ML Aggarwal Arithmetic and Geometric Progression Exe-9.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-9.3 Questions for Arithmetic and Geometric Progression (AP GP) as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Arithmetic and Geometric Progression (AP GP) Exe-9.3 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-9 | Arithmetic and Geometric Progression |

Writer / Book | Understanding |

Topics | Solutions of Exe-9.3 |

Academic Session | 2024-2025 |

### Arithmetic and Geometric Progression (AP GP) Exe-9.3

ML Aggarwal Class 10 ICSE Maths Solutions

Page-180

**Question-1 ****Find the sum of the following A.P.s :**

**(i) 2, 7, 12, … to 10 terms**

**(ii) 1/15, 1/12, 1/10, … to 11 terms**

**Answer- 1**

(i) 2, 7, 12, … to 10 terms

Here a = 2, d = 7 – 2 = 5 and n = 10

S_{10} = n/2(2a + (n – 1)d)

= 10/2 ((2 × 2) + (10 – 1)5)

= 5(4 + 45), = 5(49), = 245

(ii) First term a = 1/15

Then, d = 1/12 – 1/15

= (5 – 4)/60 , = 1/60

common difference d = 1/60, n = 11

S_{11} = 11/2(2a + (n – 1)d)

= 11/2 ((2 × (1/15)) + (11 – 1)(1/60))

= 11/2 ((2/15) + (10/60))

= 11/2 (2/15 + 1/6)

= 11/2 (4 + 5)/30

= 11/2 (9/30)

= 11/2(3/10)

= 33/20

**Question-2 ****Find the sums given below :**

**(i) 34 + 32 + 30 + … + 10**

**(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)**

**Answer -2**

**(i) 34 + 32 + 30 + … + 10**

Here, a = 34, d = 32 – 34 = -2, l = 10

T_{n} = a + (n – 1)d

10 = 34 + (n – 1)(-2)

-24 = -2 (n – 1)

-24 = – 2n + 2

2n = 24 + 2

2n = 26, n = 26/2, n = 13

S_{n} = n/2(a + 1)

= 13/2 (34 + 10)

= 13/2 (44)

= 13 (22)

= 286

**(ii) First term a = -5,**

Difference d = -8 – (-5) = -8 + 5 = -3

So, common difference d = – 3

Last term T_{n} = -230

We know that, T_{n} = a + (n – 1)d

-230 = -5 + (n – 1)(-3)

-230 = – 5 – 3n + 3

-230 = – 2 – 3n

3n = 230 -2

3n = 228

n = 228/3

n = 76

Therefore, S_{n} = n/2 (a + l)

= 76/2 (-5 + (-230))

= 38 (-5 – 230)

= 38 (235)

= – 8930

**Question-3 ****In an A.P. (with usual notations) :**

**(i) given a = 5, d = 3, a**_{n} = 50, find n and S_{n}

**(ii) given a = 7, a**_{13} = 35, find d and S_{13}

**(iii) given d = 5, S**_{9} = 75, find a and a_{9}

**(iv) given a = 8, a**_{n} = 62, S_{n} = 210, find n and d

**(v) given a = 3, n = 8, S = 192, find d.**

_{n}= 50, find n and S

_{n}

_{13}= 35, find d and S

_{13}

_{9}= 75, find a and a

_{9}

_{n}= 62, S

_{n}= 210, find n and d

**Answer-3**

**(i) a = 5, d = 3, a**_{n} = 50

a_{n} = a + (n – 1 )d

_{n}= 50

50 = 5 + (n – 1)3

50 = 5 + 3n – 3

50 = 2 + 3n

3n = 50 – 2

3n = 48, n = 48/3, n = 16

So, S_{n} = (n/2)(2a + (n – 1)d)

= (16/2) ((2 × 5) + (16 – 1) × 3)

= 8(10 + 45), = 8(55), = 440

**(ii) First term a = 7**

a_{13} = 35,

We know that, a_{n} = a + (n – 1)d

35 = 7 + (13 – 1)d

35 = 7 + 13d – d

35 = 7 + 12d

12d = 35 – 7

12d = 28

d = 28/12 … [divide by 4]

d = 7/3

So, S_{13} = (n/2)(2a + (n – 1)d)

= (13/2) ((2 × 7) + ((13 – 1) × (7/3))

