Mathematics MCQ on Sequence and Series for JEE and Engineering Exam 2022
MCQ on Sequence and Series: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.
In this post we are providing you MCQ on Sequence and Series, which will be beneficial for you in upcoming JEE and Engineering Exams.
MCQ on Sequence and Series
Q1. 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)] = ________.
a) -1
b) 1
c) 0
d) none
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 * {[b − c] / [D]} + b * {[c − a] / [D]} + c * {[a − b] / [D]}
= [1 / D] (ab − ac + bc − ab + ca −bc)
= 0
Q2. If the p^{th} term of an A.P. be q and q^{th} term is p, then its r^{th} term will be __________.
a) P – q- r
b) p+q-r
c) p+ q + r
d) p-q+r
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
Q3. The sum of integers from 1 to 100 that are divisible by 2 or 5 is _________.
a) 3045
b) 3050
c) 3055
d) 3040
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
Q4. The sum of the series [4 / 1!] + [11 / 2!] + [22 / 3!] + [37 / 4!] + [56 / 5!] + . . . is _____.
a) 6e − 5
b) 6e + 1
c) 6e − 1
d) none
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 }= 2n^{2} + n + 1
Thus, n^{th} term of the given series is T_{n }= [2n^{2} + 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
Q5. 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 _______.
a) 53/150
b) 50/150
c) 53/ 155
d) 50/155
We have Sn_{1} / Sn_{2} = [2n + 3] / [6n + 5]
{[n / 2] * [2a_{1} + (n − 1)d_{1}]} / {[n / 2] * [2a_{2} + (n − 1)d_{2}]}
= [2n + 3] / [6n + 5]
{2 * [a_{1} + ([n − 1] / 2)d1]} / {2 * [a_{2} + ([n − 1] / 2)d2]} = [2n + 3] / [6n + 5]
{a_{1} + ([n − 1] / 2)d_{1}} / {a_{2}+([n − 1] / 2)d_{2}} = [2n + 3] / [6n + 5]
Put n = 25 then,
[a_{1} + 12d_{1}] / [a_{2} + 12d_{2}]= [2 (25) + 3] / [6 (25) + 3]
T13_{1} / T13_{2 }= 53 / 155
Q6. In the expansion of log_{e} {1 / [1 − x − x^{2} + x^{3}]}, the coefficient of x is __________.
a) -1
b) 1
c) 0
d) none
log_{e} {1 / [1 − x − x^{2} + x^{3}]} = log_{e} {1 / [(1 − x) − x^{2} (1 − x)]}
= log_{e} {1 / [(1 − x) (1 − x^{2})]}
= 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) − (x^{2} / 2) − (x^{3} / 3) − (x^{4} / 4) − . . . . . . . ∞] − [x + (x^{2} / 2) + (x^{3} / 3) + (x^{4} / 4) − . . . . ∞],
Hence, coefficient of x = 2 − 1 = 1
Q7. 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?
a) 4
b) 2
c) 1
d) 0
x + y + z = 15, if x = (z^{−3})^{−1} = z^{3} 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
Q8. 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?
a) 14
b) 13
c) 12
d) 11
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
Q9. The interior angles of a polygon are in A.P. If the smallest angle be 120^{o} and the common difference be 5^{o}, then the number of sides is __________.
a) 5
b) 8
c) 9
d) 3
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
[n / 2] * [2 * 120 + (n − 1)5] = (n − 2) * 180
n^{2} − 25n + 144 = 0
(n − 9) (n − 16) = 0
n = 9, 16
But n = 16 gives
T_{16 }= a + 15d = 120^{o} + 15.5^{o }= 195^{o}, which is impossible as interior angle cannot be greater than 180^{o}.
Hence, n = 9.
Q10. If log_{3 }2, log_{3 }(2^{x} − 5) and log_{3 }(2^{x} − [7 / 2]) are in A.P., then x is equal to ______.
a) 1
b) 3
c) 2
d) 4
log_{3 }2, log_{3 }(2^{x} − 5) and log_{3 }(2^{x} − [7 / 2]) are in A.P.
⇒ 2 log_{3 }(2^{x} − 5) = log_{3 }[(2) (2^{x }− 72)]
⇒ (2^{x }− 5)^{2 }= 2^{x+1} − 7
⇒ [2^{2x }− 12] * [2^{x} − 32] = 0
⇒ x = 2, 3
But x = 2 does not hold, hence x = 3.
MCQ Link for NEET/JEE | |
JEE/NEET Physics MCQ | Click Here |
NEET/JEE Chemistry MCQ | Click Here |
NEET Biology MCQ | Click Here |
JEE Math’s MCQ | Click Here |
Notes PDF Link for NEET/JEE | |
Physics Notes PDF | Click Here |
Chemistry Notes PDF | Click Here |
Biology Notes PDF | Click Here |
Math’s Notes PDF | Click Here |
Follow on Facebook
By Team Learning Mantras