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**RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions**

RS Aggarwal Class 10 Solutions

**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

**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

**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

Om Prakash says

very nice any problems questions to answer understand the answer and solve the question

vki says

Nice 2 understand

But 2016 book should have been published

bobby kumar says

Nice full questions all 2016

bobby kumar says

Nice but 2016 new attention book