All Exams >
Bank Exams >
RBI Assistant Preparation Course >
All Questions
All questions of Sequence & Series for Bank Exams Exam
A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :- a)310 /2
- b)310 - 210
- c)243 *(35 -1)
- d)310 -25
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :
a)
310 /2
b)
310 - 210
c)
243 *(35 -1)
d)
310 -25
e)
None of these
|
Pallabi Deshpande answered |
Total number of bacteria after 10 seconds,
= 3^10 - 3^5
= 3^5 *(3^5 -1)
= 243 *(3^5 -1)
Since, just after 10 seconds all the bacterias (i.e. 3^5 ) are dead after living 5 seconds each.
After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.- a)540 m
- b)960 m
- c)1080 m
- d)1020 m
- e)1120 m
Correct answer is option 'C'. Can you explain this answer?
After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.
a)
540 m
b)
960 m
c)
1080 m
d)
1020 m
e)
1120 m
|
Vikas Choudhury answered |
The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 metres. This process will continue in the form of infinite GP with common ratio 0.8 and first term 96. The required answer is given by the formula:
a/(1-r)
Now,
[{120/(1/5)}+{96/(1/5)}]
= 1080 m.
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.- a)5
- b)6
- c)4
- d)3
- e)7
Correct answer is 'C'. Can you explain this answer?
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
a)
5
b)
6
c)
4
d)
3
e)
7
|
Rishabh Rajawat answered |
39=a+7d, 59=a+11d, on solving. 4d=20,so d=5,so 59=a+(11×5), 59=a+55,so a=4.
Find the 15th term of an arithmetic progression whose first term is 2 and the common difference is 3
- a)45
- b)38
- c)44
- d)40
Correct answer is option 'C'. Can you explain this answer?
Find the 15th term of an arithmetic progression whose first term is 2 and the common difference is 3
a)
45
b)
38
c)
44
d)
40
|
Aisha Gupta answered |
Method to Solve :
A ( first term ) :- 2
d ( common difference ) :- 3
n = 15
To find nth term we have formula as
an = a + ( n - 1 )d
a15 = 2 + 14 � 3
a15 = 2 + 42
a15 = 44
d ( common difference ) :- 3
n = 15
To find nth term we have formula as
an = a + ( n - 1 )d
a15 = 2 + 14 � 3
a15 = 2 + 42
a15 = 44
How many terms are there in 20, 25, 30......... 140- a)22
- b)25
- c)23
- d)24
Correct answer is option 'B'. Can you explain this answer?
How many terms are there in 20, 25, 30......... 140
a)
22
b)
25
c)
23
d)
24
|
|
Anaya Patel answered |
Number of terms = { (1st term - last term) / common difference} + 1
= {(140-20) / 5} + 1
⇒ (120/5) + 1
⇒ 24 + 1 = 25.
= {(140-20) / 5} + 1
⇒ (120/5) + 1
⇒ 24 + 1 = 25.
Find the 15th term of the sequence 20, 15, 10....- a)-45
- b)-55
- c)-50
- d)0
Correct answer is option 'C'. Can you explain this answer?
Find the 15th term of the sequence 20, 15, 10....
a)
-45
b)
-55
c)
-50
d)
0
|
|
Santosh Jaiswal answered |
Since above sequence is in A.P with common difference of -5 and first term 20.
Then applying formula of AP we get 15 term as
20 + (n-1) d.
15 term is 20 + (15-1) -5 i.e. -50
Then applying formula of AP we get 15 term as
20 + (n-1) d.
15 term is 20 + (15-1) -5 i.e. -50
A boy agrees to work at the rate of one rupee on the first day, two rupees on the second day, and four rupees on third day and so on. How much will the boy get if he started working on the 1st of February and finishes on the 20th of February?a) 2^20b) 2^20-1c) 2^19-1d) 2^19e) None of theseCorrect answer is option 'B'. Can you explain this answer?
|
Yash Soni answered |
Can't we find with t
How many terms are there in the GP 5, 20, 80, 320........... 20480?- a)5
- b)6
- c)8
- d)9
- e)7
Correct answer is option 'E'. Can you explain this answer?
How many terms are there in the GP 5, 20, 80, 320........... 20480?
a)
5
b)
6
c)
8
d)
9
e)
7
|
Sameer Rane answered |
Common ratio, r = 20/5 = 4;
Last term or nth term of GP = ar^[n-1].
20480 = 5*[4^(n-1)];
Or, 4^(n-1) = 20480/5 = 4^8;
So, comparing the power,
Thus, n-1 = 8;
Or, n = 7;
Number of terms = 7.