= (13/2) ((14 + (12 × 7/3))

= (13/2) (14 + 28)

= (13/2) (42) = 273

**(iii) Common difference d = 5**

S_{9} = 75

We know that, a_{n} = a + (n – 1)d

a_{9} = a + (9 – 1)5

a_{9} = a + 45 – 5

a_{9} = a + 40 … equation (i)

Then, S_{9} = (n/2) (2a + (n – 1)d)

75 = (9/2) (2a + (9 – 1)5)

75 = (9/2) (2a + (8)5)

(75 × 2)/9 = 2a + 40

150/9 = 2a + 40

2a = 150/9 – 40

2a = 50/3 – 40

2a = (50 – 120)/3

2a = -70/3

a = -70/(3 × 2)

a = – 35/3

, substitute the value of a in equation (i),

a_{9} =a + 40

= -35/3 + 40

= (-35 + 120)/3

= 85/3

**(iv) First term a = 8,**

a_{n }= 62 and S_{n} = 210

We know that, a_{n} = a + (n – 1)d

62 = 8 + (n – 1)d

(n – 1)d = 62 – 8

(n – 1)d = 54 … [equation (i)]

Then, S_{n} = (n/2) (2a + (n – 1)d)

210 = (n/2) ((2 × 8) + 54) … [from equation (i) (n – 1)d = 54]

210 = (n/2) (16 + 54)

420 = n(70)

n = 420/70

n = 6

Now, substitute the value of n in equation (i),

(n – 1)d = 54

(6 – 1)d = 54

5d = 54

d = 54/5

Therefore, d = 54/5 and n = 6

**(v) First term a = 3**

n = 8

S = 192

We know that, S_{n} = (n/2) (2a + (n – 1)d)

192 = (8/2) ((2 × 3) + (8 – 1)d)

192 = 4 (6 + 7d)

192/4 = 6 + 7d

48 = 6 + 7d

48 – 6 = 7d

42 = 7d

d = 42/7

d = 6

Hence common difference d is 6.

**Question-4**

**(i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.**

**(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.**

**Answer -4**

**(i) First term of an A.P. (a) = 5**

Last term (l) = 45

Sum = 400

l = a + (n – 1 )d

45 = 5 + (n – 1)d

⇒ (n – 1)d = 45 – 5 = 40 …(i)

So, S_{n} = (n/2) (2a + (n – 1)d)

400 = (n/2) ((2 × 5) + 40) … from equation (i) (n – 1)d = 40

800 = n(10 + 40)

800 = 50n

n = 800/50, n = 16

**(ii) First term a = 15**

Therefore, sum of first n terms of an A.P. is given by,

S_{n} = (n/2) (2a + (n – 1)d)

S_{15} = (15/2)(2a + (15 – 1)d)

750 = (15/2) (2a + 14d)

(750 × 2)/15 = 2a + 14d

100 = 2a + 14d

Dividing both the side by 2 we get,

50 = a + 7d

Now, substitute the value a,

50 = 15 + 7d

7d = 50 – 15

7d = 35

d = 35/7

d = 5

So, 20^{th} term a_{20} = a + 19d

= 15 + 19(5)

= 15 + 95, = 110

**Question -5 ****The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?**

**Answer -5**

First term of an A.P. (a) = 17

and last term (l) = 350

d= 9

l = T_{n} = a + (n – 1 )d

350 = 17 + (n – 1) × 9

350 – 17 = 9n – 9

333 + 9 = 9n

342 = 9n

n = 342/9

n = 38

So, S_{n} = (n/2) (2a + (n – 1)d)

= (38/2) ((2 × 17) + (38 – 1)d)

= 19(34 + (37 × 9))

= 19(34 + 333)

= 19 × 367

= 6973

Hence n = 38 and S_{n} = 6973

**Arithmetic and Geometric Progression Exe-9.3**

ML Aggarwal Class 10 ICSE Maths AP GP Solutions.

