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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 9 |

Chapter Name | Arithmetic Progressions |

Exercise | 9.1 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **9**** Arithmetic Progressions** Ex 9.1 Download PDF

**Arithmetic Progressions**Ex 9.1 Download PDF

**Chapter 9: Arithmetic Progressions Exercise – 9.1**

**Question: 1**

Write the first terms of each of the following sequences whose n^{th} term are

(i) a_{n} = 3n + 2

(iii) a_{n} = 3_{n}

(v) a_{n} = (-1)^{n}2^{n}

(vii) a_{n} = n^{2} – n + 1

(viii) a_{n} = n^{2} – n + 1

**Solution:**

We have to write first five terms of given sequences

(i) a_{n} = 3_{n} + 2

Given sequence a_{n} = 3_{n} + 2

To write first five terms of given sequence put n = 1, 2, 3, 4, 5, we get

a_{1} = (3 × 1) + 2 = 3 + 2 = 5

a_{2} = (3 × 2) + 2 = 6 + 2 = 8

a_{3} = (3 × 3) + 2 = 9 + 2 = 11

a_{4} = (3 × 4) + 2 = 12 + 2 = 14

a_{5} = (3 × 5) + 2 = 15 + 2 = 17

∴ The required first five terms of given sequence a_{n} = 3n + 2 are 5, 8, 11, 14, 17.

Given sequence

Put n = 1, 2, 3, 4, 5 then we get

∴ The required first five terms of given sequence

(iii) a_{n} = 3^{n}

Given sequence a_{n} = 3^{n}

To write first five terms of given sequence, put n = 1, 2, 3, 4, 5 in given sequence then,

a_{1} = 3^{1} = 3; a_{2} = 3^{2} = 9; a_{3} = 27; a_{4} = 3^{4} = 81; a_{5} = 3^{5} = 243.

Given sequence,

To write first five terms, put n = 1, 2, 3, 4, 5 in given sequence

Then, we get

∴ The required first five terms are 1/5, 4/5, 7/5, 10/5, 13/5

(v) a_{n} = (-1)^{n}2^{n}

Given sequence is a_{n} = (-1)^{n}2^{n}

To get first five terms of given sequence an, put n = 1, 2, 3, 4, 5.

a_{1} = (-1)^{1}.2^{1} = (-1).2 = -2

a_{2} = (-1)^{2}.2^{2} = (-1).4 = -4

a_{3} = (-1)^{3}.2^{3} = (-1).8 = -8

a_{4} = (-1)^{4}.2^{4} = (-1).16 = 16

a_{5} = (-1)^{5}.2^{5} = (-1).32 = -32

∴ The first five terms are – 2, 4, – 8, 16, – 32.

The given sequence is,

To write first five terms of given sequence

Put n = 1, 2, 3, 4, 5. Then, we get

∴ The required first five terms are

(vii) a_{n} = n^{2} – n + 1

The given sequence is, a_{n} = n^{2} – n + 1

To write first five terms of given sequence a_{n1 }we get put n = 1, 2, 3, 4, 5. Then we get

a_{1} = 1^{2} – 1 + 1 = 1

a_{2} = 2^{2} – 2 + 1 = 3

a_{3} = 3^{2} – 3 + 1 = 7

a_{4} = 4^{2} – 4 + 1 = 13

a_{5} = 5^{2} – 5 + 1 = 21

∴ The required first five terms of given sequence a_{n} = n^{2} – n + 1 are 1, 3, 7, 13, 21

(viii) a_{n} = 2n^{2} – 3n + 1

The given sequence is a_{n} = 2n^{2} – 3n + 1

To write first five terms of given sequence and, we put n = 1, 2, 3, 4, 5. Then we get

a_{1} = 2.1^{2} – 3.1 + 1 = 2 – 3 + 1 = 0

a_{2} = 2.2^{2} – 3.2 + 1 = 8 – 6 + 1 = 3

a_{3} = 2.3^{2} – 3.3 + 1 = 18 – 9 + 1 = 10

a_{4} = 2.4^{2} – 3.4 + 1 = 32 – 12 + 1 = 21

a_{5} = 2.5^{2} – 3.5 + 1 = 50 – 15 + 1 = 36

∴ The required first five terms of given sequence a_{n} – 2n^{2} – 3n + 1 are 0, 3, 10, 21, 36

Given sequence is,

To write first five terms of given sequence we put n = 1, 2, 3, 4, 5. Then, we get,

∴ The required first five terms of given sequence

**Question: 2**

Find the indicated terms in each of the following sequences whose nth terms are:

(i) a_{n} = 5n – 4; a_{12} and a_{15}

(iii) a_{n} = n(n – 1)(n – 2); a_{5} and a_{8}

(iv) a_{n} = (n – 1)(2 – n)(3 + n);a_{11} a_{21} a_{3}

(v) an = (-1)^{n}n;a_{3}, a_{5}, a_{8}

**Solution:**

We have to find the required term of a sequence when nth term of that sequence is given:

(i) a_{n} = 5n – 4; a_{12} and a_{15}

Given n^{th} term of a sequence an = 5_{n} – 4

To find 12^{th} terms, 15^{th} terms of that sequence, we put n = 12 15 in its n^{th} term.

Then, we get

a_{12} = 5.12 – 4 = 60 – 4 = 56

a_{15} = 5.15 – 4 = 15 – 4 = 71

∴ The required terms a_{12} = 56, a_{15} = 71

Given nth term is

To find 7^{th}, 8^{th} terms of given sequence, we put n = 7, 8.

