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 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)] = ________. 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
a) -1
b) 1
c) 0
d) none
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Q2. If the pth term of an A.P. be q and qth term is p, then its rth term will be __________. 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
a) P – q- r
b) p+q-r
c) p+ q + r
d) p-q+r
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Q3. The sum of integers from 1 to 100 that are divisible by 2 or 5 is _________. 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
a) 3045
b) 3050
c) 3055
d) 3040
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Q4. The sum of the series [4 / 1!] + [11 / 2!] + [22 / 3!] + [37 / 4!] + [56 / 5!] + . . . is _____. 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=1∞ Tn = 2e + 3e + e − 1 = 6e − 1
a) 6e − 5
b) 6e + 1
c) 6e − 1
d) none
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Q5. 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 _______. 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
a) 53/150
b) 50/150
c) 53/ 155
d) 50/155
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Q6. In the expansion of loge {1 / [1 − x − x2 + x3]}, the coefficient of x is __________. 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
a) -1
b) 1
c) 0
d) none
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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? 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
a) 4
b) 2
c) 1
d) 0
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Q8. If mth terms of the series 63 + 65 + 67 + 69 + . . . . . . . . . and 3 + 10 + 17 + 24 + . . . . . . be equal, then what is the value of m? 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
a) 14
b) 13
c) 12
d) 11
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Q9. 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 __________. 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.
a) 5
b) 8
c) 9
d) 3
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Q10. If log3 2, log3 (2x − 5) and log3 (2x − [7 / 2]) are in A.P., then x is equal to ______. 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.
a) 1
b) 3
c) 2
d) 4
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