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All questions of Sequences and Series for JEE Exam

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 Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....)
  • a)
    64
  • b)
    54
  • c)
    56
  • d)
    81
Correct answer is option 'A'. Can you explain this answer?

Anjana Sharma answered
The given number series as follow as 1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, ...
So the next number is 8^2 = 64.

The terms of an A.P. are doubled, then the resulting sequence is
  • a)
    An A.P. with common difference equal to the common difference of the original A.P​
  • b)
    An A.P. with common difference thrice the common difference of the original A.P.
  • c)
    An A.P. with common difference double the common difference of the original A.P.
  • d)
    An A.P. with common difference half the common difference of the original A.P.
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
The general form of an AP is  a,  a+d, a+2d,.....
where a is the first term and d is the common difference  
If we double the terms,  the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d  nth term= 2a+(n-1)2d               
= 2[a+(n-1)d]

The sum of the first hundred even natural numbers divisible by 5 is
  • a)
    55005
  • b)
    55000
  • c)
    50000
  • d)
    50500
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
first even no. divisible by 5= 10

d=10

Sn = n/2(2a + (n-1) d)

S100= 100/2 (2(10) + (100-1)10)

S100=50(20 + 990)

S100= 50(1010)

S100 = 50500

The first negative term of the  A.P.62,57,52…. is the
  • a)
    10th term
  • b)
    14th term
  • c)
    12th term
  • d)
    18th term
Correct answer is option 'B'. Can you explain this answer?

Naina Sharma answered
a = 62  d = 57 - 62 = -5
tn = a + (n-1)d
= 62 + (n-1)(-5)
= 62 - 5n + 5
= 67 - 5n
From the options, we take '14'
= 67 - 5(14)
= 67 - 70
= -3 (The first negative term will be at the 14th term)

How many terms of the series
24,20,16,…are required so that their sum is 72?
  • a)
    8 or 6
  • b)
    12
  • c)
    9 or 4
  • d)
    11
Correct answer is option 'C'. Can you explain this answer?

Naina Bansal answered
As given above we can clearly see that the given series forms an AP. So we can use the sum formula to calculate it.

= Sn = n / 2 [ 2a + ( n-1 ) d ]

So the values are :

a = 24

d = -4

n = ?

Sn = 72
So substitute in the formula to get the answer:

= 72 = n / 2 [ 2 ( 24 ) + ( n - 1 ) -4 ]

= 72 = n / 2 [ 48 - 4n + 4 ]

= 72 * 2 = n [ 52 - 4n ]

= 144 = 52n - 4n^2

= 4n^2 - 52n + 144 = 0 ------ Dividing by 4 throughout the equation we get,

= n^2 - 13n + 36

Factorizing the above quadratic equation we get,

= n^2 - 9n - 4n + 36 = 0

= n ( n - 9 ) -4 ( n - 9 ) = 0

= ( n - 4 ) ( n - 9 ) = 0

= n = 4 , 9

So the number of terms can be both 4 terms and 9 terms. This is because since the AP is decreasing. 
Hope it helps !!

 If A1, A2, A3,…., An are n numbers between a and b, such that a, A1, A2, A3,…, An, b are in A.P., then nth term from beginning is:
  • a)
    b
  • b)
    An-1
  • c)
    An
  • d)
    A1
Correct answer is option 'B'. Can you explain this answer?

Krishna Iyer answered
A1,A2,......, An are inserted between a and b then the series will become 
a, A1,A2,A3,......, An,b. Now a becomes the first term, A1 will be second, A2 will become third term
An will become A(n+1)th term 
therefore A(n-1) will become nth term.

The digits of a positive integer having three digits are in AP and sum of their digits is 21. The number obtained by reversing the digits is 396 less than the original number. Find the original number.
  • a)
    876
  • b)
    579
  • c)
    975
  • d)
    678
Correct answer is option 'C'. Can you explain this answer?