The sum of the first 100 numbers, 1 to 100 is divisible by- a)1,2,4,8
- b)2 and 4
- c)2
- d)none
Correct answer is option 'C'. Can you explain this answer?
The sum of the first 100 numbers, 1 to 100 is divisible by
a)
1,2,4,8
b)
2 and 4
c)
2
d)
none
|
Reddy Mounika answered |
N(n+1)/2 =100(100+1)/2
100(101)/2
= 50(101)
= 101 is a odd number
= 50 is divided by 2 so
answer is 2
100(101)/2
= 50(101)
= 101 is a odd number
= 50 is divided by 2 so
answer is 2
What is the sum of the first 15 terms of an A.P whose 11 th and 7 th terms are 5.25 and 3.25 respectively- a)56.25
- b)60
- c)52.5
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
What is the sum of the first 15 terms of an A.P whose 11 th and 7 th terms are 5.25 and 3.25 respectively
a)
56.25
b)
60
c)
52.5
d)
none of these
|
Ishani Rane answered |
a +10d = 5.25, a+6d = 3.25, 4d = 2, d = 1/4
a +5 = 5.25, a = 0.25 = 1/4, s 15 = 15/2 ( 2 * 1/4 + 14 * 1/4 )
= 15/2 (1/2 +14/2 ) = 15/2 *15/2 =225/ 4 = 56.25
In an arithmetic series consisting of 51 terms, the sum of the first three terms is 65 and the sum of the middle three terms is 129. What is the first term and the common difference of the series?- a)64, 9/8
- b)32, 8/9
- c)187/9, 8/9
- d)72, 9/8
Correct answer is option 'C'. Can you explain this answer?
In an arithmetic series consisting of 51 terms, the sum of the first three terms is 65 and the sum of the middle three terms is 129. What is the first term and the common difference of the series?
a)
64, 9/8
b)
32, 8/9
c)
187/9, 8/9
d)
72, 9/8
|
|
Aarav Sharma answered |
Given information:
- The arithmetic series consists of 51 terms.
- The sum of the first three terms is 65.
- The sum of the middle three terms is 129.
Let's find the first term:
- The sum of the first three terms can be expressed as:
S3 = 3/2 * a + 3/2 * d, where a is the first term and d is the common difference.
- We are given that S3 = 65, so we can write the equation as:
65 = 3/2 * a + 3/2 * d
Let's find the sum of the middle three terms:
- The sum of the middle three terms can be expressed as:
S_middle = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
- We are given that S_middle = 129, so we can write the equation as:
129 = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
Solving the equations:
1. Rewrite the equations:
65 = 3/2 * a + 3/2 * d
129 = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
2. Simplify the equations:
65 = 3/2 * (2a + d)
129 = 3/2 * (3a + 3d)
3. Remove the fractions by multiplying both sides of the equations by 2:
130 = 3(2a + d)
258 = 3(3a + 3d)
4. Simplify the equations:
130 = 6a + 3d
258 = 9a + 9d
5. Rearrange the first equation to solve for d:
3d = 130 - 6a
d = (130 - 6a)/3
6. Substitute the value of d in the second equation:
258 = 9a + 9((130 - 6a)/3)
258 = 9a + 3(130 - 6a)
258 = 9a + 390 - 18a
258 - 390 = -9a
-132 = -9a
a = -132/-9
a = 187/9
Conclusion:
The first term (a) of the arithmetic series is 187/9 and the common difference (d) is (130 - 6a)/3, which simplifies to 8/9. Therefore, the correct answer is option C: 187/9, 8/9.
- The arithmetic series consists of 51 terms.
- The sum of the first three terms is 65.
- The sum of the middle three terms is 129.
Let's find the first term:
- The sum of the first three terms can be expressed as:
S3 = 3/2 * a + 3/2 * d, where a is the first term and d is the common difference.
