MCQs' Sequence and Series - Arithmetic and Geometric Progressions - Quantitative

Q1: If the fourth term of an Arithmetic Progression (A.P.) series is zero, then what is the ratio of the twenty-fifth term to the eleventh term?
(a) 4
(b) 5
(c) 3
(d) 2
Ans: (c)

Sol:
Let the first term be a and the common difference be d.
The n^th term of an AP is:
t_{n =} a + (n–1)d
Since, the 4th term is zero thus
t₄ = a + 3d = 0
a = –3d
Now the 25th and 11th terms is given by:
t₂₅ = a + 24d
t₁₁ = a + 10d
Substitute a = –3d into these terms:
t₂₅ = –3d + 24d = 21d
t₁₁ = –3d + 10d = 7d
Therefore, the required ratio is given by:
t₂₅/t₁₁ = 21d/7d = 3

Q2: The 3rd term of arithmetic progression is 7 and Seventh term is 2 more than thrice of third term. The common difference is
(a) 4
(b) 3
(c) 5
(d) 6
Ans: (a)

Sol:
Let the first term be a, and common difference be d.
Third term: t₃ = a + 2d = 7 ....(1)
Seventh term: t₇ = a + 6d ....(2)
Given,
t₇ = 3 × t₃ + 2 = 3 × 7 + 2 = 21 + 2 = 23
From eq. (1)
a = 7 – 2d
Substitute into (2):
t₇ = a + 6d = (7 – 2d) + 6d = 7 + 4d
i.e., 7 + 4d = 23
⇒ 4d = 16
⇒ d = 4

Q3: If 9th and 19th term of an Arithmetic Progression are 35 and 75 respectively, then its 20th term is
(a) 80
(b) 85
(c) 90
(d) 75
Ans: (b)

Sol:
Given: 9th and 19th term of an Arithmetic Progression are 35 and 75 respectively i.e.,
a₉ = 35 & a₁₉ = 75
Here, common difference is given by:
d = (a₁₉ – a₉)/(19 – 9)
⇒ d = 4
Therefore, 20th term is given by:
a₂₀ = a₁₉ + d
⇒ a₂₀ = 75 + 4
⇒ a₂₀ = 79

Q4: Find the 9th term of the A.P.: 8, 5, 2, –1, –4, ……
(a) –24
(b) –10
(c) –16
(d) –4
Ans: (c)

Sol:
Given A.P: 8, 5, 2, -1, -4, ...
Here, a = 8, d = 5 – 8 = -3
Thus, 9th term is given by 
t₉ = a + (9 – 1)d
t₉ = 8 + 8(-3) 
t₉ = 8 – 24 
t₉ = -16

Q5: In an arithmetic progression (A.P.), the seventh term is x and the (x + 7)^th term is 0. Then, what is the x^th term?
(a) 6
(b) 7
(c) 8
(d) 10
Ans: (b)

Sol:
First term = a
Common difference = d
Given,
t₇ = a + 6d = x ...(1)
t _{x + 7} = a + (x + 6)d = 0 ...(2)
or,
a = x – 6d   [From Equ. (1)]
t_{x + 7} = (x – 6d) + (x + 6)d = 0
Thus,
x – 6d + xd + 6d = 0
⇒ x + xd = 0
⇒ d = -1
Now, a = x – 6d = x – 6(1) = x + 6
Therefore, the xᵗʰ term is given by:
t_x = a + (x – 1)d
= (x + 6) + (x – 1)(-1)
= x + 6 – x + 1 = 7
Hence, the x^th  term is 7.

Q6: Find the 17th term of an AP series if 15th and 21st terms are 30.5 and 39.5 respectively.
(a) 33.5
(b) 35.5
(c) 36.0
(d) 38.0
Ans: (a)

Sol:
According to question, we have 
t₁₅= 30.5 and t₂₁ = 39.5
 ⇒ a + 14d = 30.5 and a + 20d = 39.5
On subtracting both the equations, we get 
a + 20d – (a + 14d) = 39.5 – 30.5 
⇒ 20d – 14d = 9
 ⇒ 6d = 9 
⇒ d = 1.5
Now, a = 30.5 – 14d = 30.5 – 14(1.5) = 9.5
Therefore, 17th term is given by: 
t₁₇  = a + 16d 
⇒ t₁₇ = 9.5 + 16(1.5) 
⇒ t₁₇ = 33.5

Q7: Find the value of ‘x’ for the following data 1 + 7 + 13 + 19 + ……. + x = 225.
(a) 56
(b) 63
(c) 49
(d) 42
Ans: (c)

Sol:
Given data; 1 + 7 + 13 + 19 + ....... x = 225
Clearly, it is an A.P with a = 1 & d = 6. 
We know that, 
Sum of n terms of A.P

