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Class 10 Exam  >  Class 10 Tests  >  Practice Test: Arithmetic Progressions - Class 10 MCQ

Practice Test: Arithmetic Progressions - Class 10 MCQ


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15 Questions MCQ Test - Practice Test: Arithmetic Progressions

Practice Test: Arithmetic Progressions for Class 10 2024 is part of Class 10 preparation. The Practice Test: Arithmetic Progressions questions and answers have been prepared according to the Class 10 exam syllabus.The Practice Test: Arithmetic Progressions MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Arithmetic Progressions below.
Solutions of Practice Test: Arithmetic Progressions questions in English are available as part of our course for Class 10 & Practice Test: Arithmetic Progressions solutions in Hindi for Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Practice Test: Arithmetic Progressions | 15 questions in 25 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study for Class 10 Exam | Download free PDF with solutions
Practice Test: Arithmetic Progressions - Question 1
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If the sum of first n terms of an AP be 3n2 + n and it's common difference is 6, then its first term is :

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Practice Test: Arithmetic Progressions - Question 2
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If 7th and 13th terms of an A.P. be 34 and 64, respectively, then it's 18th term is :

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Practice Test: Arithmetic Progressions - Question 3
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The sum of all 2-digit odd positive numbers is :

Detailed Solution for Practice Test: Arithmetic Progressions - Question 3

Here a = 11 and d = 2, tn= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But tn = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45  in sum of the n terms we obtain
⇒ s45 = (45/2)(2×11 + (45 -1)2)
⇒ s45 = (45/2)(110)
⇒ s45 = 45×55.
⇒  s45 = 2475.
∴ sum of all two digit odd positive numbers = 2475.

Practice Test: Arithmetic Progressions - Question 4
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The fourth term of an A.P. is 4. Then the sum of the first 7 terms is :
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Practice Test: Arithmetic Progressions - Question 5
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In an A.P., s1 = 6, s7 = 105, then sn : sn-3 is same as :
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Practice Test: Arithmetic Progressions - Question 6
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In an A.P. s3 = 6, s6 = 3, then it's common difference is equal to :
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Practice Test: Arithmetic Progressions - Question 7
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The number of terms common to the two A.P. s 2 + 5 + 8 + 11 + ...+ 98 and 3 + 8 + 13 + 18 +...+198
Detailed Solution for Practice Test: Arithmetic Progressions - Question 7

For first A.P
2+5+8+11+......+98
a=2,an​=98,d=3
an​=a+(n−1)d
98=2+(n−1)3
98=2+3n−3
3n=99
n=33
Number of term =33
For first A.P
3+8+13+18+......+198
a=3,an​=198,d=5
an​=a+(n−1)d
198=3+(n−1)5
198=3+5n−5
5n=200
n=40
No of terms =40
Common terms=40−33=7

Practice Test: Arithmetic Progressions - Question 8
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The first, second and last terms of an A.P. are a,b and 2a. The number of terms in the A.P. is :
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Practice Test: Arithmetic Progressions - Question 9
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Sum of first 5 terms of an A.P. is one fourth of the sum of next five terms. If the first term = 2, then the common difference of the A.P. is :
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Practice Test: Arithmetic Progressions - Question 10
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If sn denotes the sum of first n terms of an A.P., whose common difference is d, then sn – 2sn-1 + sn-2 (n >2) is equal to :
Detailed Solution for Practice Test: Arithmetic Progressions - Question 10

sn​−2sn−1​+sn−2 ​= (sn​−sn−1​)−(sn−1​−sn−2​)

= an​−an−1 ​[∵(sn​−sn−1​)= an​]

= [a+(n−1)d]−[a+(n−2)d]

= a+nd−d−a−nd+2d

= d

Practice Test: Arithmetic Progressions - Question 11
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The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is :
Detailed Solution for Practice Test: Arithmetic Progressions - Question 11
The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
tn = 97
nth term of an AP is tn = a + (n – 1)d
97 = 10 + (n – 1)3
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
Therefore, n = 90/3 = 30
Recall sum of n terms of AP, 
Sn = n/2[2a + (n-1)d]

S30 = 30/2[2(10) + (30-1)3]
= 15[20 + 87] = 15 x 107 = 1605
Hence sum of 2-digit numbers which when divided by 3 yield 1 as remainder is 1605.
Practice Test: Arithmetic Progressions - Question 12
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If {an} = {2.5, 2.51, 2.52,...} and {bn} = {3.72, 3.73, 3.74,...} be two AP's, then a100005 – b100005 =
Detailed Solution for Practice Test: Arithmetic Progressions - Question 12

Observing both the AP’s we see that the common difference of both the AP’s is same ,so difference between their corresponding terms will be same ie,a1-b1=2.5-3.72=-1.22
a2-b2=2.51-3.73=-1.22
 So , a100005-b100005=-1.22

Practice Test: Arithmetic Progressions - Question 13
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If A1 and A2 be the two A.M.s between two numbers a and b, then (2A1 – A2) (2A2 – A1) is equal to :
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Practice Test: Arithmetic Progressions - Question 14
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 In an Arithmetic Progression, if a = 28, d = -4, n = 7, then an is:
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Explanation: For an AP,

an = a+(n-1)d

= 28+(7-1)(-4)

= 28+6(-4)

= 28-24

an=4

Practice Test: Arithmetic Progressions - Question 15
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The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
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