## RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

**Exercise 11A**

**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

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**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)

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**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

First term = 6, common difference =

=

=

a = 6, d =

The n^{th} term is given by

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

T_{14} = 6 + (37 – 1) = 6+ 63 = 69

Hence, 37^{th} term is 69

**Question 6:**

The given AP is

The first term = 5,

common difference =

∴ a = 5, d =

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 = ;

Suppose there are n terms in given AP, we have

Then,

T_{n } = 3 ⇒ a + (n-1) d = 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:**

Principal = Rs. 1000, rate of interest = 28% p.a, time = T years

Put T = 1 in (1)

∴ S.I at the end of the first year = Rs. 80

Put T = 2, SI at the end of two years = Rs. 80 x 2 = Rs. 160

Put T = 3, S.I. at the end of third year = Rs. 80 x 3 = Rs. 240

Thus, simple interests = Rs. 80, Rs. 160, Rs.240…. form an AP whose first term is Rs. 80 and common difference is Rs. (160 – 80) = Rs. 80

Put T = 30, S.I. at the end of 30 years = a + 29d = (80 + 29 x 80) = Rs. 80 x 30 = Rs. 2400

### Exercise 11B

**Question 1:**

If T_{1},T_{2},T_{3 }are consecutive terms of an AP, then

T_{2} – T_{1} = T_{3} – T_{2 }or 2T_{2} = T_{1} + T_{3 }

∴ x+2, 2x, 2x + 3 are in AP, if

2(2x) = x +2 + 2x + 3

⇒ 4x = 3x + 5 ⇒ x = 5

**Question 2:**

Let the required numbers be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) × a × (a + d) = a(a^{2} – d^{2})

But sum = 15 and product = 80

∴ 3a = 15 ⇒ a = 5

and a(a^{2} – d^{2}) = 5 x (25 – d^{2}) = 80 [∵ a = 5]

⇒ (25 – d^{2}) = 16

⇒ d^{2} = 25 – 16 ⇒ d^{2} = 9

⇒ d = 3

Thus, a = 5 and d = 3

Hence, the required numbers are (2, 5, 8)

**Question 3:**

Let the required number be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a+ (a + d) = 3a

Product of these numbers = (a – d) × a × (a – d) = a(a^{2} – d^{2})

But sum = 27 and product = 405

∴ 3a = 27 ⇒ a = 9

and a(a^{2} – d^{2}) = 405

⇒ 9 x (81 – d^{2}) = 405 [∵ a = 5]

⇒ 729 – 9d^{2 }= 405

⇒ 9d^{2 }= 729 – 405 = 324

⇒ d^{2} = 36

d = ±6

a = 9 and d = 6

Hence the required numbers are (3, 9, 15)

**Question 4:**

Let the required numbers be (a – d), a, (a + d)

Sum of these number = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) × a × (a + d) = a(a^{2} – d^{2})

∴ 3a = 3 ⇒ a = 1

and a(a^{2} – d^{2}) = 1 (1 – d^{2}) = -35

⇒ 1 – d^{2 }= -35

⇒ d^{2} = 36

⇒ d = 6

But, sum = 3 and product = – 35

Thus, a = 1 and d = 6

Hence, the required numbers are (-5, 1, 7)

**Question 5:**

Let the required number be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) × a × (a + d) = a(a^{2} – d^{2})

But sum = 24 and product = 440

∴ 3a = 24 ⇒ a = 8

and a(a^{2} – d^{2}) = 8 (64 – d^{2}) = 440

⇒ 64 – d^{2 }= 55

⇒ d^{2} = 9

⇒ d = 3

Thus, a = 8 and d = 3

Hence the required numbers are (5, 8, 11)

**Question 6:**

Let the required numbers be (a – d), a, (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

∴ sum of these squares = (a – d )^{2} + a^{2} + (a + d )^{2} = 3a^{2} + 2d^{2}

Sum of three numbers = 21, sum of squares of these numbers = 165

∴ 3a = 21

a = 7

and 3a^{2} + 2d^{2 }= 165 ⇒ 3(7)^{2 }+ 2d^{2 }= 165 ⇒ 2d^{2 }= 18

⇒ d^{2} = 9

⇒ d = ±3

Thus, a = 7 and d = ±3

Hence, the required numbers are (4, 7, 10) or (10, 7, 4)

**Question 7:**

Let the required angles be (a – 3d)°, (a – d) °, (a + d) ° and (a + 3d) °

Common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d

Common difference = 10°

∴ 2d = 10° = d = 5°

Sum of four angles of quadrilateral = 360°

(a – 3d)°, (a – d) °, (a + d) ° and (a + 3d) ° = 360°

4a = 360°

a = 90°

∴ a = 90° d = 5°

First angle = (a – 3d)° = (90 – 3 × 5) ° = 75°

Second angle = (a – d)° = (90 – 5) ° = 85°

Third angle = (a + d)° = (90 + 5°) = 95°

Fourth angle = (a + 3d)° = (90 + 3 × 5)° = 105°

**Question 8:**

Let the required number be (a – 3d), (a – d), (a + d) and (a + 3d)