⇒ n(1 + 3n – 3) = 225
⇒ n(3n – 2) = 225
⇒ 3n² – 2n – 225 = 0
⇒ 3n² – 27n + 25n – 225 = 0
⇒ 3n(n – 9) + 25(n – 9) = 0
⇒ (3n + 25)(n – 9) = 0
⇒ n = 9, n = -25/3 (Reject)
⇒ n = 9
Also, sum of n terms is given by:
S_n= n/2 (a + l)
⇒ n/2 (1 + x) = 225
⇒ 9/2 (1 + x) = 225
⇒ (1 + x) = 50 
⇒ x = 449

Q8: The 4 arithmetic means between –2 and 23 are
(a) 3, 13, 8, 18
(b) 18, 8, 3, 13
(c) 3, 8, 13, 18
(d) None of these
Ans:(c)

Sol:
We are to insert 4 arithmetic means between –2 and 23. 
So, the sequence will have 6 terms total: 
–2, _, _, _, _, 2
This is an Arithmetic Progression (A.P.) Let the common difference be d.
Let the sequence be: 
–2, a₂, a₃, a₄, a₅, 23
The 6th term in an A.P. is: 
a₆ = a₁ + 5d = –2 + 5d = 23 
⇒ 5d = 25 
⇒ d = 5
Now compute the terms:
⇒ a₁ = –2
⇒ a₂ = –2 + 5 = 3 
⇒ a₃ = 3 + 5 = 8 
⇒ a₄ = 8 + 5 = 13 
⇒ a₅ = 13 + 5 = 18
⇒ a₆ = 18 + 5 = 23
The four arithmetic means are: 3, 8, 13, 18

Q9: Insert 4 numbers between 2 and 22 such that the resulting sequence is an Arithmetic Progression (A.P.).
(a) 4, 8, 12, 16
(b) 5, 9, 13, 17
(c) 4, 10, 15, 19
(d) 6, 10, 14, 18
Ans: (d)

Sol:
We are to insert 4 numbers between 2 and 22 → total 6 terms in the A.P. 
Let the 6 terms be:
2, a₂, a₃, a₄, a₅, .....22
Let the common difference be d. Then:
First term, a = 2
Sixth term, a + 5d = 22
2 + 5d = 22
⇒ 5d = 20
⇒ d = 4
Now compute the terms:
a₂ = 2 + d= 6
a₃ = 2+ 2d = 10
a₄ = 2+ 3d = 14 
a₅ = 2+ 4d = 18
Therefore, the required numbers are 6, 10, 14, 18.

Q10: A roadside tea stall merchant borrows ₹5,000 at 2.76% simple interest per annum. The principal and interest are to be paid in 10 monthly instalments, where each instalment is double the preceding one. Find the value of the last instalment.
(a) ₹4,608
(b) ₹1,024
(c) ₹9,207
(d) ₹4,096
Ans: (b)

Sol:
According to question, we have
Principle P = ₹9,000
Rate R = 2.76% per annum
Time T = 10 months = 10/5 =  12/6 = year
Simple Interest (S.I.) is given by

Total amount to be repaid = ₹9,000 + ₹207 = ₹9,207

Let the first instalment be = ₹x.

Each instalment is double the previous one → geometric progression with: 
First term a = x 
Common ratio r = 2 
Number of terms n = 10
Sum of a G.P

Therefore, the last instalment is given by: 
Last (10th) instalment = x.2⁹ = 9.512 = ₹4,608

Q11: The 7th term of the series 6, 12, 24, ….. nth term is_____.
(a) 384
(b) 834
(c) 438
(d) 854
Ans: (a)

Sol:
The series is 6, 12, 24...... which is a Geometric Progression (G.P.) because each term is multiplied by the same ratio. 
First term, a = 6
Common ratio, r = 12/6 = 2
The nth term of a G.P. is given by
t_n = a × r^n–1
For the 7th term: 
t₇ = 6 × 2^7–1 = 6 × 2⁶ = 6 × 64 = 384

Q12: The last term of the series x², x, 1, ….. to 31 terms is
(a) x²⁸
(b) 1/x
(c) 1/x²⁸
(d) 1/x³⁰
Ans: (c)

Sol:
The series is:
x², x, 1,.......

This is a Geometric Progression (G.P.) with:
 First term a = x2
Common ratio r = x/x² = 1/x
Therefore, the 31st term is given by;

Q13: In a G.P. 5th term is 27 and 8th term is 729. Find its 11th term
(a) 729
(b) 6561
(c) 2187
(d) 19683
Ans: (d)

Sol:
Let the first term of G.P. be ‘a’ and common ratio ‘r’.
According to question, we have
t₅ = ar⁴ = 27 ...(1)
t₈ = ar⁷ = 729 ...(2)
Dividing eq., (2) by eq., (1), we get
ar⁷/ ar⁴ = 729/27
⇒ r³ = 27 
⇒ r³ = 3 ³
 ⇒ r = 3
Put the value of ‘r’ in eq (1), we get 
a (3)⁴ = 27 
⇒ a = 1/3
Therefore, the 11th term will be

