CBSE Class 6 Maths (Ganita Prakash) Important Questions Chapter 3 - Number Play

1. What is the sum of the first 10 natural numbers?  

Ans:  \[S = \frac{n(n + 1)}{2} = \frac{10(10 + 1)}{2} = \frac{10 \times 11}{2}\]= 55

2. If a number is increased by 8 and the result is 15, what is the number?  

Ans: Let the number be x.  

x + 8 = 15

x = 15 - 8 = 7 

3. What is the 5th term in the sequence 2, 4, 6, 8, ...?  

Ans: The sequence is an arithmetic progression where the first term a = 2 and the common difference d = 2 

The nth term is given by $ a_n = a + (n-1)d $.  

So, $ a_5$ = 2 + (5-1) $\times $2 = 2 + 8 = 10 .

4. What is the sum of the first 5 odd numbers?  

Ans: The first 5 odd numbers are 1, 3, 5, 7, and 9.  

Sum =  1 + 3 + 5 + 7 + 9 = 25   

Alternatively, S = $n^{2}$ = $5^{2}$ = 25).

5. What is the sum of the first 4 even numbers?  

Ans: The first 4 even numbers are 2, 4, 6, and 8.  

Sum = 2 + 4 + 6 + 8 = 20 

Alternatively, S = n(n + 1) = 4 $\times $5 = 20.

6. If the 3rd term of an arithmetic sequence is 12 and the common difference is 4, what is the first term?  

Ans: Let the first term be a

The 3rd term is a + 2d = 12 

So,  a + 2 $\times $4 = 12 

a + 8 = 12  

a = 12 - 8 = 4.

7. How many numbers are there between 1 and 100 that are multiples of 5?  

Ans: The multiples of 5 from 1 to 100 are 5, 10, 15, ..., 100.  

This is an arithmetic sequence where a = 5 , d = 5, and $a_n $= 100 

n = $\dfrac{100 - 5}{5} $+ 1 = 20 . There are 20 multiples of 5.

8. If the 4th term of a sequence is 20 and the common difference is 3, what is the 1st term?  

Ans: Let the first term be a 

The 4th term is a + 3d = 20 

So, a + 3 $\times $3 = 20 

a + 9 = 20 

a = 20 - 9 = 11 

9. Find the sum of the first 6 squares of natural numbers.  

Ans: The squares of the first 6 natural numbers are $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2} $.  

Sum = 1 + 4 + 9 + 16 + 25 + 36 = 91 

Alternatively, using the formula S = $\dfrac{n(n + 1)(2n + 1)}{6} $

S =$ \dfrac{6(6 + 1)(2 \times 6 + 1)}{6} $= $\dfrac{6 \times 7 \times 13}{6} $= 91

10. What is the 7th term in the sequence: 5, 10, 15, 20, ...?  

Ans: This sequence has a first term a = 5 and a common difference d = 5

The 7th term is given by $a_n $= a + (n - 1)d 

So, $a_7 $= 5 + (7 - 1) $\times $5 = 5 + 30 = 35 

11. What is the sum of the first 15 natural numbers?  

Ans: S = $\dfrac{n(n + 1)}{2}$ = $\dfrac{15(15 + 1)}{2} $= $\dfrac{15 \times 16}{2} $= 120 

12. If a number is decreased by 4 and the result is 9, what is the number?  

Ans: Let the number be x 

x - 4 = 9 

x = 9 + 4 = 13 

13. What is the 6th term in the sequence 3, 6, 9, 12, ...?  

Ans: The sequence is an arithmetic progression where a = 3 and d = 3 

The 6th term is $a_6$ = a + (6 - 1)d = 3 + 15 = 18 

14. What is the sum of the first 7 odd numbers?  

Ans: The first 7 odd numbers are 1, 3, 5, 7, 9, 11, and 13.  

Sum = 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 

Alternatively, S =$ n^{2} $= $7^{2}$= 49).

15. What is the sum of the first 10 even numbers?  

Ans: The first 10 even numbers are 2, 4, 6, ..., 20.  

Sum = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110 

Alternatively, S = n(n + 1) = 10 $\times $11 = 110

Solve Some Extra Questions to Score High in Test Paper

1. If the 2nd term of an arithmetic sequence is 8 and the common difference is 3, what is the first term?  

Ans: Let the first term be a 

The 2nd term is a + d = 8 

So, a + 3 = 8 

a = 8 - 3 = 5

2. How many numbers are there between 1 and 50 that are multiples of 10?  

Ans: The multiples of 10 from 1 to 50 are 10, 20, 30, 40, and 50.  

This forms an arithmetic sequence where a = 10, d = 10 

There are n =$ \dfrac{50 - 10}{10} $+ 1 = 5 multiples of 10.

3. If the 5th term of a sequence is 25 and the common difference is 5, what is the 1st term?  

Ans: Let the first term be a.  

The 5th term is a + 4d = 25   

So, a + 4 $\times $5 = 25 

a + 20 = 25 

a = 25 - 20 = 5

4. Find the sum of the first 8 squares of natural numbers.  

Ans: The squares of the first 8 natural numbers are $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, 8^{2}$.  

Sum = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204 

Alternatively, using the formula S = $\dfrac{n(n + 1)(2n + 1)}{6} $:  

S =$ \dfrac{8(8 + 1)(2 \times 8 + 1)}{6}$ = 204.

5. What is the 9th term in the sequence: 4, 8, 12, 16, ...?  

Ans: The sequence has a first term a = 4 and a common difference d = 4.  

The 9th term is given by $a_9$ = a + (9 - 1)d .  

So, $ a_9$ = 4 + (8 $\times$ 4) = 4 + 32 = 36 

5 Important Formulas of Class 6 Chapter 3 You Shouldn’t Miss!

To score high on your test paper you must practise our CBSE Class 6 Maths Chapter 3 Number Play Important Questions which also contains some short question answers to help you to solve all the problems easily.

Along with that, we are also providing extra questions which will test your knowledge and understanding of the chapter. This kind of preparation will help you identify your strengths and areas for improvement.

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