MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Arithmetic Progressions Class 10 MCQ Questions with Answers

Question 1.

30th term of the A. P., 10, 7,4,…………, is:

(A) 97
(B) 77
(C) -77
(D) -87
Answer:
(B) 77

Explanation:
In the given AP,a = 10 and d = 7-10
Thus, the 30th term is t30= 10 + (30 -1) (-3) = -77

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 2.

11th term of the A.P.,-3,-i, 2,… is:

(A) 28
(B) 22
(C) -38
(D) -48
Answer:
(B) 22

Explanation:
In the given A.P., a = -3 and
d = – \(\frac{1}{2}\) + 3 = \(\frac{5}{2}\)
Thus, the 11th term is t11 =-3 + (11-1)\(\left(\frac{5}{2}\right)\) = 22

Question 3.

In an A.P., if d = -4, n = 7, an = 4, then a is;

(A) 6
(B) 7
(C) 20
(D) 28
Answer:
(D) 28

Explanation:
In the given A.P., d = -4, n = 7,
an = 4
an =a + (n – 1)d ⇒ 4 = a + (7 – 1)(-4) ⇒ a = 28

Question 4.

In an A.P., if a = 3.5, d = 0, n = 101, then an will be

(A) 0
(B) 3.5
(C) 103.5
(D) 104.5
Answer:
(B) 3.5

Explanation:
In the given A.P., a = 3.5, d = 0, n = 101 than an will be
(A) 0
(B) 3.5
(C) 103. 5
(D) 104.5

Explanation:
In the given A.p., = 3.5, d = 0,
n = 101
an = a + (n – 1) d ⇒an = 3.5 + (101 -1) 0 ⇒an = 3.5

Question 5.

The list of numbers – 10, – 6, – 2, 2,…………is:

(A) an A.P., with d = – 16
(B) an A.P., with d = 4
(C) an A.P., with d = – 4
(D) not an A.P., 0
Answer:
(B) an A.P., with d = 4

Explanation:
In the given numbers
-10,-6,-2, 2,…
(-6)-(-10) = 4
(-2)-(-6) = 4
2 – (-2) = 4
Since, (-6) – (-10) = (-2) – (-6) = 2 – (-2) = 4, thus, the given numbers are in AP with d = 4.

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 6.

The 11th term of the A.P., -5, \(-\frac{5}{2}\), 0, \(\frac{5}{2}\),… is:

(A) -20
(B) 20
(C) -30
(D) 30
Answer:
(B) 20

Explanation:
In the given A.P.,
a = -5, d = \(-\frac{5}{2}\)-(-5) = \(\frac{5}{2}\) , n = 11
tn = a + (n – 1)d ⇒ t11 = -5 + -(11 1) \(\left(\frac{5}{2}\right)\) ⇒t11 = 20

Question 7.

The first four terms of an A.P., whose first term is – 2 and the common difference is -2, are:

(A) -2,0,2,4
(B)-2,4,-8,16
(C) – 2, – 4, – 6, – 8
(D)-2,-4,-8,-16 [1
Answer:
(C) – 2, – 4, – 6, – 8

Explanation:
In the given AP, a = -2, d = -2,
tn =a + (n -1)d
t1 = (-2) + (1 – 1) (-2) = -2
t2 = (-2) + (2 – 1)(-2) = -4
t3 = (-2) + (3 -1)(-2) = -6
t4= (-2) + (4 -1)(-2) = -8

Question 8.

The 21st term of the A.P., whose first two terms are -3 and 4 is :

(A) 17
(B) 137
(C) 143
(D) -143
Answer:
(B) 137

Explanation:
In the given A.P., t1 = —3 and t2 = 4
⇒ d =t2 – t1 = 4 – (-3) = 7
tn = a +(n – 1)d
⇒ t21 = (-3) + (21 – 1) (7) = 137

Question 9.

The famous mathematician associated with finding the sum of the first 100 natural numbers is :

(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid
Answer:
(C) Gauss

Explanation:
The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.

Question 10.

If the first term of an A.P. is -5 and the common difference is 2, then the sum of the first 6 terms is :

(A) 0
(B) 5
(C) 6
(D) 15
Answer:
(A) 0

Explanation:
In the given A.P., a = -5 and d = 2 Thus,
Sn = \(\frac{n}{2}\) [2a+(n – 1)d]
Sn = \(\frac{5}{2}\)[2x(-5)+(6-1)x2] = 0

Question 11.