Q14: If 4th, 7th and 10th terms of a Geometric Progression are p, q and r respectively, then:
(a) p² = qr + r²
(b) p² = qr
(c) q² = pr
(d) pqr + pr + 1 = 0
Ans: (b)

Sol:

According to question, we have
a₄ = ar³ = p ...(1)
a₇ = ar⁶ = q ...(2)
a₁₀ = ar⁹ = r ...(3)
Multiplying eq. (1) and eq (3), we get
(ar³)(ar⁹) = pr
⇒ a² r¹² = pr
⇒ (ar⁶)² = pr
⇒ q² = pr

Q15: Find the sum of the series:
…… up to 6 terms
(a) 63/32
(b) 32/63
(c) 26/53
(d) 53/26
Ans: (a)

Sol:
Given series:
This is a Geometric Progression (G.P.) where;

First term a = 1

Common ratio r = 1/2

Number of terms, n = 6

We know that,

Q16: The sum of 3 numbers of a G.P is 39 and their product is 729. The numbers are-
(a) 3, 27, 9
(b) 9, 3, 27
(c) 3, 9, 27
(d) None of these
Ans: (c)

Sol:
Let the three numbers of Geometric Progression be;
a/r , a , a r
According to question, we have
a/r + a + ar = 39
a + ar + ar² = 39r 
⇒ a(1 + r + r2) = 39r
Also, product of numbers is given by
(a/r) ·a· ar = a³ = 729
⇒ a = ³√729 = 9
Substitute the value of a in above equation, we get
9(1 + r + r²) = 39 r
3(1 + r + r²) = 13 r
3 + 3 r + 3 r² = 13 r
⇒ 3 r² – 10 r + 3 = 0
Use quadratic formula:

So,
r = 18/6 = 3
r = 2/6 = 1/3
Case 1: r = 3, a = 9
Numbers

Case 2: r = 1/3
Then the numbers are:
a/r = 9/(1/3) = 27 
a = 9
 a r = 9 · (1/3) = 3
So the numbers are: 27, 9, 3 
Therefore, the required numbers are 3, 9, 27.

Q17: The product of three numbers which are in GP is 512. Then the second number is:
(a) 3
(b) 2
(c) 6
(d) 8
Ans: (d)

Sol:
Let the three numbers in GP be:
a/r, a, ar
i.e., a³ = 512

Therefore, the second number is a = 8

Q18: The sum of three numbers in G.P is 70. If the two extremes multiplied each by 4 and the mean by 5, the product are in A.P. The numbers are
(a) 12, 18, 40
(b) 10, 20, 40
(c) 40, 20, 15
(d) None of these
Ans: (b)

Sol:
Go by options:
Option (a): 12, 18, 40
Clearly, 12, 18, 40 are not in G.P. since common ratio is not constant i.e., 18/12 ≠ 40/18
Option (b): 10, 20, 40
Clearly, they are in G.P.
Sum of numbers = 10 + 20 + 40 = 70
Now, on multiplying two extremes each by 4 and the means by 5, we get 40, 100 and 160
Therefore, 100 – 40 = 60 and 160 – 100 = 60
Clearly, they are in A.P.
Hence, 10, 20, 40 satisfies all the given conditions.

Q19: The numbers x, 8, y in Geometric Progression (G.P) and the numbers x, y, -8 are in Arithmetic Progression (A.P). Find the values of x and y respectively.
(a) 4, 16
(b) 16, 4
(c) 4, 8
(d) 8, 4
Ans: (a)

Sol:

Since, x, 8, y are in G.P., the middle term squared equals the product of the first and third terms: 
8² = x × y ⇒ 64 = x y  ...(1)

Since, x, y, –8 are in A.P., twice the middle term equals the sum of the first and third terms: 
2y = x + (–8) ⇒ 2y = x – 8   ...(2)

From eq. (2), we have 
x = 2y + 8

Therefore, eq.(1) will becomes 
64 = y(2y + 8) = 2y² + 8y 
2y² + 8y – 64 = 0 
y² + 4y – 32 = 0


Therefore, the value of x will be; 
For y = 4; x = 2(4) + 8 = 8 + 8 = 16

For y = –8; x = 2(–8) + 8 = –16 + 8 = –8

Therefore, (16, 4), (–8, –8) are the requiredpairs.

Hence, out of the given options, option (a) is correct.

Q20: If the A.M and G.M. for two numbers are 6.50 and 6 respectively then the two numbers are
(a) 6 and 7
(b) 9 and 4
(c) 10 and 3
(d) 8 and 5
Ans: (b)

Sol:
Let the two numbers be a and b.
Step 1: Use A.M. formula
A.M = (a + b)/2 = 6.5
⇒ a + b = 13
Step 2: Use G.M. formula
G.M = √(ab) = 6
⇒ ab = 36
We now have,
 a + b = 13
ab = 36
Clearly, option (b) satisfies both the above conditions i.e.,
9 + 4 = 13 and 9 × 4 = 36
Hence, option (b) is correct.