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Cbse Class 10 Maths Chapter 5 Arithmetic Progression Objective Questions

Here you get the CBSE Class 10 Mathematics chapter 5- Arithmetic Progression Objective Questions. You will find only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is also provided with a detailed solution, which the students can refer to at a later stage.

All these questions are crucial as it helps the students to prepare for CBSE Class 10 Mathematics Board Examination. As per the latest modified exam pattern, the question papers will contain more of objective type questions. This topic is also very important from the standpoint of competitive exams. Find the CBSE Class 1o Maths Objective Questions below :

5.1 General Term of AP (7 MCQs From The Topic)

5.2 Introduction to AP (6 MCQs Listed From The Topic)

5.3 Sum of Terms in AP (7MCQs From This Topic)

Using formula an=a+ (n−1) d to find n th  term of arithmetic progression, we get

Therefore, there are 34 terms in the given arithmetic progression.

Common difference, d= (31/2) −18=−5/2

Using formula a n =a+ (n−1) d to find n th  term of arithmetic progression, we get

Therefore, there are 27 terms in the given arithmetic progression.

Solution: We are given that a 11 =38 and a 16 =73 where, a 11  is the 11 th  term and  a 16  is the 16 th  term of an AP.

Using formula a n =a+ (n−1) d to find nth term of arithmetic progression, we get

equation (ii) – equation (i) gives,

Substituting (iii) in (i) we get, a = -32

Therefore, 31 st  term of AP is 178.

Solution: It is given that 3rd and 9th term of AP are 4 and -8 respectively.

Where, a 3  and a 9  are third and ninth terms respectively.

Using formula a n =a+(n−1)d to find n th  term of arithmetic progression, we get

Substituting in equation (ii) , we have

Therefore, first term a = 8 and Common Difference d =−2

We know  a n  = a+ (n−1) d (where a n  is the n th  term)

Solution: Let’s first calculate 54 th  of the given AP.

Using formula a n =a+ (n−1) d,   to find n th   term of arithmetic progression, we get

We want to find which term is 132 more than its 54 th  term. Let’s suppose it is n th  term which is 132 more than 54 th  term.

Therefore, 65 th  term is 132 more than its 54 th  term.

Solution: Let first term of first AP = a

It is given that their common difference is same. Let their common difference be d.

It is given that difference between their 100 th  terms is 100. Using formula a n  = a + (n−1) d,  to find n th  term of arithmetic progression, we can say that

a+ (100−1) d− (a′ +(100−1)d)=a+99d−a′−99d=100

We want to find difference between their 1000 th  terms which means we want to calculate:

a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′−999d=a−a′

Putting equation (1) in the above equation we get,

a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′-999d=a−a′=100

Therefore, difference between their 1000 th  terms would be equal to 100

Solution: We have an AP starting at 105 because it is the first three digit number divisible by 7.

AP will end at 994 because it is the last three digit number divisible by 7.

Therefore, we have an AP 105,112,119…994

Common difference, d = 112 – 105 = 7

Using formula a n =a+ (n−1) d, to find  n th  term of arithmetic progression, we can say that

994 is the 128 th  term of AP. Therefore, there are 128 terms in AP. In other words, we can also say that there are 128 three digit numbers divisible by 7.

Solution: Let’s first consider AP: 63, 65, 67…..

Using formula a n =a+ (n−1) d,   to find  n th  term of arithmetic progression, we can say that

a n =63+ (n−1) (2)                      (1)

Using formula a n =a+ (n−1) d,   to find  n th  term of arithmetic progression, we can say that

a n =3+(n−1)(7)                    (2)

According to the given condition, we can write

Therefore, 13th terms of both the AP’s are equal.

Solution: Given, (x+1), 3x, (4x+2) are in AP.

Hence, the difference of two consecutive terms will be same.

The n th  term of an AP is given by

Solution: Let S 10  be the sum of first 10 terms and S5 be the sum of first 5 terms.

⇒ (10/2) [2a + (10-1) d] = 4 × (5/2) [2a + (5-1) d]

⇒ (10/2) [2a+ 9d] = 4 × (5/2) [2a+ 4d]

Answer: (iii) -1, -1.25, -1.5, -1.75…… and

Solution: Consider each list of numbers:

Difference between the first two terms = 4 – 2 = 2

Difference between the third and second term = 8 – 4 = 4

Since a 2 –a 1 ≠a 3 –a 2 , this sequence is not an AP

Difference between the first two terms = 3 – 2 = 1

Difference between the third and second term = 5 – 3 = 2

Since a 2 –a 1 ≠a 3 –a 2 , this sequence is not an AP

Difference between the first two terms = -1.25 – (-1) = -0.25

Difference between the third and second term = -1.5 – (-1.25) = -0.25

Difference between the fourth and third term = -1.75 – (-1.5) = -0.25

Since a 2 –a 1 =a 3 −a 2 =a 4 −a 3 , this sequence is an AP

Difference between the first two terms = -1 – 1 = -2

Difference between the third and second term = -3 – (-1) = -2

Difference between the fourth and third term = -5 – (-3) = -2

Since a 2 –a 1 =a 3 −a 2 =a 4 −a 3 , this sequence is an AP

Answer: (C)The difference between two consecutive terms should be constant.

