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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 11 |

Chapter Name | Arithmetic Progressions |

Exercise | 11 A |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A** **PDF**

**Question 1:**

The given progression is 3, 9, 15, 21 …..

Clearly (9 – 3) = (15 – 9) = (21 – 15) = 6 which is constant

Thus, each term differs from its preceding term by 6

So, the given progression is an AP

Its first term = 3 and the common difference = 6

**Question 2:**

The given progression is 16, 11, 6, 1, -4 ….

Clearly (11 – 16) = (1 – 6) = (-4 – 1) = – 5 which is constant

Thus, each term differs from its preceding term by – 5

So the given progression is an AP

Its first term = 16 and the common difference = – 5

**Question 3:**

(i) The given AP is 1, 5, 9, 13, 17…..

Its first term = 1 and common difference = (5 – 1) = 4

∴ a = 1 and d = 4

The n^{th} term of the AP is given by

T_{n }= a + (n-1) d

T_{20} = 1 + (20-1) x 4 = 1+ 76 = 77

Hence, the 20^{th} term is 77

(ii) The given AP is 6, 9, 12, 15 ……

Its first term = 6 and common difference = (9 – 6) = 3

∴ a = 6, d = 3

The n^{th} term of the AP is given by

T_{n }= a + (n-1) d

T35 = 6 + (35-1) x 3 = 6+ 102 = 108

Hence, the 35^{th} term is 108

(iii) The given AP is 5, 11, 17, 23 …..

Its first term = 5, and common difference = (11 – 5) = 6

∴ a = 5, d = 6

The n^{th} term of AP is given by

T_{n }= a + (n-1) d

T_{n}= 5 + (n-1) x 6 = 5+ 6n – 6 = 6n – 1

(iv) The given AP is (5a – x), 6a, (7a + x) …..

Its first term = (5a – x) and common difference = 6a – 5a – x = a + x

The n^{th} term of AP is given by

T_{n }= a + (n-1) d

T_{11 }= (5a – x) + (11-1) (a + x)

= 5a – x + 10x + 10x

= 15a + 9x = 3(5a +3x)

Hence the 11^{th} term is 3(5a + 3x)

**Question 4:**

(i) The given AP is 63, 58, 53, 48 ….

First term = 63, common difference = 58 – 63 = – 5

∴ a = 63, d = – 5

The n^{th} term of AP is given by

T_{n }= a + (n-1) d

T_{10} = 63 + (10-1) (-5) = 63- 45 = 18

Hence the 10^{th} term is 18

(ii) The given AP is 9, 5, 1, -3….

First term = 9, common difference = 5 – 9 = -4

∴ a = 9, d= – 4

The n^{th} term of AP is given by

T_{n }= a + (n-1) d

T_{14} = 9 + (14-1) (-4) = 9- 52 = -43

Hence, the 14^{th} term is – 43

(iii) The given AP is 16, 9, 2, -5

First term = 16, common difference = 9 – 16 = – 7

∴ a = 16, d = -7

The n^{th} term of AP is given by

T_{n }= a + (n-1) d

T_{n} = 16 + (n-1) (-7) ⇒ 16- 7n + 7 = (23 – 7n)

Hence, the n^{th }term is (23 – 7n).

**Question 5:**

The given AP is 6,734,912,1114……..

First term = 6, common difference = (734−6)

= (314−6)

= 74

a = 6, d = 74

The n^{th} term is given by

T_{n }= a + (n-1) d

T_{14} = 6 + (37 – 1) (74) = 6+ 63 = 69

Hence, 37^{th} term is 69

**Question 6:**

The given AP is 5,412,4,312,3……..

