## RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A

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

**More Resources**

**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)

Read More:

**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 \(6,7\frac { 3 }{ 4 } ,9\frac { 1 }{ 2 } ,11\frac { 1 }{ 4 } ……..\)

First term = 6, common difference = \(\left( 7\frac { 3 }{ 4 } -6 \right) \)

= \(\left( \frac { 31 }{ 4 } -6 \right) \)

= \(\frac { 7 }{ 4 } \)

a = 6, d = \(\frac { 7 }{ 4 } \)

The n^{th} term is given by

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

T_{14} = 6 + (37 – 1) \((\frac { 7 }{ 4 }) \) = 6+ 63 = 69

Hence, 37^{th} term is 69

**Question 6:**

The given AP is \(5,4\frac { 1 }{ 2 } ,4,3\frac { 1 }{ 2 } ,3……..\)

The first term = 5,

common difference = \(\left( 4\frac { 1 }{ 2 } -5 \right) =\left( \frac { 9 }{ 2 } -5 \right) =-\frac { 1 }{ 2 } \)

∴ a = 5, d = \(-\frac { 1 }{ 2 } \)

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 = \(\frac { 1 }{ 2 } \) ; \(\left( 1-\frac { 5 }{ 6 } \right) =\frac { 1 }{ 6 } \)

Suppose there are n terms in given AP, we have

Then,

T_{n } = 3 ⇒ a + (n-1) d = 3

⇒ \(\frac { 5 }{ 6 } +(n-1)\frac { 1 }{ 6 } =3\)

⇒ \(\frac { 5 }{ 6 } +\frac { 1 }{ 6 } n-\frac { 1 }{ 6 } =3\)

⇒ 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

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.