 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), mth term = (2m + 61) and mth term of (ii) series = (7m − 4)

Under condition, ⇒ 7m − 4 = 2m + 61

⇒ 5m = 65

⇒ m = 13

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

Solution:

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

⇒ 2 log3 (2x − 5) = log3 [(2) (2x − 72)]

⇒ (2x − 5)2 = 2x+1 − 7

⇒ [22x − 12] * [2x − 32] = 0

⇒ x = 2, 3

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

Question 8: If the pth, qth and rth 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,

pth term = A + (p − 1)D = a …… (i)

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

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

So, [a (q − r) + b (r − p) + c (p − q)] = a * {[b − c] / [D]} + b * {[c − a] / [D]} + c * {[a − b] / [D]}

= [1 / D] (ab − ac + bc − ab + ca −bc)

= 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 13th 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,

[a1 + 12d1] / [a2 + 12d2]= [2 (25) + 3] / [6 (25) + 3]

T131 / T132 = 53 / 155

Question 10: The interior angles of a polygon are in A.P. If the smallest angle be 120o and the common difference be 5o, 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 = 120o, d = 5, therefore

[n / 2] * [2 * 120 + (n − 1)5] = (n − 2) * 180

n2 − 25n + 144 = 0

(n − 9) (n − 16) = 0

n = 9, 16

But n = 16 gives

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

Hence, n = 9.

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

Solution:

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

Required sum = [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

[x z] / [x + z] = 4 / 2 or

2xz / [x + z] = 4

Hence, x, 4, 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, if x = (z−3)−1 = z3 are in A.P.

Sum =9+15+a=52(9+a)

⇒ 24 + a = 5 / 2 (9 + a)

⇒ 48 + 2a = 45 + 5a

⇒ 3a = 3

⇒ a = 1 ..(i) and

[1 / x] + [1 / y] + [1 / z] = 5 / 3, if 9, x, y, z, a are in H.P.

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

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

⇒ a = 1

 

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