∴ The required terms

(iii) a_{n} = n(n – 1)(n – 2); a_{5} and a_{8}

Given n^{th} term is a_{n} = n(n – 1)(n – 2)

To find 5^{th}, 8^{th} terms of given sequence, put n = 5, 8 in an then, we get

a_{5} = 5(5 – 1).(5 – 2) = 5.4.3 = 60

a_{8} = 8.(8 – 1).(8 – 2) = (8.7.6) = 336

∴ The required terms are a_{5} = 60 and a_{8} = 336

(iv) a_{n} = (n – 1)(2 – n)(3 + n);a_{11} a_{21} a_{3}

The given n^{th} term is a_{n} = (n + 1)(2 – n)(3 + n)

To find a_{1}, a_{2}, a_{3} of given sequence put n = 1, 2, 3 is an

a_{1} = (1 – 1)(2 – 1)(3 + 1) = 0.1.4 = 0

a_{2} = (2 – 1)(2 – 2)(3 + 2) = 1.0.5 = 0

a_{3} = (3 – 1)(2 – 3)(3 + 3) = 2.-1.6 = -12

∴ The required terms a_{1} = 0, a_{2} = 0, a_{3} = -12

(v) a_{n} = (-1)^{n}n; a_{3}, a_{5}, a_{8}

The given n^{th} term is, a_{n} = (-1)^{n}.n

To find a_{3}, a_{5}, a_{8} of given sequence put n = 3, 5, 8, in a_{n}.

a_{3} = (-1)^{3}.3 = -1.3 = -3

a_{5} = (-1)^{5}.5 = -1.5 = -5

a_{8} = (-1)^{8} = 1.8 = 8

∴ The required terms a_{3} = -3, a_{5} = -5, a_{8} = 8

**Question: 3**

Find the next five terms of each of the following sequence given by:

(i) a_{1} = 1, 1_{n} = a_{n-1} + 2, n ≥ 2

(ii) a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 3, n > 2

(iv) a_{1} = 4, a_{n} = 4 a_{n-1} + 3, n > 1

**Solution:**

We have to find next five terms of following sequences.

(i) a_{1} = 1, a_{n} = a_{n-1} + 2, n ≥ 3

Given first term (a_{1}) = 1,

n^{th} term a_{n} = a_{n-1} + 2, n ≥ 2

To find 2^{nd}, 3^{rd}, 4^{th}, 5^{th}, 6^{th} terms, we use given condition n ≥ 2 for nth term a_{n} = a_{n-1} + 2

a_{2} = a_{2-1} + 2 = a_{1} + 2 = 1 + 2 = 3 (∴ a_{1} = 1)

a_{3} = a_{3-1} + 2 = a_{2} + 2 = 3 + 2 = 5

a_{4} = a_{4-1} + 2 = a_{3} + 2 = 5 + 2 = 7

a_{5} = a_{5-1} + 2 = a_{4} + 2 = 7 + 2 = 9

a_{6} = a_{6-1} + 2 = a_{5} + 2 = a + 2 = 11

∴ The next five terms are,

a_{2} = 3, a_{3} = 5, a_{4} = 7, a_{5} = a, a_{6} = 11

(ii) a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 3, n > 2

Given,

First term (a_{1}) = 2

Second term (a_{2}) = 2

n^{th} term (a_{n}) = a_{n-1} – 3

To find next five terms i.e., a_{3}, a_{4}, a_{5}, a_{6}, a_{7} we put n = 3, 4, 5, 6, 7 is a_{n}

a_{3} = a_{3-1} – 3 = 2 – 3 = – 1

a_{4} = a_{4-1} – 3 = a_{3} – 3 = – 1 – 3 = – 4

a_{5} = a_{5-1} – 3 = a_{4} – 3 = -4 – 3 = -7

a_{6} = a_{6-1} – 3 = a_{5} – 3 = -7 – 3 = -10

a_{7} = a_{7-1} – 3 = a_{6} – 3 = -10 – 3 = -13

∴ The next five terms are, a_{3} = -1, a_{4} = – 4, a_{5} = -7, a_{6} = -10, a_{7} = -13

Given, first term (a_{1}) = – 1

To find next five terms i.e., a_{2}, a_{3}, a_{4}, a_{5}, a_{6} we put n = 2, 3, 4, 5, 6 is a_{n}

∴ The next five terms are,

(iv) a1 = 4, an = 4 an-1 + 3, n > 1

Given,

First term (a_{1}) = 4

nth term (a_{n}) = 4 a_{n-1} + 3, n > 1

To find next five terms i.e., a_{2}, a_{3}, a_{4}, a_{5}, a_{6} we put n = 2, 3, 4, 5, 6 is a_{n}

Then, we get

a_{2} = 4a_{2-1} + 3 = 4. a_{1} + 3 = 4.4 + 3 = 19 (∴ a_{1} = 4)

a_{3} = 4a_{3-1} + 3 = 4.a_{2} + 3 = 4(19) + 3 = 79

a_{4} = 4a_{4-1} + 3 = 4.a_{3} + 3 = 4(79) + 3 = 319

a_{5} = 4a_{5-1} + 3 = 4.a_{4} + 3 = 4(319) + 3 = 1279

a_{6} = 4a_{6-1} + 3 = 4.a_{5} + 3 = 4(1279) + 3 = 5119

∴ The required next five terms are,

a_{2} = 19, a_{3} = 79, a_{4} = 319, a_{5} = 1279, a_{6} = 5119

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