Trisha Vashisht answered
Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21                        ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396          ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975

The first 4 terms of the sequence   are
  • a)
    2, 6, 13, 27
  • b)
    2, 5, 11, 23
  • c)
    2, 6, 11, 24
  • d)
    2, 5, 12, 37
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
a1 = 2 
a2 = 2a1 + 1
=> 2(2) + 1 = 5
a3 = 2a2 + 1
=> 2(5) + 1 = 11
a4 = 2a3 + 1
=> 2(11) + 1 = 23
Hence, the required series is : 2,5,11,23………

What is the 10th A.M between 2 and 57 if 10 A.M s are inserted between these numbers?
  • a)
    54
  • b)
    53
  • c)
    52
  • d)
    55
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
2 and 57 have 10 terms between them so including them there would be 12 terms
an = 57, a = 2, n = 12
an = a + (n-1)d
=> 57 = 2 + (12 - 1)d
=> 55 = 11d
d = 5
T10 = a + 10d
=> 2 + 10(5) 
= 52

If the sums of n, 2n and 3n terms of an AP are S1, S2 and S3 respectively, then 
  • a)
  • b)
  • c)
    3
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Pooja Shah answered
Given x 2 - 3x = 0 
Factor x out in the expression on the left. 
x (x - 3) = 0 
For the product x (x - 3) to be equal to zero we nedd to have 
x = 0 or x - 3 = 0 
Solve the above simple equations to obtain the solutions. 
x = 0 
or 
x = 3 

The sum of the series 
  • a)
    49/25
  • b)
    49/24
  • c)
    47/24
  • d)
    49/29
Correct answer is option 'B'. Can you explain this answer?

Knowledge Hub answered
 (1 + 1/52) + (1/2 + 1/52) + (1/22 + 1/54)+......
= (1 + 1/2 + 1/22 +....) + (1/52 + 1/54 + 1/56 +......)
= r = (1/4)/(1/2) , r = (1/54)/(1/52)
⇒ r = 1/2 + 1/25
⇒ S = 1 + [(1/2)/(1-1/2)],    S = [(1/25)/(1-1/25)]
⇒ S = 1 + 1,    S = 1/24
Total sum = 2 + 1/24
⇒ 49/24 

If the 10 times of the 10th term of an AP is equal to 15 times to the 15th term, then the 25th term is:
  • a)
    25 times of 25th term
  • b)
    zero
  • c)
    1
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Nandini Patel answered
According to question ,
10 x 10th term =15 x 15th term 
let a is the first term and d is the common difference .
10 (a+9d)=15 (a+14d)

5a+120d=0

a +24d=0
now ,
25th term =a+(25-1) d=a+24d=0
hence 25th term=0

How many terms of the G.P. 4 + 16 + 64 + … will make the sum 5460?
  • a)
    9
  • b)
    7
  • c)
    8
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Nandini Iyer answered
Sum (Sn) = a x (rn -1)/(r-1)
5460 = 4 x (4n -1)/3
16380 = 4n+1 - 4
16384 = 4n+1
47 = 4n+1
7 = n + 1
n = 6

The 10th term of an G.P. is and the 5th term is.What is the Common Ratio?
  • a)
    1/4
  • b)
    1/16
  • c)
    1/2
  • d)
    1/8
Correct answer is option 'C'. Can you explain this answer?

Praveen Kumar answered
ar9 = 1/(2)8 -----------------------(1)
ar3 = 1/(2)2 ------------------------------(2)
Comparing eq (1) & (2)
1/(2)8 r9 = 1/(2)^2 r4
1/r5 = (2)5
r5 = 1/(2)5
r = 1/2

  • a)
    4
  • b)
    6
  • c)
    8
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
 6 + 6 + 6.........∞
61/2 * 61/4 * 61/8………..∞
= 61/2 + 1/(2*2) + 1/(2*2*2)   (sum of infinte G.P.= a/(1−r))
= (1/2)/(6(1-½))
= (1/2)/(61/2)
= 6

Which term of the following sequence is 64?
2 , 2√2, 4, .....
  • a)
    12
  • b)
    10
  • c)
    11
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
Given sequence : 2, 2√2, 4….
First term a1 = a = 2 and 2nd term a2 = 2√2, then
Common ratio r = a2/a = (2√2)/2
Let an = 64
∴ ar(n-1) = 64
⇒ 2.(√2)(n-1) = 32
⇒ (2)(n-1)/2 = 32
∴ (2)(n-1)/2 = (2)5
⇒ (n − 1)/2 = 5,
⇒ n = 11

 If a, b, c are in A.P., then
  • a)
    b - a = b - c
  • b)
    b is the arithmetic mean of a and c.
  • c)
    2a = b + c
  • d)
    a - b = c - b
Correct answer is option 'B'. Can you explain this answer?

Anirban Sharma answered
Understanding Arithmetic Progression (A.P.)
When three numbers a, b, and c are in Arithmetic Progression (A.P.), it means they have a specific relationship. A.P. is characterized by a constant difference between consecutive terms.
Key Definition
- In A.P., the middle term is the average of the two outer terms. Hence, for numbers a, b, and c:
- b = (a + c) / 2
This implies that b is the arithmetic mean of a and c.
Analyzing the Options
Let's break down each option to see their validity:
- Option a) b - a = b - c
This implies a + c = 2b, which is true.
- Option b) b is the arithmetic mean of a and c
This directly aligns with the definition of A.P., making it correct.
- Option c) 2a = b + c
Rearranging gives b = 2a - c. This is not a general rule for A.P.
- Option d) a - b = c - b
Rearranging gives a + c = 2b, which is true but doesn't define A.P. directly.
Conclusion
The correct answer is Option b) because it succinctly captures the essence of A.P., where b is indeed the arithmetic mean of a and c. This fundamental property is central to understanding and solving problems related to Arithmetic Progression.
In summary, when a, b, and c are in A.P., b perfectly represents the mean of the outer terms a and c, confirming that Option b is the correct choice.

The sum of n terms of the sequence 8, 88, 888,…. is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Infinity Academy answered
8(1) + 8(11) + 8(111) + 8(1111) +........up to n times
= 8{ 1 + 11 + 111 + 1111 + ....... up to n times}
= 8/9{9 + 99 + 999 + 9999 + ..... up to n times}
= 8/9 {(10-1 )+ (100-1) + (1000-1) + (10000-1) + ........up to n times}
= 8/9 {(10 + 100 + 1000 + 10000 +...... n times) - (n×1)}
= 8/9{( (10×(10n -1))/(10-1)) -n}
= 80/81(10n - 1) - n
=  8/81(10n+1 - 10 -9n)

The G.M. of 5 and 8 is
  • a)
    40
  • b)
    2√10
  • c)
    10√2
  • d)
    20
Correct answer is option 'B'. Can you explain this answer?

Hansa Sharma answered
The geometric mean of two numbers, say x, and y is the square root of their product x * y.
 Mean = [x * y]1/2
= [5 * 8]1/2
= [40]1/2
= 2(10)1/2

For what value of n,   is the arithmetic mean (AM) of a and b?
  • a)
    1
  • b)
    1/2
  • c)
    -1/2
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?

Neha Joshi answered
(aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is mean between a & b
Mean of a & b =  ( a + b)/2
=> (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = ( a + b)/2
=> 2aⁿ⁺¹ + 2bⁿ⁺¹  = aⁿ⁺¹  + bⁿ⁺¹  + baⁿ + bⁿa
=> aⁿ⁺¹ + bⁿ⁺¹ = baⁿ + bⁿa
=> aⁿ(a - b) = bⁿ(a - b)
=> aⁿ = bⁿ
=> (a/b)ⁿ = 1
=> n = 0    or a  = b

7th term of Geometric Progression 2, 6, 18, ... is
  • a)
    5832
  • b)
    2919
  • c)
    1458
  • d)
    729
Correct answer is option 'C'. Can you explain this answer?

Saptarshi Sen answered
The 7th term of a geometric progression (GP) can be found using the formula:
Tn=a⋅rn−1
Where:
  • Tn​ is the n-th term,
  • a is the first term,
  • r is the common ratio,
  • n is the term number.
In the given GP: 2, 6, 18, ...
  • The first term a=2
  • The common ratio r=6/2=3
Now, substitute into the formula for the 7th term:
T7 = 2⋅37−1 = 2⋅36 = 2⋅729 = 1458
So, the 7th term is 1458

Consider the sequence 8 A + 2 B,6 A + B,4 A 2 A−B,……., Which term of this sequence will have a coefficient of A which is twice the coefficient of B?
  • a)
    10th 
  • b)
    14th 
  • c)
    16th 
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Harsh Mehta answered
Understanding the Sequence
The sequence provided is:
- 8A + 2B
- 6A + B
- 4A^2 - B
- ...
To determine which term has a coefficient of A that is twice the coefficient of B, we need to analyze the pattern in the coefficients of A and B in the sequence.
Identifying the Coefficients
Let's denote the coefficient of A in the nth term as A_n, and the coefficient of B as B_n. The terms are:
- 1st term: A_1 = 8, B_1 = 2
- 2nd term: A_2 = 6, B_2 = 1
- 3rd term: A_3 = 4 (from 4A^2), B_3 = -1
Pattern Analysis
To find a general formula, we can observe:
- A decreases by 2 for the first two terms and then changes form in the third term.
- B decreases by 1 and then becomes negative.
Let's analyze the coefficients further:
- The potential formula is not straightforward, but we can check manually for the provided options.
Finding Terms Where A = 2B
We need to find when A_n = 2B_n. Testing the first few terms:
- For the 1st term: 8 ≠ 2(2)
- For the 2nd term: 6 ≠ 2(1)
- For the 3rd term: 4 ≠ 2(-1)
Continue this process for higher terms until reaching higher values.
Conclusion
By observing the trends, it becomes clear that none of the terms in the sequence will satisfy the condition where the coefficient of A is precisely double that of B. Therefore, the answer is:
Correct Answer: None of these

If |3x−1|,3,|x−3| are the first three terms of an arithmetic progression, then the sum of the first five terms can be
  • a)
    5
  • b)
    10
  • c)
    20
  • d)
    30
Correct answer is option 'A'. Can you explain this answer?

Nipun Tuteja answered
Case-I: x<1/3 ⇒ −3x+1,3,−x+3 are in A.P.
⇒ 6 = −3x + 1 − x + 3
⇒ 4x = −2 ⇒ x = −1/2
⇒ terms are 5/2, 3, 7/2, 4, 9/2
⇒ sum =35/2
Case-II: 1/3 ≤ x < 3
3x − 1,3, −x + 3 are in A.P.
⇒ 6 = 3x − 1 − x + 3 = 2x + 2 ⇒ x = 2
⇒ terms are 5, 3, 1, –1, –3
⇒ sum = 5
Case-III: x ≥ 3
⇒ terms are 3x − 1,3,x − 3
⇒ 6 = 3x − 1 + x − 3 = 4x − 4
⇒ 4x = 10 ⇒ x = 5/2 not possible

  • a)
    ac/b
  • b)
    bc/a
  • c)
    ab/c
  • d)
    cb/a
Correct answer is option 'C'. Can you explain this answer?

Anshu Joshi answered
 x  = a + a/r  +  a/r² + .....................+ ∞
First term = a  , Common Ratio = 1/r
=> x =  a/(1 - 1/r)  =  ra/(r - 1)
y  = b - b/r  + b/r² + .....................+ ∞
 
First term = b  , Common Ratio = -1/r
=> y =  b/(1 - (-1/r))  =  rb/(r + 1)
z=c+c/r²+c/r⁴+..........................+ ∞
 
First term = c  , Common Ratio = 1/r²
z = c/(1 - (1/r²) = r²c/(r² - 1)
 
LHS = xy/z    
 
=   (ra/(r - 1) )( rb/(r + 1)) / ( r²c/(r² - 1))
= r²ab(r² - 1)/ r²c/(r² - 1)
= ab/c

If the pth term of an A.P. be 1/q and qth term be 1/p, then the sum of its pqth terms will be
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Sai Kulkarni answered
Given ap = 1/q
a + (p − 1)d = 1/q
aq + (pq − q)d=1…(1)
Similarly, we get
ap + (pq − p)d = 1…(2)
From (1) and (2), we get
aq + (pq − q)d = ap + (pq − p)d
aq − ap = d[pq − p − pq + q]
a(q − p) = d(q − p)
∴ a = d
Equation (1) becomes,
dq + pqd − dq = 1

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