The sum of first 16 terms of the A.P., 10,6,2,… is :

(A) -320
(B) 320
(C) -352
(D) -400
Answer:
(A) -320

Explanation:
In the given A.P.,a = 10,d = 6- 10 = -4
Thus,
Sn = \(\frac{n}{2}\) [2a + (n-1)d]
S16 = \(\frac{n}{2}\) [2 × 10 + (16 – 1) (-4)]
= -320

Question 12.

In an A.P., if a = 1, an = 20 and Sn = 399, then n is :

(A) 19
(B) 21
(C) 38
(D) 42
Answer:
(C) 38

Explanation:
In the given A.P., a = 1, an = 20 and Sn = 399

an = a+ (n – 1)d
⇒20 – 1 + (n -1)d
(n – 1)d = 19
Sn = \(\frac{n}{2}\) [2a + (n -1)d]
⇒399 = \(\frac{n}{2}\) [2 +19]
⇒n = 38

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 13.

The sum of tirst live multiples of 3 is :

(A) 45
(B) 55
(C) 65
(D) 75
Answer:
(A) 45

Explanation:
In the given AP, a = 3, d = 3 and n = 5
Thus,
Sn = \(\frac{n}{2}\) [2a + (n -1)d]
5 = \(\frac{5}{2}\) [2 3 + (5 -1) 3] =45

Question 14.

The sum of first five positive integers divisible by 6 is:

(A) 180
(B) 90
(C) 45
(D) 30
Answer:
(B) 90

Explanation:
Positive integers divisible by 6 are 0,12,18,24,30 Since difference is same, its an AP We need to find sum of first 5 integers We can use formula
Sn = \(\frac{5}{2}\)[(2a + (n – 1)d)
n = 5, d = 6, a =6
S5 = \(\frac{n}{2}\)(2 x 6 + (5 – 1) x 6)
S5 = \(\frac{5}{2}\) (12 + 24)
S5 =\(\frac{5}{2}\) × 36 = 90.

Assertion and Reason Based MCQs

Direction: In the. following question, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) BothAandR are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): If the second term of an A.P., is 13 and the fifth term is 25, then its 7th term is 33.
Reason (R): If the common difference of an A.P. is 5, then a18 – a13 is 25.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
In the given A.P.,t2 = 13 and t5 = 25
a + d = 13
a + 4d – 25
Solving these equations, we get a = 9 and d = 4 Thus,
tn = a +(n -1)dayt7 = 9 +(7 – 1)4 = 33
∴ Assertion is correct:
In case o freason
In the given A.P.,d = 5 thus,
a18 -a13 = a + 17d – a 12d = 5d = 25
∴Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 2.

Assertion (A): If a18 – a14 = 32, then the common difference of an A.P., is – 8.
Reason (R): If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then its 18th term will be 0.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
In the given A.P., a18 – a14 = 32
Thus,
a18 – a14 = 32
⇒ a + 17d – a – 13d = 32
⇒ 4d = 32
⇒ d = 8
∴ Assertion is incorrect.
In case of reason:
According to question,
7t7 = 11t11
⇒ 7(a +6d) = 11(a + 10d)
⇒ 4a + 68d = 0
⇒ 4(a + 17d) = 0
⇒ (a + 17d) = 0
t18 = 0
∴ Reason is correct.
Hence, assertion is incorrect but reason is correct.

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 3.

Assertion (A): The fourth term from the end of the A.P., -11, -8, -5,…, 49 is 40.
Reason (R): If the nth term of an A.P., – 1,4,9,14,… is 129, then the value of n is 100.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
In the given A.P., the last term l= 49 and common difference d = -8 + 11 = 3
4th term from last is t4 = 49 – (4 – 1) x 3 = 40
∴ Assertion is correct.
In case of reason:
Given, a = – 1 and d = 4 – (- 1) = 5
an = -1 + (n- 1) x 5 – 129
or, (n – 1)5 = 130
(n – 1) = 26
n = 27
Hence, 27th term = 129.
∴ Reason is incorrect.
Hence, assertion is correct but reason is incorrect.

Question 4.