Solution: Let a1, a 2 , a 3 , a 4 ,   a 5 , a 6,  a 7,  a 8 … be a sequence.

For this sequence to be an AP, the difference between any two consecutive terms should be constant.

This difference is called the common difference of the AP and is denoted by d.

So an AP can also be represented in this form as well a, a+d, a+2d…

(i) 4, 4 + √3 , 4 + 2√3, 4 + 3√3, 4 + 4√3

(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333

Solution: Consider each series (i) 4, 4 + √3, 4 + 2 √3, 4 + 3 √3, 4 + 4 √3

Difference between first two consecutive terms = 4 + √3 – 4 = √3

Difference between third and second consecutive terms = 4 + 2 √3 – 4 + √3 =  √3

Difference between fourth and third consecutive terms = 4 + 3 √3– 4 + 2 √3 =  √3

Since a 2 –a 1  = a 3 −a 2  = a 4 −a 3

(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333

Difference between first two consecutive terms = 0.33-0.3 = 0.03

Difference between third and second consecutive terms = 0.333 – 0.33 = 0.003

Difference between first two consecutive terms = (6/5)–(3/5) = 3/5

Difference between third and second c onsecutive terms = (9/5)–(6/5) = 3/5

Difference between fourth and third consecutive terms = (12/5) – (9/5) = 3/5

Since  a 2 –a 1  = a 3 –a 2  = a 4 –a 3

Difference between first two consecutive terms = -1/5 – (- 1/5) = 0

Difference between third and second consecutive terms = -1/5 – (-1/5) = 0

Difference between fourth and third consecutive terms = -1/5 – (-1/5) = 0

Since a 2 –a 1  = a 3 –a 2  = a 4 –a 3

Applying formula, S n = n/2 (a+l) to find sum of n terms of AP, we get

Applying formula, S n =n/2(2a+ (n−1) d) to find sum of n terms of AP and putting value of n, we get

Using formula a n  = a + (n−1) d, to find nth term of arithmetic progression, we can say that

Applying formula, S n =n/2(2a+ (n−1) d) to find sum of n terms of AP and putting value of n, we get

Therefore, there are 38 terms and their sum is equal to 6973.

Solution: It is given that 22nd term is equal to 149.

Using formula a n =a+ (n−1) d,   to find n th  term of AP, we can say that

Applying formula, S n  = n/2(2a+ (n−1) d) to find Sum of n terms of AP and putting value of a, we get

Therefore, sum of first 22 terms of AP is equal to 1661.

Solution: It is given that second and third terms of AP are 14 and 18 respectively.

Using formula a n =a+ (n−1) d,   to find n th term of arithmetic progression, we can say that

⇒14=a+ d                                   (1)

⇒18=a+2d                                 (2)

These are equations consisting of two variables. We can solve them by the method of substitution.

Using equation (1), we can say that a=14−d

Putting value of a in equation (2), we can say that

Putting value of d in equation number (1), we can say that

Applying formula, S n = n/2(2a+ (n−1) d) to find sum of n terms of AP, we get

S 15 =51/2(20+ (51−1) 4) =51/2(20+200) =51×110 =5610

Therefore, sum of first 51 terms of an AP is equal to 5610.

Solution: Let the numbers are a−3d,a−d,a+d,a+3d

(a−3d) 2 + (a−d) 2  + (a+3d) 2  + (a+3d) 2 =120

Solution: Here, t 1  = 100, common difference, d = t 2  – t 1  = 97 – 100 = -3

t n  = t 1  + (n-1) d ⇒100 + (n-1)(-3) = 100 – 3(n-1) = 103 – 3n.

Let t n  be the first negative term. i.e,

t n  < 0 ⇒ 103 – 3n < 0 ⇒ n > 103/3 > 34

That is, the 35 th  term will be negative.

Sum of the first 34 terms = (34/2) (first term + last term)

Solution: The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.

The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ….

The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.

So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.

we know that in an A.P., the nth term a n =a 1 + (n−1)×d

Sum of the AP will therefore, be (101+198)/ 2 ×300= 164, 850

1. In an AP, if d = –4, n = 7, a n  = 4. Calculate a. (a) 20 (b) 7 (c) 28 (d) 6

2.What is the sum of first five multiples of 3? (a)75 (b) 55 (c) 65 (d) 45

In this chapter, concepts discussed include pattern in succeeding term obtained by adding a fixed number to the preceding te rms. Students will also see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems.

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