Important Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Arithmetic progression is an arrangement of numbers that are placed in such a way that the difference between every consecutive number remains the same. For example 11, 14, 17, 20, 23. is an arrangement of numbers with a common difference of 3. Here each number (11,… 23) will be called a ‘term’ and so arithmetic progression can also be seen as a list of numbers in which every term has emerged by adding a specific number or value to the previous term except the initial term. This specific number will be called the common difference of this arithmetic progression and it can be ve, -ve, or ‘0’. It is basically to identify the pattern and use it to find a mathematical solution to a day-to-day problem or any mathematical equation.

Those who desire to practice the important questions in class 10 maths Chapter 5 Arithmetic Progression should pay attention to the following topics:

• Concept of arithmetic progression
• Terms in arithmetic progression
• Common difference in arithmetic progression

Important Questions Chapter 5 – Arithmetic Progressions

Q1. Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .

Solution:

Given AP: 11, 8, 5, 2, …

First term, a = 11

Common difference, d = a2 − a1 = 8 − 11 = −3

Let −150 be the nth term of this AP.

As we know, for an AP,

an = a + (n − 1) d

-150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

⇒ -164 = -3n

⇒ n = 164/3

Clearly, n is not an integer but a fraction.

Therefore, – 150 is not a term of the given AP.

Q2. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

Given that,

first term, a = 5

last term, l = 45

Sum of the AP, Sn = 400

As we know, the sum of AP formula is;

Sn = n/2 (a + l)

400 = n/2 (5 + 45)

400 = n/2 (50)

Number of terms, n = 16

As we know, the last term of AP can be written as;

Last term, l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

Therefore, the Common difference is d = 40/15 = 8/3.

Q3. What is the common difference of an A.P. in which a21 – a7 = 84?

Solution:

a21 – a7 = 84 …[Given

∴ (a + 20d) – (a + 6d) = 84 …[an = a + (n – 1)d

20d – 6d = 84

14d = 84 ⇒ d 84/14 = 6

Q4. Which term of the progression 4, 9, 14, 19, … is 109?

Solution:

Here, d = 9 -4 = 14 -9 = 19 – 14 = 5

∴ Difference between consecutive terms is constant.

Hence it is an A.P.

Given: First term, a = 4, d = 5, an = 109 (Let)

∴ an = a + (n – 1) d … [General term of A.P.

∴ 109 = 4 + (n – 1) 5

⇒ 109 – 4 = (n – 1) 5

⇒ 105 = 5(n − 1) ⇒ n – 1 = 105/5 = 21

⇒ n = 21 + 1 = 22 ∴ 109 is the 22nd term

Q5. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term. 

Solution:

Let 1st term = a, Common difference = d

a4 = 0 a + 3d = 0 ⇒ a = -3d … (i)

To prove: a25 = 3 × a11

a + 24d = 3(a + 10d) …[From (i)

⇒ -3d + 24d = 3(-3d + 10d)

⇒ 21d = 21d

From above, a25 = 3(a11) Hence proved

Q6. Find how many two-digit numbers are divisible by 6?

Solution:

12, 18, 24, …,96

Here, a = 12, d = 18 – 12 = 6, an = 96

a + (n – 1)d = an

∴ 12 + (n – 1)6 = 96

⇒ (n − 1)6 = 96 – 12 = 84

⇒ n – 1 = 84/6 = 14

⇒ n = 14 + 1 = 15

∴ There are 15 two-digit numbers divisible by 6.

Q7. How many natural numbers are there between 200 and 500, which are divisible by 7? 

Solution:

203, 210, 217, …, 497

Here a = 203, d = 210 – 203 = 7, an = 497

∴ a + (n – 1) d = an

203 + (n – 1) 7 = 497

(n – 1) 7 = 497 – 203 = 294

n – 1 = 294/7 = 42 ∴ n = 42 + 1 = 43

∴ There are 43 natural nos. between 200 and 500 which are divisible by 7.

Q8. How many two-digit numbers are divisible by 3?

Solution:

Two-digit numbers divisible by 3 are:

12, 15, 18, …, 99

Here, a = 12, d = 15 – 12 = 3, an = 99

∴ a + (n – 1)d = an

12 + (n – 1) (3) = 99

(n – 1) (3) = 99 – 12 = 87

n – 1 = 87/3 = 29

∴ n = 29 + 1 = 30

∴ There are 30 such numbers.

Q9. How many three-digit natural numbers are divisible by 7?

Solution:

“3 digits nos.” are 100, 101, 102, …, 999

3 digits nos. “divisible by 7” are:

105, 112, 119, 126, …, 994

a = 105, d = 7, an = 994, n = ?

As a + (n – 1)d = 994 = an

∴ 105 + (n – 1)7 = 994

(n − 1)7 = 994 – 105

(n – 1) = 889/7 = 127

∴ n = 127 + 1 = 128

Q10. Find the sum of all three digit natural numbers, which are multiples of 11.

Solution:

To find: 110 + 121 + 132 + … + 990

Here a = 110, d = 121- 110 = 11, an = 990

∴ a + (n – 1)d = 990

110 + (n – 1).11 = 990

(n – 1). 11 = 990 – 110 = 880

(n – 1) = 880 = 80

n = 80 + 1 = 81

As Sn = n/2 (a1 + an)

∴ S81 = 81/2 (110 + 990)

= 81/2 (1100) = 81 × 550 = 44,550