Sum of these numbers = (a – 3d) + (a – d)+ (a + d) + (a + 3d)

∴ 4a = 28 ⇒ a = 7

Sum of the squares of these numbers

= (a – 3d )^{2} + (a – d )^{2} + (a + d )^{2} + (a + 3d )^{2}= 4(a^{2} + 5d^{2)}

∴ 4(a^{2} + 5d^{2}) = 216

⇒ a^{2} + 5d^{2 }= 54 [∵ a = 5]

⇒ 5d^{2} = 54 – 49

⇒ 5d^{2} = 5

⇒ d^{2} = 1

⇒ d = ±1

Hence, the required numbers (4, 6, 8, 10)

### Exercise 11C

https://www.youtube.com/watch?v=AAz6NRBhLCU

**Question 1:**

Here a = 2, d = (7 – 2) = 5, and n = 19

Using the formula

Hence, the sum of first 19 terms of the given AP is 893

**Question 2:**

Here, a = 1, d = (3 – 1) = 2 and n = 26

Using the formula

Hence, the sum of first 26 terms of the given AP is 676

**Question 3:**

Here, a = 9, d = (7 – 9) = – 2 and n = 18

Using the formula

Hence the sum of first 18 terms of the given AP is – 144

**Question 4:**

Here a = 5, d = (13 – 5) = 8, and l = 181

Let the total number of terms be n, then

Hence, the required sum is 2139

**Question 5:**

Here a = 5, d = (9 – 5) = 4, and l = 81

Let the total number of terms be n, then

Hence sum of first 20^{th} terms of the given AP is 860

**Question 6:**

Here, a = 15, d = (11 – 15) = -4, and l = – 13

Let the total number of term be n, then

Hence, the sum of first 8^{th} term of given AP is 8

**Question 7:**

All 2 – digit whole numbers, which are divisible by 3 are 12, 15, 18, 21, … 99

This is an AP in which a = 12, d = (15 – 12) = 3, and l = 99

Let the number of these terms be n, then

**Question 8:**

All the even number between 5 and 100 are 6, 8, 10, 12, …, 98

This is an AP in which a = 6, d = (8 – 6) = 2, l = 98

**Question 9:**

All natural number divisible by 6 and less than 100 are 6, 12, 18, 24, ….96

This in AP in which a = 6, d = (12 – 6) = 6 and l = 96

**Question 10:**

All natural numbers between 100 and 500 divisible by 7 are 105, 112, 119, 126, … 497

This is an AP in which a = 105, d = (112 – 105) = 7, l = 497

Let the number of term be n, then

**Question 11:**

All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, …, 693

This is an AP in which a = 306, d = (315 – 306) = 9, l = 693

Let the number of these terms be n, then

**Question 12:**

All three digit natural numbers divisible by 13 are 104, 117, 130, 143,…, 988

This is an AP in which a = 104, d = (117 – 104) = 13, l = 988

**Question 13:**

First 15 multiples of 8 are 8, 16, 24, … to 15^{th} term

**Question 14:**

Odd natural numbers between 0 and 50 are 1, 3, 5, … 49

∴ a = 1, d = 3 – 1= 2, l = 49

Let the number of terms be n

**Question 15:**

First 100 even natural numbers divisible by 5 are

10, 20, 30, … to 100 term

First term of AP = 10

Common difference d = 20 – 10 = 10

Number of terms = n = 100

**Question 16:**

All 3 digit numbers which are divisible by 13 are 104, 117, 130, 143, … 988

This is an AP in which a = 104, d = 117 – 104 = 13 and l = 988

Let the number of these term be n, then

**Question 17:**

Let a be the first term and d be the common difference of the given AP, then

**Question 18:**

Let a and d be the first term and common difference of an APrespectively.

n = 20

**Question 19:**

Here a = 21, d = (18 – 21) = -3

Let the required number of terms be n, then

sum of first 15 terms = 0

**Question 20:**

Here a = 63, d = 60 – 63 = -3

Let the sum of n terms be 693, then

sum of first 22 terms = sum of first 21 terms = 693

∴ This means that 22^{nd} term is zero

T_{22} = 0

**Question 21:**

Here a = 64, d = 60 – 64 = – 4

Let the sum of n terms be 544, then

Sum of first 16 terms = sum of first 17 terms = 544

This means that 17^{th} term is zero

**Question 22:**

Here a = 18, d = 15 – 18 = -3

Let the sum if n terms be 45, then

Sum of first three terms = sum of first 10 terms = 45

This means the sum of all terms from 4^{th} to 10^{th} is zero

**Question 23:**

nth term = (4n + 1) —-(1)

putting n = 1 in (1), we get

Hence the sum of 15 and n terms of AP are 495, (2n^{2} + 3n) respectively.