The first term = 5,

common difference = (412−5)=(92−5)=−12

∴ a = 5, d = −12

The n^{th} term is given by

T_{n }= a + (n-1) d

T_{14} = 5 + (25 – 1) (-1/2) = 5- 12 = -7

Hence the 25^{th} term is – 7

**Question 7:**

In the given AP, we have a = 6 and d = (10 – 6) = 4

Suppose there are n terms in the given AP, then

T_{n } = 174 ⇒ a + (n-1) d = 174

⇒ 6 + (n-1) 4 = 174

⇒ 6 + 4n – 4 = 174

⇒ 2 + 4n = 174 ⇒ n = 172/4 ⇒ 43

Hence there are 43 terms in the given AP

**Question 8:**

In the given AP we have a = 41 and d = 38 – 41 = – 3

Suppose there are n terms in AP, then

T_{n } = 8 ⇒ a + (n-1) d = 8

⇒ 41 + (n-1) (-3) = 8

⇒ 41 – 3n + 3 = 8

⇒ -3n = – 36 ⇒ n = 12

Hence there are 12 terms in the given AP

**Question 9:**

In the given AP, we have a = 3 and d = 8 – 3 = 5

Suppose there are n terms in given AP, then

T_{n } = a + (n-1) d = 88

⇒ 3 + (n-1) 5 = 88

⇒ 3 + 5n – 5 = 88

⇒ 5n = 90

⇒ n = 12

Hence, the 18^{th} term of given AP is 88

**Question 10:**

In the given AP, we have a = 72 and d = 68 – 72 = – 4

Suppose there are n terms in given AP, we have

T_{n } = 0 ⇒ a + (n-1) d = 0

⇒ 72 + (n-1) (-4) = 0

⇒ 72 – 4n + 4 = 0

⇒ 4n = 76

⇒ n = 19

Hence, the 19^{th} term in the given AP is 0**Question 11:**

In the given AP, we have a = 12 ; (1−56)=16

Suppose there are n terms in given AP, we have

Then,

T_{n } = 3 ⇒ a + (n-1) d = 3

⇒ 56+(n−1)16=3

⇒ 56+16n−16=3

⇒ 4 + n = 18

⇒ n = 14

Thus, 14^{th} term in the given AP is 3**Question 12:**

We know that T_{1 }– (5x + 2), T_{2 }– (4x – 1) and T_{3 }– (x + 2)

Clearly,

T_{2} – T_{1} = T_{3} – T_{2}

⇒ (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)

⇒ 4x – 1 – 5x – 2 = x + 2 – 4x + 1

⇒ -x – 3 = -3x + 3

⇒ -x + 3x = 6

⇒ 2x = 6 ⇒ x = 3

Hence x = 3**Question 13:**

T_{n } = (4n – 10)

⇒ T_{1 } = (4 x 1 – 10) = -6 and T_{2 } = (4 x 2 – 10) = -2

Thus, we have

(i) First term = -6

(ii) Common difference = (T_{2} – T_{1}) = (-2+6) = 4

(iii) 16^{th} term = a + (16-1) d, where a = -6 and d = 4

= (-6 + 15 x 4) = 54**Question 14:**

In the given AP, let first term = a and common difference = d,

Then, T_{n } = a + (n-1) d

⇒ T_{4 } = a + (4 – 1)d, T_{10 } = a + (10 – 1)d

⇒ T_{4 } = a + 3d, T_{10 } = a + 9d

Now, T_{4 } = 13 ⇒ a + 3d = 13 – – – (1)

T_{10 } = 25 ⇒ a + 9d = 25 – – – (2)

Subtracting (1) from (2), we get

⇒ 6d = 12 ⇒ d = 2

Putting d = 2 in (1), we get

a + 3 x 2 = 13

⇒ a = (13 – 6) = 7

Tthus, a = 7, and d = 2

17^{th} term = a + (17 – 1)d, where a= 7, d = 2

(7 + 16 x 2) = (7 + 32) = 39

∴ a = 7, d = 2,**Question 15:**

In the given AP, let first term = a and common difference = d

Then, T_{n } = a + (n-1) d

⇒ T_{8 } = a + (8 – 1)d, T_{12 } = a + (12 – 1)d

⇒ T_{8 } = a + 7d, T_{12 } = a + 11d

Now, T_{8 } = 37 ⇒ a + 7d = 37 – – – (1)

T_{12 } = 57 ⇒ a + 11d = 57 – – – (2)