Assertion (A): If nlh temrof an A.P. is (2n + 1), then the sum of its first three terms is 15.
Reason (R): The sum of first 16 terms of the A.P. 10, 6,2,… is – 320.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
∴ an = (2n + 1)
∴ a1 = 2 × 1 + 1 = 3
l = a3 = 2 ×3 + 1 = 7
Since, Sn = \(\frac{n}{2}\)[a + l]
Hence, S3 = \(\frac{3}{2}\)[3 + 7]
S3 = 15
∴ Assertion is correct.
In case of reason:
Hwere, a = 10, d = 6 – 10 = -4 and n = 16
Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
S16 = \(\frac{16}{2}\) [2 × 10 + (16 – 1)(-4)]
= 8[20 + 15 ×(-4)]
= 8[20 – 60]
= 8 × (-40)
= – 320
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 5.

Assertion (A): If the sum of first k terms of an A.P. is 3k2– k and its common difference is 6, then the first term is 5.
Reason (R): If the «th term of an A.P. is 7 – 3n, then the sum of 25 terms is – 800.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
Let the sum of k terms of A.P. is Sk = 3k2 – k Now kih term of A.P.
ak = Sk – Sk-1
ak = (3k2-k)-[3(k – 1)2 – (k – 1)]
= 3k2 – k – [3k2 – 6k + 3 – k + 1]
= 3k2 – k – 3k2 + 7k – 4
= 6k – 4
Hence, first term a = 6 x 1- 4 = 2
∴ Assertion is incorrect.
In case of reason:
Here n = 25 and an = 7 – 3n
Taking n = 1,2,3,…………, we get
a1 = 7 – 3 x 1 = 4
a2 = 7 – 3 x 2 = 1
a3 = 7 – 3 x 3 = – 2
∴ GivenA.P. is 4, 1, -2,……….. .
Here, a = 4 and d = 1 – 4 = – 3
Since, Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
Now, S25 = \(\frac{25}{2}\) [2 x 4 +(25 – 1)(- 3)]
= \(\frac{25}{2}\)[ 8 + 24 (-3)]
= \(\frac{25}{2}\) (8 – 72)
= – 800
∴ Reason is correct.
Hence, assertion is incorrect but reason is correct :

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the below questions:
Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 1

Question 1.

Which of the following terms are in AP for the given situation

(A) 51,53,55….
(B) 51,49,47….
(C) -51,-53,-55….
(D) 51,55,59…
Answer:
(B) 51,49,47….

Explanation:
a = 51
d = – 2
AP = 51,49,47 ……

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 2.

What is the minimum number of days he needs to practice till his goal is achieved ?

(A) 10
(B) 12
(C) 11
(D) 9
Answer:
(C) 11

Explanation:
Goal = 31 second
n = number of days
an = 31
a +(n – 1) d = 31
51 +(n – 1)(- 2) = 31
51 – 2n + 2 = 31
-2n = 31 – 53
-2n = -22
n = 11

Question 3.

Which of the following term is not in the AP of the above given situation

(A) 41
(B) 30
(C) 37
(D) 39
Answer:
(B) 30

Question 4.

If nth term of an AP is given by an = 2n + 3 then common difference of an AP is

(A) 2
(B) 3
(C) 5
(D) 1
Answer:
(A) 2

Question 5.

The value of x, for which 2x, x + 10,3x + 2 are three consecutive terms of an AP

(A) 6
(B) -6
(C) 18
(D) -18
Answer:
(A) 6

Explanation:
Since, 2x,x + 10,3x + 2 are in AP, this common difference will remain same.
x + 10 – 2x = (3x + 2) – (x + 10)
10 – x = 2x – 8
2x = 18
x = 6

II. Read the following text and answer the below questions: Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month.
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 2

Question 1.

The amount paid by him in 30th installment is

(A) 3900
(B) 3500
(C) 3700
(D) 3600
Answer:
(A) 3900

Explanation:
a = 1000
d = 100
a80= a + (n -1 )d
= 1000 – (30 – 1)100
= 1000 + 2900

Question 2.

The amount paid by him in the 30 installments is

(A) 37000
(B) 73500
(C) 75300
(D) 75000
Answer:
(B) 73500

Explanation:
Sum of 30 installments
= \(\frac{n}{2}\)[2a + (n – 1 ) d ]
= \(\frac{30}{2}\) [2 x 1000 + (30-1)100]
= 15[2000 + 2900]
= 15 x 4900
= 73500
Total Amount paid in 30 installments = ₹ 73500

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 3.

What amount does he still have to pay after 30th installment ?

(A) 45500
(B) 49000
(C) 44500
(D) 54000
Answer:
(C) 44500

Question 4.

If total installments are 40 then amount paid in the last installment ?