- We are given that S3 = 65, so we can write the equation as:
65 = 3/2 * a + 3/2 * d
Let's find the sum of the middle three terms:
- The sum of the middle three terms can be expressed as:
S_middle = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
- We are given that S_middle = 129, so we can write the equation as:
129 = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
Solving the equations:
1. Rewrite the equations:
65 = 3/2 * a + 3/2 * d
129 = 3/2 * (a + 2d) + 3/2 * (a + d) + 3/2 * a
2. Simplify the equations:
65 = 3/2 * (2a + d)
129 = 3/2 * (3a + 3d)
3. Remove the fractions by multiplying both sides of the equations by 2:
130 = 3(2a + d)
258 = 3(3a + 3d)
4. Simplify the equations:
130 = 6a + 3d
258 = 9a + 9d
5. Rearrange the first equation to solve for d:
3d = 130 - 6a
d = (130 - 6a)/3
6. Substitute the value of d in the second equation:
258 = 9a + 9((130 - 6a)/3)
258 = 9a + 3(130 - 6a)
258 = 9a + 390 - 18a
258 - 390 = -9a
-132 = -9a
a = -132/-9
a = 187/9
Conclusion:
The first term (a) of the arithmetic series is 187/9 and the common difference (d) is (130 - 6a)/3, which simplifies to 8/9. Therefore, the correct answer is option C: 187/9, 8/9.
If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?- a)36
- b)18
- c)54
- d)24
- e)27
Correct answer is option 'C'. Can you explain this answer?
If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?
a)
36
b)
18
c)
54
d)
24
e)
27
|
Gowri Chakraborty answered |
5th term of GP = ar5-1 = 16*r4 = 81;
Or, r = (81/16)1/4 = 3/2;
4th term of GP = ar4-1 = 16*(3/2)3 = 54.
If(12+22+32+…..+102)=385,then the value of (22+42+62 + …+202) is :
- a)770
- b)1155
- c)1540
- d)(385*385)
Correct answer is option 'C'. Can you explain this answer?
If(12+22+32+…..+102)=385,then the value of (22+42+62 + …+202) is :
a)
770
b)
1155
c)
1540
d)
(385*385)
|
|
Gowri Chakraborty answered |
(1^2+2^2+3^2+....10^2)=385
(2^2+4^2+6^2+....+20^2)=2^2(1^2+2^2+ 3^2+....+10^2 )
=4(385)
=1540
The sum of the three numbers in A.P is 21 and the product of their extremes is 45. Find the numbers.- a)9, 7 and 5
- b)5, 7 and 9
- c)Both (1) and (2)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The sum of the three numbers in A.P is 21 and the product of their extremes is 45. Find the numbers.
a)
9, 7 and 5
b)
5, 7 and 9
c)
Both (1) and (2)
d)
None of these
|
Arya Roy answered |
The correct answer is c.
Let the numbers are be a - d, a, a + d
Let the numbers are be a - d, a, a + d
Then a-d+a+a+d= 21
3a = 21
a = 7
and (a - d) (a + d) = 45
a^2 - d^2 = 45
d^2=4
d = 
Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2. In both the cases numbers are the same.
If Sn denotes the sum of the first n terms in an Arithmetic Progression and S1: S4 = 1: 10Then the ratio of first term to fourth term is:- a)1 : 3
- b)2 : 3
- c)1 : 4
- d)1 : 5
Correct answer is option 'C'. Can you explain this answer?
If Sn denotes the sum of the first n terms in an Arithmetic Progression and S1: S4 = 1: 10
Then the ratio of first term to fourth term is:
a)
1 : 3
b)
2 : 3
c)
1 : 4
d)
1 : 5
|
|
Gowri Chakraborty answered |
Use Sn = (n/2)[ 2a + (n-1)d] and Tn = a + (n – 1) d
S1/S4 = 1/10 = a/ (4/2) [2a + 3d]
6a = 6d or a = d
Therefore T1/T4 = a/ (a+ 3d) = a/4a = 1/4
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is- a)600
- b)765
- c)640
- d)680
- e)690
Correct answer is option 'D'. Can you explain this answer?
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
a)
600
b)
765
c)
640
d)
680
e)
690
|
|
Arya Roy answered |
1st Method:
1st term = 5;
3rd term = 15;
Then, d = 5;
16th term = a+15d = 5+15*5 = 80;
Sum = {n*(a+L)/2} = {No. of terms*(first term + last term)/2}.
Thus, sum = {16*(5+80)/2} = 680.
2nd Method (Thought Process):
Sum = Number of terms * Average of that AP.
Sum = 16* {(5+80)/2} = 16*45 = 680.
Chapter doubts & questions for Sequence & Series - RBI Assistant Preparation Course 2025 is part of Bank Exams exam preparation. The chapters have been prepared according to the Bank Exams exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Bank Exams 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Sequence & Series - RBI Assistant Preparation Course in English & Hindi are available as part of Bank Exams exam.
Download more important topics, notes, lectures and mock test series for Bank Exams Exam by signing up for free.
RBI Assistant Preparation Course
713 videos|966 docs|289 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup