Chapter 5 Arithmetic Progressions- MCQ Online Test 1 Class 10 Maths
1. Find the value of ‘k’ for which the numbers x, 2x + k, 3x + 6 are in A.P?
a. 4
b. 5
c. 3
d. 2
2. What will be the next term of the A.P. √18, √32 and √50?
a. √72
b. √18
c. √36
d. √12
3. If a, b and c are in A.P., then the relation between them is given by
a. 2b = a + c
b. 3b = a + c
c. b = 2a + c
d. 2b = a + c
4. What are the next two terms of the AP k, 2k + 1, 3k + 2, 4k + 3, ………… ?
a. 5k + 4
b. 6k + 5
c. Both a and b
d. None of these
5. Which of the following is not an A.P.?
a. a, 2a , 3a ….
b. 2, 4, 8, 16…..
c. Both a and b
d. None of these
6. If a, 7, b, 23, c are in A.P. then find the value of ‘c’?
a. 25
b. 31
c. 23
d. 17
7. Find the first term of an A.P., if it’s Sn = n2+ 2n?
a. 2
b. 3
c. 5
d. 6
8. If the common difference of an A.P. is 5, then find the value of a20−a13?
a. 25
b. 29
c. 35
d. 40
9. If 9 times the 9th term of an A.P. is equal to 11 times the 11th term, then find its 20th term?
a. 0
b. 4
c. 5
d. 9
10. The 7th term from the end of the A.P. – 11, – 8, – 5……… if the last number of AP is 49?
a. 18
b. 31
c. 15
d. 34
Chapter - 5 Arithmetic Progressions Quiz-1 | Math Class 10th
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Question 1 of 10
1. Question
Find the value of ‘k’ for which the numbers x, 2x + k, 3x + 6 are in A.P?
Correct
If the numbers x, 2x+k, 3x+6 are in A.P.
then 2x+k−x = 3x + 6 − 2x−k
x + k = x+6−k
2k = 6
k = 3Incorrect
If the numbers x, 2x+k, 3x+6 are in A.P.
then 2x+k−x = 3x + 6 − 2x−k
x + k = x+6−k
2k = 6
k = 3 -
Question 2 of 10
2. Question
What will be the next term of the A.P. √18, √32 and √50?
Correct
Incorrect
-
Question 3 of 10
3. Question
If a, b and c are in A.P., then the relation between them is given by
Correct
We know that if a, b, c … are in AP then 2nd term – 1st term = 3rd term – 2nd term
b – a = c – b
2b = a + cIncorrect
We know that if a, b, c … are in AP then 2nd term – 1st term = 3rd term – 2nd term
b – a = c – b
2b = a + c -
Question 4 of 10
4. Question
What are the next two terms of the AP k, 2k + 1, 3k + 2, 4k + 3, ………… ?
Correct
Given: k,2k+1,3k+2,4k+3,……………
Here d = 2k+1−k = k+1
Therefore, the next two terms are
4k +3 +k +1= 5k + 4 and 5k+4+k+1 = 6k + 5Incorrect
Given: k,2k+1,3k+2,4k+3,……………
Here d = 2k+1−k = k+1
Therefore, the next two terms are
4k +3 +k +1= 5k + 4 and 5k+4+k+1 = 6k + 5 -
Question 5 of 10
5. Question
Which of the following is not an A.P.?
Correct
In 2, 4, 8, 16…
d = 4−2 = 2
And d = 8 – 4 = 4
Also d = 16 – 8 = 8
Here, common difference is not same for all terms, therefore, it is not an AP.Incorrect
In 2, 4, 8, 16…
d = 4−2 = 2
And d = 8 – 4 = 4
Also d = 16 – 8 = 8
Here, common difference is not same for all terms, therefore, it is not an AP. -
Question 6 of 10
6. Question
If a, 7, b, 23, c are in A.P. then find the value of ‘c’?
Correct
Let d be the common difference. Then
a5 = c = a + 4d ………. (i)
a2= a + d = 7 ………. (ii)
a4 = a + 3d = 23 ………. (iii)
Solving eq. (ii) and (iii),
we get a = −1 and d = 8
∴c = a + 4d = −1 + 4 × 8 = 31Incorrect
Let d be the common difference. Then
a5 = c = a + 4d ………. (i)
a2= a + d = 7 ………. (ii)
a4 = a + 3d = 23 ………. (iii)
Solving eq. (ii) and (iii),
we get a = −1 and d = 8
∴c = a + 4d = −1 + 4 × 8 = 31 -
Question 7 of 10
7. Question
Find the first term of an A.P., if it’s Sn = n2+ 2n?
Correct
Given: Sn = n2 + 2n
Putting n = 1, we get
S = a = (1)2 + 2 × 1 = 1 + 2 = 3Incorrect
Given: Sn = n2 + 2n
Putting n = 1, we get
S = a = (1)2 + 2 × 1 = 1 + 2 = 3 -
Question 8 of 10
8. Question
If the common difference of an A.P. is 5, then find the value of a20−a13?
Correct
Given: a20 − a13 and d = 5
a20−a13 = a + (20−1) d − [a + (13−1) d]= a + (20−1) ×5 − [a + (13−1) ×5]
a20 − a13 = a + 95 − [a + 60]
= a + 95 – a – 60 = 35Incorrect
Given: a20 − a13 and d = 5
a20−a13 = a + (20−1) d − [a + (13−1) d]= a + (20−1) ×5 − [a + (13−1) ×5]
a20 − a13 = a + 95 − [a + 60]
= a + 95 – a – 60 = 35 -
Question 9 of 10
9. Question
If 9 times the 9th term of an A.P. is equal to 11 times the 11th term, then find its 20th term?
Correct
According to question,
9 × a9 =11 × a11
9[a+(9−1)d] = 11[a+(11−1)d]
9[a+8d] = 11[a+10d]
9a + 72d =11a + 110d
2a + 38d = 0
a + 19d = 0
a + (20−1) d = 0
a20 = 0Incorrect
According to question,
9 × a9 =11 × a11
9[a+(9−1)d] = 11[a+(11−1)d]
9[a+8d] = 11[a+10d]
9a + 72d =11a + 110d
2a + 38d = 0
a + 19d = 0
a + (20−1) d = 0
a20 = 0 -
Question 10 of 10
10. Question
The 7th term from the end of the A.P. – 11, – 8, – 5……… if the last number of AP is 49?
Correct
Reversing the given A.P., we have
49, 46, 43 …−11
Here, a = 49, d = 46 – 49 = −3 and n=7
∴ an= a + (n−1) d
a7= 49 + (7−1) × (−3)
= 49 + 6 × (−3)
a7 = 49 −18 = 31Incorrect
Reversing the given A.P., we have
49, 46, 43 …−11
Here, a = 49, d = 46 – 49 = −3 and n=7
∴ an= a + (n−1) d
a7= 49 + (7−1) × (−3)
= 49 + 6 × (−3)
a7 = 49 −18 = 31