Subtracting (1) from (2), we get

⇒ 4d = 20 ⇒ d = 5

Putting d = 5 in (1), we get

a + 7 x 5 = 37

⇒ a = 2

Tthus, a = 2, and d = 5

So the required AP is 2, 7, 12..**Question 16:**

In the given AP, let the first term = a, and common difference = d

Then, T_{n } = a + (n-1) d

⇒ T_{7 } = a + (7 – 1)d, and T_{13 } = a + (13 – 1)d

⇒ T_{7 } = a + 6d, T_{13 } = a + 12d

Now, T_{7 } = -4 ⇒ a + 6d = -4 – – – (1)

T_{13 } = -16 ⇒ a + 12d = -16 – – – (2)

Subtracting (1) from (2), we get

⇒ 6d = -12 ⇒ d = -2

Putting d = -2 in (1), we get

a + 6 (-2) = -4

⇒ a – 12 = -4

⇒ a = 8

Tthus, a = 8, and d = -2

So the required AP is 8, 6, 4, 2, 0……**Question 17:**

In the given AP let the first term = a, And common difference = d

Then, T_{n } = a + (n-1) d

⇒ T_{10 } = a + (10 – 1)d, T_{17 } = a + (17 – 1)d, T_{13 } = a + (13 – 1)d

⇒ T_{10 } = a + 9d, T_{17 } = a + 16d, T_{13 } = a + 12d

Now, T_{10 } = 52 ⇒ a + 9d = 52 – – – (1)

and T_{17 } = T_{13 }+ 20 ⇒ a + 16d = a + 12d + 20

⇒ 4d = 20 ⇒ d = 5

Putting d = 5 in (1), we get

a + 9 x 5 = 52 ⇒ a = 52-45 ⇒ a = 7

Thus, a = 7 and d = 5

So the required AP is 7, 12, 17, 22….**Question 18:**

Let the first term of given AP = a and common difference = d

Then, T_{n } = a + (n-1) d

⇒ T_{4 } = a + (4 – 1)d, T_{25 } = a + (25 – 1)d, T_{11 } = a + (11 – 1)d

⇒ T_{4 } = a + 3d, T_{25 } = a + 24d, T_{11 } = a + 10d

Now, T_{4 } = 0 ⇒ a + 3d = 0 ⇒ a = -3d

∴ T_{25 } = a + 24d = (-3d +24d) ⇒ 21d

and T_{11 } = a + 10d = (-3d +10d) ⇒ 7d

∴ T_{25 } = 21d = 3 x 7d = 3 x T_{11}

Hence 25^{th} term is triple its 11^{th} term**Question 19:**

The given AP is 3, 8, 13, 18…..

First term a = 3, common difference a = 8 – 3 = 5

∴ T_{n } = a + (n-1) d = 3 + (n – 1) x 5 = 5n – 2

T_{20 } = 3 + (20-1) 5 = 3 + 19 x 5 = 98

Let n^{th} term is 55 more than the 20^{th} term

∴ (5n – 2) – 98 = 55

Or 5n = 100 + 55 = 155

n = 155/5 = 31

∴ 31^{st} term is 55 more than the 20^{th} term of given AP**Question 20:**

The given AP is 5, 15, 25….

a = 5, d = 15 – 5 = 10

We have, T_{n } = 130+T_{31}

⇒ a + (n-1) d = 130 + 5 + (31 – 1) x 10

⇒ 5 + (n-1) 10 = 130 + 5 + (31 – 1) x 10

⇒ 5 + 10n – 10 = 135 + 300

⇒ 10n – 5 = 435 or 10n = 453 + 5

∴ n = 440/10 = 44

Thus, the required term is 44^{th}**Question 21:**

First AP is 63, 65, 67….

First term = 63, common difference = 65 – 63 = 2

∴ nth term = 63 + (n – 1) 2 = 63 + 2n – 2 = 2n + 61

Second AP is 3, 10, 17 ….

