Class |
10 |

Chapter |
Arithmetic Progressions |

Subject |
Math |

Category |
Important Question Answer |

**Class 10 Math Chapter 5 Important Question Answer**

**Q1. If the sum of first 7 terms of A. P. is 49 and sum of first 17 terms is 289, then find the sum of n terms of A. P. Most Important**

**Ans**. Given S_{7} = 49 and S_{17} = 289, S_{n} = ?

Using S_{n} = [2a + (n – 1)d]

S_{7} = [2a + 6d] = 49 → 2a + 6d = 14 — (i)

S_{17} = [2a + 16d] = 289 → 2a + 16d = 34 —— (ii)

Subtracting eq (i) from (ii) we get 10d = 20

d = 2

Putting value of d in eq (i) we get

2a + 6(2) = 14

a = 1

Now,

S_{n} = [2a + (n – 1)d]

= [2×1 + (n – 1)2] = [2+ 2n – 2] = [2n] = n^{2}

**Q2. If the sum of first 6 terms is 12 and sum of first 10 terms is 60, then find the sum of its n terms.**

**Ans**. Given S_{6} = 12 and S_{10} = 60, S_{n} = ?

Using S_{n} = [2a + (n – 1)d]

S_{6} = [2a + 5d] = 12 → 2a + 5d = 4 — (i)

S_{10} = [2a + 9d] = 60 → 2a + 9d = 12 —— (ii)

Subtracting eq (i) from (ii) we get 4d = 8

d = 2

Putting value of d in eq (i) we get

2a + 5(2) = 4

2a = -6

a = – 3

Now,

S_{n} = [2a + (n – 1)d]

= [2×(-3) + (n – 1)2]

= [ -6 + 2n – 2] = [2n – 8]

**Q3. If the sum of first 10 terms of an A. P. is -60 and sum of first 15 terms is-165, then find the sum of its n terms.**

**Ans**. Given S_{10} = -60 and S_{15} = -165, S_{n} = ?

Using S_{n} = [2a + (n – 1)d]

S_{10 }= [2a + 9d] = -60 → 2a + 9d = -12 — (i)

S_{15} = [2a + 14d] = -165 → 2a + 14d = -22 —— (ii)

Subtracting eq (i) from (ii) we get 5d = -10

d = -2

Putting value of d in eq (i) we get

2a + 9(-2) = -12

2a = 6

a = 3

Now,

S_{n} = [2a + (n – 1)d]

= [2×3 + (n – 1)(-2)]

= [6 – 2n + 2] = [8 – 2n]

**Q4. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference:**

**Ans**. a = 5, l = 45, S_{n} = 400, n = ?, d = ?

Using, S_{n} = (a + l)

400 = (5 + 45)

= 8

n = 16

Now, S_{n} = [2a + (n – 1)d]

400 = [2×5 + 15d]

50 = 10 + 15d

40 = 15d

d = =

**Q5. How many terms of A. P. : 9, 17, 25, ………. Should be taken so that their sum is 636 ?**

**Ans**. a = 9, d = 17 – 9 = 8, S_{n }= 636, n = ?

Using, S_{n} = [2a + (n – 1)d]

636 = [2×9 + (n – 1)8]

636 = [8n + 10]

636 = 4n^{2} + 5n

4n^{2} + 5n – 636 = 0

4n^{2} + 53n – 48n – 636 = 0

n(4n – 53) – 12(4n – 53) = 0

(n – 12)(4n – 53) = 0

n = 12 or n =

As n cannot be in fraction. Therefore, n = 12 is correct.

**Q6. Which term of A.P. 3, 15, 27, 39………… will be 132 more than its 54th term?**

**Ans**. a = 3, d = 15 – 3 = 12

a_{n} = a + (n – 1)d

a_{54} = 3 + (54 – 1)12 = 3 + 53×12 = 639

According to question,

We have to find the term which is equal to a_{54 }+ 132 i.e 639 + 132 = 771

a_{n} = a + (n – 1)d

771 = 3 + (n – 1)12

768 = (n – 1)12

64 = n – 1

65 = n

Therefore, 65^{th} term will be 132 more than 54^{th} term of given A.P.