**Question 24:**

**Question 25:**

(i) The nth term is given by

(ii) Putting n = 1 in (1) , we get

T_{1} = (6 x 1 – 4) = 2

∴ First term = 2

(iii) Putting n = 2 in (1), we get T_{2} = (6 x 2 – 4) = 8

∴ d = (T_{2} – T_{1}) = 8 – 2 = 6

**Question 26:**

It is given that

Now, 20^{th} term

=(sum of first 20 term) – (sum of first 19 terms)

Putting = 20 in (1) we get

Putting n = 19 in (1), we get

Hence, the 20^{th} term is 99

**Question 27:**

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

**Question 28:**

Let a be the first term and d be the common difference of the given AP, then we get

Multiplying (1) by 2 and subtracting the result from (2), we get

∴ 11d = 22 ⇒ d = 2

Putting d = 2 in (1), we get a + 8 = 9 a = 1

Thus, a = 1, d = 2

**Question 29:**

First term a = 4

Last term l = 81

Common difference = 7

**Question 30:**

First term of an AP, a = 22

Last term = n^{th} term = – 11

Thus, n = 12, d = -3

**Question 31:**

First term ‘a’ of an AP = 2

The last term l = 29

∴ common difference = 3

**Question 32:**

**Question 33:**

TV production every year forms an AP

Let a be the TV production in first year and d be the common difference

Production in nth year = a + (n – 1)d

production in 6^{th} year = a + 5d = 8000

a + 5d = 8000 —-(1)

Production in 9^{th} year = a + (9 – 1)d = 11300

a + 8d = 11300 —-(2)

Subtracting (1) from (2), we get

Thus,

(i) TV production in first year = 2500

(ii) Production in 8^{th} year = 10200

(iii) Total production in 6 years = 31500

**Question 34:**

Let the value of first prize be Rs. a

Subsequent prizes are Rs (a – 200), Rs (a – 400) and Rs (a – 600)

Total value of these prizes

Hence the first prize is Rs. 1000 and subsequent prizes are worth Rs. 800, Rs. 600 and Rs. 400

**Question 35:**

Number of logs in 1^{st} row (from bottom) = 20

Number of logs in 2^{nd} row = 19

Number of logs in 3^{rd} row = 18

Let there are n number of rows

20 + 19 + 18 + … to n terms = 200

For n = 16, number of logs in nth row

= a + (n – 1)d

= 20 + (16- 1)(-1)

= 20 -15

= 5

n = 25, ∵ number of logs in 25^{th} row is negative which is not admissible

Thus, there are 16 rows and number of logs at the top is 5

### Exercise 11D

**Question 1:**

**Question 2:**

**Question 3:**

**Question 4:**

(2p – 1), 7, 3p are in AP

∴ 7 – (2p – 1) = 3p – 7 or 7-2p + 1=3p – 7

⇒ 5p = 15

∴ p = 3

**Question 5:**

Thus, common difference = 3

**Question 6:**

**Question 7****:**

First term of AP = a = p

Common difference = d = q

n^{th} term = a + (n – 1)d

10^{th} term = p + (10 – 1)q

= p + 9q

**Question 8:**

**Question 9:**

**Question 10:**

The given AP is 21, 18, 15, ….

First term = 21, common difference = 18 – 21= – 3

Let n^{th} term be zero

a + (n – 1)d = 0 or 21 + (n – 1)(-3) = 0

21 – 3n + 3 = 0

3n = 24

or n = 8

Hence, 8^{th} term of given series is 0

**Question 11:**

The given AP is 25, 20, 15, …

∴ first term a = 25, common difference = 20 – 25 = – 5

Let for lowest value of n, n^{th} term is negative

So, the first negative term is the 7^{th} term

**Question 12:**

The given AP is 5, 7, 9, … 201

last term l = 201, common difference d = 7 – 5 = 2

6^{th} term from the end = l – (n – 1)d

=201 – (6 – 1) x 2

= 201 – 10 = 191

**Question 13:**

Sum of n natural numbers = 1 + 2 + 3 + … + n

Here a = 1, d = 2 – 1 = 1

**Question 14:**

Sum of even natural numbers = 2 + 4 + 6 + … to n terms

a = 2, d = 4 – 2 = 2

**Question 15:**

Sum of n odd natural numbers = 1 + 3 + 5 + … to n terms

a = 1, d = 3 – 1 = 2

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

Anish says

Do the formative assessment solutions also please.

Devnandan kumar yadav says

Please solve the all question of exercise 11D

Archi Shaw says

Please solve exercise 11D all the questions.

Archi Shaw says

Please solve question no 26 of ex 11D