Maths Quiz on Sequences and Series for JEE Mains/Advance

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Maths Quiz on Sequences and Series
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JEE Maths Quiz on Sequences and Series

JEE Maths Quiz on Sequences and Series : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Sequences and Series

Quiz on Sequences and Series

 

Q1. 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) 1
b) 2
c) 4
d) 9

View Answer

a) 1

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

Q2. If x, 1, z are in A.P. and x, 2, z are in G.P., then x, 4, z will be in __________.
a) G.P.
b) H.P.
c) Both
d) None

View Answer

b) H.P.

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.

Q3. 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 __________.
a) 9
b) 25
c) 16
d) 90

View Answer

a) 9

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 + 15.5= 195o, which is impossible as interior angle cannot be greater than 180o.

Hence, n = 9.

Q4. 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 _______.
a) 55 / 155
b) 27 / 112
c) 53 / 147
d) 57 / 143

View Answer

a) 55 / 155

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 / T13= 53 / 155

Q5. 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)] = ________.
a) 0
b) 1
c) 2
d) 3

View Answer

a) 0

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

Q6. If log2, log(2x − 5) and log(2x − [7 / 2]) are in A.P., then x is equal to ______.
a) 2
b) 3
c) 5
d) 7

View Answer

b) 3

Solution:

log2, log(2x − 5) and log(2x − [7 / 2]) are in A.P.

⇒ 2 log(2x − 5) = log[(2) (2− 72)]

⇒ (2− 5)= 2x+1 − 7

⇒ [22x − 12] * [2x − 32] = 0

⇒ x = 2, 3

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

Q7. If mth terms of the series 63 + 65 + 67 + 69 + . . . . . . . . . and 3 + 10 + 17 + 24 + . . . . . . be equal, then what is the value of m?
a) 17
b) 15
c) 13
d) 11

View Answer

c) 13

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

Q8. The sum of integers from 1 to 100 that are divisible by 2 or 5 is _________.
a) 3050
b) 3020
c) 3200
d) 3030

View Answer

a) 3050

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

Q9. In the expansion of loge {1 / [1 − x − x2 + x3]}, the coefficient of x is __________.
a) 2
b) 1
c) -2
d) -1

View Answer

b) 1

Solution:

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

= loge {1 / [(1 − x) (1 − x2)]}

= loge {[1 / (1 − x)(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

Q10. The sum of the series [4 / 1!] + [11 / 2!] + [22 / 3!] + [37 / 4!] + [56 / 5!] + . . . is _____.
a) – 6e − 1
b) 6e − 6
c) 6e + 1
d) 6e − 1

View Answer

d) 6e − 1

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 + . . . . + (T− Tn−1)] − Tn

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

So, T= 2n2 + n + 1

Thus, nth term of the given series is T= [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= 2e + 3e + e − 1

= 6e − 1

 



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