**Q7. Find the 20th term from the last of the A. P. 3, 8, 13, …………, 253.**

**Ans**. Now AP will be 253, 248, ……. 3

a = 253, d = – 5, a_{20 }= ?

a_{n} = a + (n – 1)d

a_{20 }= 253 + 19(-5) = 253 – 95 = 158

**Q8. Find the 31st term of an A. P. whose 11th term is 38 and 16th term is 73.**

**Ans**. a_{31 }= ?, a_{11 }= 38, a_{16 }= 73

a_{n} = a + (n – 1)d

a_{11 }= a + 10d = 38 — (i)

a_{16 }= a + 15d = 73 — (ii)

Subtracting eq (i) from (ii) we get 5d = 35

d = 7

Putting value of d in equation (i)

a + 10×7 = 38

a = – 32

Now, a_{n} = a + (n – 1)d

a_{31 }= – 32 + 30×7 = – 32 + 210 = 178

**Q9. The sum of first n natural numbers is __________.**

**Ans**. Natural numbers form an AP : 1, 2, 3, …… n

a = 1, d = 1, S_{n} = ?

Using, S_{n} = [2a + (n – 1)d]

S_{n} = [2×1 + (n – 1)1] = [n + 1]

**Q10. The sum of first 50 natural numbers is ________ .**

**Ans**. Natural numbers form an AP : 1, 2, 3, ……

a = 1, d = 1, S_{50} = ?

Using, S_{n} = [2a + (n – 1)d]

S_{50} = [2×1 + (50 – 1)1] = 25(51) = 1275

**Q11. Write the next four terms of the A. P. 1, -1, -3, -5, ……………****\**

**Ans**. a = 1, d = -1 – 1 = -2,

Next term comes when we add d in previous term.

-5 + (-2) = – 7

-7 + (-2) = -9

-9 + (-2) = – 11

-11 + (-2) = -13

Therefore, required next four terms are : -7, -9, -11, -13

**Q12. Find the sum of first 10 terms of A.P. 2, 7, 12, 17, …………..**

**Ans**. a = 2, d = 7 – 2 = 5, S_{10} = ?

Using, S_{n} = [2a + (n – 1)d]

S_{n} = [2×2 + (10 – 1)5] = 5[4 + 45] = 5×49 = 245

**Q13. Find 11th term of A.P. 7, 13, 19 …………. Most Important**

**Ans**. a = 7, d = 13 – 7 = 6, a_{11} = ?

a_{n} = a + (n – 1)d

a_{11 }= a + 10d = 7 + 10×6 = 67

**Q14. Find the common difference of the A.P. : 3, 1, −1, -3, ……………. Most Important**

**Ans**. d = a_{2 }– a_{1} = 1 – 3 = – 2

**Q15. Find common difference of AP. 7, 5, 3, 1 ……………**

**Ans**. d = 5 – 7 = – 2

**Q16. How many three digit numbers are divisible by 7 ?**

**Ans**. Three digit numbers divisible by 7 form an AP : 105, 112 …….. 994

a = 105, d = 7, a_{n} = 994, n = ?

a_{n} = a + (n – 1)d

994 = 105 + (n – 1)7

889 = (n – 1)7

127 = (n – 1)

n = 128

**Q17. How many two-digit numbers are divisible by 3?**

**Ans**. Two digit number divisible by 3 form an AP : 12, 15, ……. 99

a = 12, d = 3, a_{n} = 99, n = ?

a_{n} = a + (n – 1)d

99 = 12 + (n – 1)3

87 = (n – 1)3

29 = (n – 1)

n = 30

**Q18. Find the sum of first 40 positive integers divisible by 6.**

**Ans**. First 40 positive integers divisible by 6 form an AP : 6, 12, ….

a = 6, d = 6, S_{40} = ?

S_{n} = [2a + (n – 1)d]

S_{40 }= [2×6 + (40 – 1)6] = 20[12 + 234]

= 20 × 246 = 4920

**Q****19****. How many multiples of 4 lie between 10 and 250 ?**

**Ans**. Multiples of 4 will form an AP : 12, 16, …… 248

a = 12, d = 4, a_{n }= 248

a_{n} = a + (n – 1)d

248 = 12 + (n – 1)4

236 = (n – 1)4

59 = (n – 1)

n = 60

**Q2****0****. How many terms of sequence 1, 4, 7, 10, ………. should be taken so that their sum is 176 ?**

**Ans**. Given Sn = 176, a = 1, d = 4 – 1= 3

n = ?

Using, S_{n} = [2a + (n – 1)d]

176 = [2 + (n – 1)3]

352 = n[3n – 1]

3n^{2} – n – 352 = 0 (We have to find two numbers whose sum is – 1 and product is – (3×352) )

3n^{2} – 33n + 32n – 352 = 0

3n(n – 11) + 32(n – 11) = 0

(3n + 32)(n – 11) = 0

As n cannot be negative.

Therefore, n – 11= 0

n = 11.

Also Read |
Class 10 Math NCERT Solution |

Also Read |
Class 10 Important Questions [Latest] |