First term = 3, common difference = 10 – 3 = 7

nth term = 3 + (n – 1) 7 = 3 + 7n – 7 = 7n – 4

The two nth terms are equal

∴ 2n + 61 = 7n – 4 or 5n = 61 + 4 = 65

⇒ n = 65/4 = 13.**Question 22:**

Three digit numbers which are divisible by 7 are 105, 112, 119,….994

This is an AP where a= 105, d = 7 and l = 994

Let n^{th} term be 994

∴ a + (n – 1)d =994 or 105 + (n – 1)7 = 994

⇒ 105 + 7n – 7 = 994 or 7n = 94 – 98 = 896

∴ n = 896/7 = 128.

Hence, there are 128 three digits number which are divisible by 7.**Question 23:**

Here a = 7, d = (10 – 7) = 3, l = 184

And n = 8

Now, nth term from the end = [ l – (n-1) d ]

= [ 184 – (8-1) 3 ]

= [ 184 – 7 x 3]

= 184-21

= 163

Hence, the 8^{th} term from the end is 163**Question 24:**

Here a = 17, d = (14 – 17) = -3, l = -40

And n = 6

Now, n^{th} term from the end = [ l – (n – 1) d ]

= [ -40 – (6-1)(-3) ]

= [ -40 + 5 x 3]

= -40+15

= -25

Hence, the 6^{th} term from the end is – 25**Question 25:**

The given AP is 10, 7, 4, ….. (-62)

a = 10, d = 7 – 10 = -3, l = -62

Now, 11^{th} term from the end = [ l – (n – 1) d ]

= [ -62 – (11-1)(-3) ]

= -62 + 30

= -32**Question 26:**

Let a be the first term and d be the common difference

p^{th} term = a +(p – 1)d = q (given) —–(1)

q^{th} term = a +(q – 1) d = p (given) —–(2)

subtracting (2) from (1)

(p – q)d = q – p

(p – q)d = -(p – q)

d = -1

Putting d = -1 in (1)

a – (p – 1) = q ∴ a = p + q -1

∴ (p + q)th term = a+ (p + q -1)d

= (p + q -1) – (p + q -1) = 0**Question 27:**

Let a be the first term and d be the common difference

T_{10 } = a + 9d, T_{15 } = a + 14d

10T_{10} = 15T_{15}

⇒ 10(a + 9) d = 15(a + 14d)

⇒ 2(a + 9) d = 3(a + 14d)

⇒ a + 24d = 0

∴ T_{25 } = 0**Question 28:**

Let a be the first term and d be the common difference

∴ n^{th} term from the beginning = a + (n – 1)d —–(1)

n^{th} term from end = l – (n – 1)d —-(2)

adding (1) and (2),

sum of the n^{th} term from the beginning and n^{th} term from the end = [a + (n – 1)d] + [l – (n – 1)d] = a + l**Question 29:**

Number of rose plants in first, second, third rows…. are 43, 41, 39… respectively.

There are 11 rose plants in the last row

So, it is an AP . viz. 43, 41, 39 …. 11

a = 43, d = 41 – 43 = -2, l = 11

Let n^{th} term be the last term

∴ l = a + (n-1) d

⇒ 11 = 43 + (n-1) x (-2)

43 – 2n + 2 = 11 or 2n = 45 -11 = 34

∴ n = 34/2 = 17

Hence, there are 17 rows in the flower bed.**Question 30:**

Total amount = ₹ 2800

and number of prizes = 4

Let first prize = ₹ a

Then second prize = ₹ a – 200

Third prize = a – 200 – 200 = a – 400

and fourth prize = a – 400 – 200 = a – 600

But sum of there 4 prizes are ₹ 2800

a + a – 200 + a – 400 + a – 600 = ₹ 2800

⇒ 4a – 1200 = 2800

⇒ 4a = 2800 + 1200 = 4000

⇒ a = 1000

First prize = ₹ 1000

Second prize = ₹ 1000 – 200 = ₹ 800

Third prize = ₹ 800 – 200 = ₹ 600

and fourth prize = ₹ 600 – 200 = ₹ 400

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