RD Sharma Test: Arithmetic Progressions - Class 10 MCQ

# RD Sharma Test: Arithmetic Progressions - Class 10 MCQ

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## 25 Questions MCQ Test Mathematics (Maths) Class 10 - RD Sharma Test: Arithmetic Progressions

RD Sharma Test: Arithmetic Progressions for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The RD Sharma Test: Arithmetic Progressions questions and answers have been prepared according to the Class 10 exam syllabus.The RD Sharma Test: Arithmetic Progressions MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RD Sharma Test: Arithmetic Progressions below.
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RD Sharma Test: Arithmetic Progressions - Question 1

### Progressions with equal common difference are known as

Detailed Solution for RD Sharma Test: Arithmetic Progressions - Question 1

Progressions with equal common difference are known as Arithmetic Progression.

RD Sharma Test: Arithmetic Progressions - Question 2

### The first term of an A.P., if its Sn = n2+2n is

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RD Sharma Test: Arithmetic Progressions - Question 3

### If the second term of an AP is 13 and its fifth term is 25, then its 7th term is

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RD Sharma Test: Arithmetic Progressions - Question 4

In an A.P., if am = 1/n and an = 1/m, then amn =

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∴ amn = a + (mn - 1) d = 1/mn + (mn - 1)

RD Sharma Test: Arithmetic Progressions - Question 5

The first term of an AP is 5, the last term is 45 and the sum is 400. The number of terms is

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RD Sharma Test: Arithmetic Progressions - Question 6

The value of ‘k’ for which the numbers x, 2x + k, 3x + 6 are in A.P. is

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RD Sharma Test: Arithmetic Progressions - Question 7

If the common difference of an A.P. is 5, then the value of a20 − a13 is

Detailed Solution for RD Sharma Test: Arithmetic Progressions - Question 7

Given: a20 - a13 and d = 5
⇒ a20 - a13 = a + (20 - 1) d - [a + (13 - 1) d] = a + (20 - 1) x 5 - [a + (13 - 1) x 5]
⇒ a20 - a13 - a = a + 95 - [a + 60]
= a+95 - a - 60 = 35

RD Sharma Test: Arithmetic Progressions - Question 8

Two APs have the same common difference. The difference between their 100th terms is 100, then the difference between their 1000th terms is

Detailed Solution for RD Sharma Test: Arithmetic Progressions - Question 8

The formula for nth term of an AP is aₙ = a + (n - 1) d

Here, aₙ is the nth term, a is the first term, d is the common difference and n is the number of terms.

For first A.P., a₁₀₀ = a₁ + (100 - 1) d

a₁₀₀ = a₁ + 99d ------ (1)

a₁₀₀₀ = a₁ + (1000 - 1)d

a₁₀₀₀ = a₁ + 999d ------ (2)

For second A.P.,

b₁₀₀ = b₁+ (100 - 1)d

b₁₀₀ = b₁ + 99d ------ (3)

b₁₀₀₀ = b₁ + (1000 - 1)d

b₁₀₀₀ = b₁ + 999d ------ (4)

Given that, difference between 100th term of these A.P.s = 100

Thus, from equations (1) and (3) we have

(a₁ + 99d ) - (b₁ + 99d ) = 100

a₁ - b₁ = 100 ...(5)

Difference between 1000th terms of these A.P.s

Thus, from equations (2) and (4) we have

(a₁ + 999d ) - (b₁ + 999d ) = a₁ - b₁

But a₁ - b₁ = 100 [From equation(5)]

Hence, the difference between the 1000th terms of these A.P. will be 100.

RD Sharma Test: Arithmetic Progressions - Question 9

The common difference of the A.P whose Sn = 3n2+ 2n is

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Thus, initial term of the A.P. is 5 and the common difference is 6.

RD Sharma Test: Arithmetic Progressions - Question 10

The sum of (a + b), (a – b), (a – 3b), …….. to 22nd term is

Detailed Solution for RD Sharma Test: Arithmetic Progressions - Question 10

First Term =  a+b
Second Term =  a-b
Common Difference is a-b-a-b = -2b.

Summation of 22 terms of an A.P. is
n/2 [ 2a + (n-1)d ]
22/2 [2 ( a+b) + (22-1)-2b ]
11  [ 2a+2a-b + (21)-2b  ]
11  [  2a+2b - 42b ]
11  [ 2a - 40b ]
22a - 40b.

So, the summation of given A.P. for 22 terms is 22a - 40b

RD Sharma Test: Arithmetic Progressions - Question 11

The next term of the A.P. √18 , √32 and √50 is

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RD Sharma Test: Arithmetic Progressions - Question 12

If 9 times the 9th term of an A.P. is equal to 11 times the 11th term , then its 20th term is

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RD Sharma Test: Arithmetic Progressions - Question 13

The 17th term of an AP exceeds its 10th term by 7, then the common difference is

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RD Sharma Test: Arithmetic Progressions - Question 14

The sum of three terms of an A.P. is 72, then its middle term is

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Let the middle term be a, then first term is a−d and next term is a+d

RD Sharma Test: Arithmetic Progressions - Question 15

The sum of odd numbers between 0 and 50 is

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Odd numbers between 0 and 50 are 1, 3, 5, 7, ………, 49 Here a = 1,d = 3−1 = 2 and

RD Sharma Test: Arithmetic Progressions - Question 16

If a, b and c are in A.P., then the relation between them is given by

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If a, b and c are in A.P., then

RD Sharma Test: Arithmetic Progressions - Question 17

The 7th term from the end of the A.P. – 11, – 8, – 5, ……., 49 is

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RD Sharma Test: Arithmetic Progressions - Question 18

The number of three digit numbers divisible by 7 is

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Three digits numbers divisible by 7 are 105, 112, 119,.........., 994
Here, a= 105, d = 112 - 105 = 7, an = 994

RD Sharma Test: Arithmetic Progressions - Question 19

The first and last terms of an A.P. are 1 and 11. If their sum is 36, then the number of terms will be

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RD Sharma Test: Arithmetic Progressions - Question 20

If 1 + 4 + 7 + ……. + k = 287, then the value of ‘k’ is

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RD Sharma Test: Arithmetic Progressions - Question 21

If the angles of a right angled triangle are in A.P. then the angles of that triangle will be

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Let the three angles of a triangle be a - d, a and a + d.

Therefore, one angle is of 60° and other is 90° (given). Let third angle be x° . then

Therefore the angles of the right angled triangle are 30°, 60°, 90°.

RD Sharma Test: Arithmetic Progressions - Question 22

The 10th term of an A.P. 2, 7, 12, …….. is

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RD Sharma Test: Arithmetic Progressions - Question 23

If a= 4 and an = 4an−1+3, n  >1, then the value of a4 is

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RD Sharma Test: Arithmetic Progressions - Question 24

The number of terms of the A.P. 5, 8, 11, 14, ……. to be taken so that the sum is 258 is

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RD Sharma Test: Arithmetic Progressions - Question 25

A sum of Rs.700 is to be used to award 7 prizes. If each prize is Rs.20 less than its preceding prize, then the value of the first prize is

Detailed Solution for RD Sharma Test: Arithmetic Progressions - Question 25

Let the first prize be a.
The seven prizes form an AP with first term a and common difference d = −20
Now the sum of all seven prizes = Rs. 700

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## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests