Chapter 5 Arithmetic Progressions- MCQ Online Test 3 Class 10 Maths
1. Find the common difference of the A.P whose sn= 3n2 + 7n?
a. 5
b. 6
c. 8
d. 9
2. The first and last terms of an A.P. are 1 and 11. If their sum is 36, then find the number of terms?
a. 8
b. 5
c. 6
d. 2
3. Find the number of multiples of 4 lie between 10 and 250?
a. 65
b. 60
c. 45
d. 50
4. Find the common difference of an A.P. in which a18 − a14 = 32
a. 32
b. 8
c. 16
d. 24
5. In an A.P. it is given that a = 5, d = 3 and an= 50, then find the value of ‘n’?
a. 25
b. 16
c. 30
d. 14
6. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms?
a. 20
b. 16
c. 28
d. 32
7. In an A.P., if sn= 3nn+ 2n then find the value of‘d’?
a. 8
b. 5
c. 6
d. 3
8. What is the sum of first n natural numbers?
a. n(n+1)
b. 2n(n+1)
c. n(n+1)/2
d. (2n-1)2n
9. The first term of an A.P. is ‘m’ and its common difference is ‘n’, then find its 10th term?
a. m + 9n
b. m – 9n
c. 2m + 9n
d. m + 3n
10. Find the 31st term of an AP whose first two terms are – 3 and 4?
a. 200
b. 207
c. 300
d. 315
Chapter - 5 Arithmetic Progressions Quiz-3 | Math Class 10th
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Question 1 of 10
1. Question
Find the common difference of the A.P whose sn = 3n2 + 7n?
Correct
Given: sn= 3n2 + 7n
Putting n = 1, 2, 3 we get
s1 = a = 3 × (1)2 + 7 × 1 = 3 + 7 = 10
s2 = 3 × (2)2 + 7× 2 = 12 + 14 = 26
s3 = 3 × (3)2 + 7× 3 = 27 + 21 = 48Incorrect
Given: sn= 3n2 + 7n
Putting n = 1, 2, 3 we get
s1 = a = 3 × (1)2 + 7 × 1 = 3 + 7 = 10
s2 = 3 × (2)2 + 7× 2 = 12 + 14 = 26
s3 = 3 × (3)2 + 7× 3 = 27 + 21 = 48 -
Question 2 of 10
2. Question
The first and last terms of an A.P. are 1 and 11. If their sum is 36, then find the number of terms?
Correct
Incorrect
-
Question 3 of 10
3. Question
Find the number of multiples of 4 lie between 10 and 250?
Correct
Multiples of 4 lie between 10 and 250 are
12, 16…… 248
Here a = 12, d = 16 – 12 = 4 and an = 248
∴ an = a + (n−1) d
248 = 12 + (n−1) × (4)
236 = (n−1) × (4)236 = (n−1) × (4)
236/4 = n – 1
n – 1 = 59
n = 60Incorrect
Multiples of 4 lie between 10 and 250 are
12, 16…… 248
Here a = 12, d = 16 – 12 = 4 and an = 248
∴ an = a + (n−1) d
248 = 12 + (n−1) × (4)
236 = (n−1) × (4)236 = (n−1) × (4)
236/4 = n – 1
n – 1 = 59
n = 60 -
Question 4 of 10
4. Question
Find the common difference of an A.P. in which a18−a14 = 32
Correct
Given: a18−a14 = 32
a + (18 − 1) d − [a + (14 − 1) d] = 32a + (18−1) d − [a + (14−1) d] = 32
a + 17d – a − 13d = 32
4d = 32
d = 8Incorrect
Given: a18−a14 = 32
a + (18 − 1) d − [a + (14 − 1) d] = 32a + (18−1) d − [a + (14−1) d] = 32
a + 17d – a − 13d = 32
4d = 32
d = 8 -
Question 5 of 10
5. Question
In an A.P. it is given that a = 5, d = 3 and an = 50, then find the value of ‘n’?
Correct
Given: a = 5, d = 3 and an = 50
∴an =a + (n−1) d
50 = 5 + (n−1) × 3
45 = (n−1) × 3
45/3 = n – 1
n – 1 = 15
n = 16Incorrect
Given: a = 5, d = 3 and an = 50
∴an =a + (n−1) d
50 = 5 + (n−1) × 3
45 = (n−1) × 3
45/3 = n – 1
n – 1 = 15
n = 16 -
Question 6 of 10
6. Question
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms?
Correct
Incorrect
-
Question 7 of 10
7. Question
In an A.P., if sn= 3n2 + 2n then find the value of‘d’?
Correct
Given: = 3n2+ 2n
s1= 3(1)2+ 2 × 1 = 3 + 2 = 5
a = 5s2= 3(2)2 + 2 X 2 = 3 X 4 + 4 = 12 + 4 = 16
a1+a2 = 16
a2 = 11
∴ d = a2 − a1 = 11− 5 = 6Incorrect
Given: = 3n2+ 2n
s1= 3(1)2+ 2 × 1 = 3 + 2 = 5
a = 5s2= 3(2)2 + 2 X 2 = 3 X 4 + 4 = 12 + 4 = 16
a1+a2 = 16
a2 = 11
∴ d = a2 − a1 = 11− 5 = 6 -
Question 8 of 10
8. Question
What is the sum of first n natural numbers?
Correct
Incorrect
-
Question 9 of 10
9. Question
The first term of an A.P. is ‘m’ and its common difference is ‘n’, then find its 10th term?
Correct
Given: a=m, d=n
∴a10= m + (10−1) n
m + 9nIncorrect
Given: a=m, d=n
∴a10= m + (10−1) n
m + 9n -
Question 10 of 10
10. Question
Find the 31st term of an AP whose first two terms are – 3 and 4?
Correct
Given: a = −3, d = 4− (−3) = 7
∴ a31= −3 + (31−1) × 7
= −3 + 30 × 7= −3 + 210
= 207
Incorrect
Given: a = −3, d = 4− (−3) = 7
∴ a31= −3 + (31−1) × 7
= −3 + 30 × 7= −3 + 210
= 207