(A) 4900
(B) 3900
(C) 5900
(D) 9400
Answer:
(A) 4900

Explanation:
Amount paid in 40th installment, a40
Formula a + (n -1 )d
a40 = 1000 + (40 – 1)100
= 1000 + 3900
= 4900

Question 5.

The ratio of the 1st installment to the last installment is

(A) 1: 49
(B) 10: 49
(C) 10: 39
(D) 39:10
Answer:
(B) 10: 49

III. Read the following text and answer the below questions: Jaspal Singh takes a loan from a bank for his car. Jaspal Singh repays his total loan of ₹ 118000 by paying every month starting with the first installment of ₹ 1000. If he increases the installment by ₹ 100 every month.
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 3

Question 1.

If the given problem is based on A.P., then what is the first term and common difference ?

(A) 1000,100
(B) 100,1000
(C) 100,100
(D) 1000,1000
Answer:
(A) 1000,100

Explanation:
The number involved in this case form an A.P. in which first term (a) = 1000 and common difference (d) = 100.

Question 2.

The amount paid by him in 25th installment is:

(A) ₹ 3300
(B) ₹ 3200
(C) ₹ 3400
(D) ₹ 3500
Answer:
(C) ₹ 3400

Explanation:
The amount paid by him in 25th installment is:
T25 = a + 24d
= 1000 + 24 × 100
= 1000 + 2400
= ₹ 3400.

Question 3.

The amount paid by him in 30th installment is

(A) ₹ 3900
(B) ₹ 3500
(C) ₹ 3000
(D) ₹ 3600
Answer:
(A) ₹ 3900

Explanation:
The amount paid by him in 30th installment,
T30 = a + 29d
= 1000 + 29 × 100
= 1000 + 2900 = ₹ 3900.

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 4.

The total amount paid by him in 25th and 30th installment is:

(A) ₹ 7500
(B) ₹ 7300
(C) ₹ 7800
(D) ₹ 7600
Answer:
(B) ₹ 7300

Explanation:
Total amount paid by him in 25th and 30th installment = ₹ (3400 + 3900) = ₹ 7300.

Question 5.

The difference amount paid by him in 26th and 28th installment is:

(A) ₹400
(B) ₹100
(C) ₹ 500
(D) ₹ 200
Answer:
(D) ₹ 200

Explanation:
The amount paid by him in 26th installment,
T26 = a + 25 d
= 1000 + 25 x 100
= 1000 + 2500 = ₹ 3500
The amount paid by him in 28th installment,
T28 = a + 27d
= 1000 + 27 × 100
= 1000 + 2700
= ₹ 3700
∴ The difference amount paid by him in 26th and 28th installment is:
= ₹ (3700-3500)
= ₹ 200.

IV. Read the following text and answer the below questions:
A ladder has rungs 25 cm apart, (see the below).
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 4
The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and the bottom rungs are \(2 \frac{1}{2}\) m apart.

Question 1.

The top and bottom rungs are apart at a distance:

(A) 200 cm
(B) 250 cm
(C) 300 cm
(D) 150 cm
Answer:
(B) 250 cm

Explanation:
Since the top and the bottom rungs are apart by \(2 \frac{1}{2}\) m = \(\frac{5}{2}\)m
= \(\frac{5}{2}\) x 100m
= 250 cm

Question 2.

Total number of the rungs is:

(A) 20
(B) 25
(C) 11
(D) 15
Answer:
(C) 11

Exploration:
The distance between the two rungs is 25 cm.
Hence, the total number of rungs = \(\frac{250}{25}\) +1 = 11.

Question 3.

The given problem is based on A.P. find its first term.

(A) 25
(B) 45
(C) 11
(D) 13
Answer:
(A) 25

Explanation:
The length of the rungs increases from 25 to 45 and total number of rungs is 11. Thus, this is in the form of an A.P., whose first term is 25.

Question 4.

What is the last term of A.P. ?

(A) 25
(B) 45
(C) 11
(D) 13
Answer:
(B) 45

Explanation:
Total number of terms, n = 11 and the last term, T11 = 45.

MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Question 5.

What is the length of the wood required for the rungs ?

(A) 385
(B) 538
(C) 532
(D) 382
Answer:
(A) 385

Explanation:
The required length of the wood,
S11 = \(\frac{11}{2}\)[25 + 45]
= \(\frac{11}{2}\)× 70
= 385 cm.

MCQ Questions for Class 10 Maths with Answers

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