# 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

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