**Question -6 ****Solve for x : 1 + 4 + 7 + 10 + … + x = 287.**

**Answer-6**

First term a = 1

Difference d = 4 – 1 = 3

n = x

x = a = (n – 1)d

x – 1 = (n – 1)d

S_{n} = (n/2) (2a + (n – 1)d)

287 = (n/2) ((2 × 1)+ (n – 1)3)

= n (2 + 3n – 3)

574 = n(2 + 3n – 3)

574 = 2n + 3n^{2} – 3n

574 = – n + 3n^{2}

3n^{2} – n – 574 = 0

3n^{2} – 42n + 41 – 574 = 0

3n(n – 14) + 41(n – 14) = 0

(n – 14) (3n + 41) = 0

If n – 14 = 0

n = 14

or 3n + 41 = 0

3n = -41

n = -41/3

taking positive number so n = 14

Then, = a + (n – 1)d

= 1 + (14 – 1) 3

= 1 + (13)3

= 1 + 39 , = 40

Hence, x = 40

**Question -7**

**(i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.**

**(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.**

**Answer-7**

**(i) A.P. is 25, 22, 19, …**

Sum = 116

Here, a = 25, d = 22 – 25 = -3

Let number of terms be n, then

S_{n} = (n/2) (2a + (n – 1)d)

116 = (n/2) (2a + (n – 1)d)

on cross multiplication,

232 = n ((2 × 25) + (n – 1) (-3))

232 = n (50 – 3n + 3)

232 = n (53 – 3n)

232 = 53n – 3n^{2}

3n^{2} – 53n + 232 = 0

3n^{2} – 24n – 29n + 232 = 0

3n (n – 8) – 29 (n – 8) = 0

(n – 8) (3n – 29) = 0

If n – 8 = 0

n = 8

or 3n – 29 = 0

3n = 29

n = 29/3

fraction, not possible

So, n = 8

Then, T = a + (n – 1)d

= 25 +(8 – 1) (-3)

= 25 + 7 (-3)

= 25 – 21, = 4

**(ii) First term a = 24**

Common difference d = 21 – 24 = – 3

Sum = 78

S_{n} = (n/2) (2a + (n – 1)d)

78 = (n/2) (2a + (n – 1)d)

on cross multiplication,

156 = n ((2 × 24) + (n – 1) (-3))

156 = n (48 – 3n + 3)

156 = n (51 – 3n)

156 = 51n – 3n^{2}

3n^{2} – 51n + 156 = 0

3n^{2} – 12n – 39n + 156 = 0

3n (n – 4) – 39 (n – 4) = 0

(n – 4) (3n – 39) = 0

If n – 4 = 0

n = 4

or 3n – 39 = 0

3n = 39

n = 39/3

n = 13

So, n = 4

Then, T = a + (n – 1)d

= 24 +(4 – 1) (-3)

= 24 + 3 (-3)

= 24 – 9

= 15

n = 13

Then, T = a + (n – 1)d

= 24 +(13 – 1) (-3)

= 24 + 12 (-3)

= 24 – 36

= -12

So, (12 + 9 + 6 + 3 + 0 + (-3)+ (-6) + (-9) + (-12)) = 0

Hence, the sum of 5^{th} term to 13^{th} term = 0

**Question-8 ****Find the sum of first 22 terms, of an A.P. in which d = 7 and a**_{22} is 149.

_{22}is 149.

**Answer-8**

Sum of first 22 terms of an A.P. whose d = 7

a_{22} = 149 and n = 22

a_{22} = (n – 1)d

149 = a + (22 – 1)7

149 = a + (22)7

149 = a + 147

a = 149 – 147

a = 2

So, S_{22 }= (n/2) (2a + (n – 1)d)

= (22/2) ((2 × 2) + (22 – 1)7)

= 11(4 + (21)7)

= 11 (4 + 147)

= 11 (151)

= 1661

**Question-9 **In an AP the fourth and sixth term are 8 and 14 respectively find the

(i) first term

(ii) common difference

(iii) sum of first 20 term (2019)

**Answer-9**

T_{4} = 8 and T_{6} = 14

⇒ a + 3d = 8 … (i)

⇒ a + 5d = 14 … (ii)

Subtracting (i) from (ii),

(a + 5d) – (a + 3d) = 14 – 8

5d – 3d = 6

2d = 6

d = 6/2

d = 3

** Hence, common difference d = 3**

Substituting the value of d in (i),

a + 3(3) = 8

a = 8 – 9

a = -1

** Hence, the first term = -1**

(iii) The sum of first 20 terms

n = 20

S_{20} = n/2 × [2a + (n – 1)d]

= 20/2 × [2(-1) + (20 – 1)(3)]

= 10 × [-2 + (19)(3)]

= 10 × [-2 + 57]

= 10 × 55

= 550

Hence The sum of first 20 terms is 550

**Question -10 **

**(i) Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.**

**(ii) If the 4th term of an A.P. is 22 and 15th term is 66, find the ** **first term Common difference Hence find the sum of 8 term of the AP (2018)**

**Answer -10**

**(i) **Sum of first 51 terms of an A.P. in which

T_{2} = 14, T_{3} = 18

common difference d = T_{3} – T_{2}

= 18 – 14, = 4

Where, a = T_{1} = 14 – 4 = 10

n = 51

We know that,

S_{51 }= (n/2) (2a + (n – 1)d)

= (51/2) ((2 × 10) + (51 – 1)4)

= (51/2) (20 + (50 × 4))

= (51/2) (20 + 200)

= (51/2) × 220

= 5610

**(ii) T**_{4} = 22, T_{15} = 66

_{4}= 22, T

_{15}= 66

⇒ a + 3d = 22 … (i)

⇒ a + 14d = 66 … (ii)

Subtracting (i) from (ii), we get

(a + 14d) – (a + 3d) = 66 – 22

14d – 3d = 44

11d = 44

d = 44/11

d = 4

So, common difference d = 4

Substituting the value of d in (i), we have

a + 3(4) = 22

a = 22 – 12

a = 10

Hence, the first term a = 10

Now, the sum of first 8 terms of the A.P is

n = 8

S_{n} = n/2 × [2a + (n – 1)d]

S_{8} = 8/2 × [2(10) + (8 – 1)(4)]

= 10 × [20 + (7)(4)]

= 10 × [20 + 28]

= 10 × 48

= 480

### ML Aggarwal AP GP Class 10 ICSE Maths Solutions

Arithmetic and Geometric Progression Exe-9.3

**Question- 11 ****If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.**

**Answer- 11**

S_{6} = 36

S_{16} = 256

S_{n} = (n/2) (2a + (n – 1)d)

S_{6} = (6/2) (2a + (6 – 1)d) = 36

3 (2a + 5d) = 36

Divide both the side by 3,

2a + 5d = 12 … equation (i)

Now, S_{16} = (16/2) (2a + (16 – 1)d) = 256

8 (2a + 15d) = 256

Divide both the side by 8,

2a + 15d = 32 … equation (ii)

Then, subtract equation (ii) from equation (i)

(2a + 5d) – (2a + 15d) = 12 – 32

2a + 5d – 2a – 15d = -20

-10d = -20

d = -20/-10

d = 2

now substitute the value of d in equation (i) to find a,

2a + 5d = 12

2a + 5(2) = 12

2a + 10 = 12

2a = 12 – 10

2a = 2

a = 1

So, S_{10} = (n/2) (2a + (n – 1)d)

= (10/2) ((2 × 1) + (10 – 1)2)

= 5 (2 + 18)

= 5 (20)

= 100

Hence, the sum of first 10 terms is 100

**Question -12 ****Show that a**_{1}, a_{2}, a_{3}, … form an A.P. where a_{n} is defined as a_{n} = 3 + 4n. Also find the sum of first 15 terms.

_{1}, a

_{2}, a

_{3}, … form an A.P. where a

_{n}is defined as a

_{n}= 3 + 4n. Also find the sum of first 15 terms.

**Answer -12**

a_{n} = 3 + 4n

a_{1} = 3 + 4 x 1 = 3 + 4 = 7

a_{2} = 3 + 4 x 2 = 3 + 8 = 11

a_{3} = 3 + 4 x 3 = 3 + 12 = 15

a_{4} = 3 + 4 x 4 = 3 + 16 = 19

So, The numbers are 7, 11, 15, 19, ….

Then, first term a = 7, common difference d = 11 – 7 = 4

We know that,

S_{15} = (n/2) (2a + (n – 1)d)

= (15/2) ((2 × 7) + (15 – 1) × 4)

= (15/2) (14 + (14 × 4))

= (15/2) (14 + 56)

= (15/2) × 70

= 525

Hence, the sum of first 15 terms is 525.

**Question -13 **** Sum of first six terms of an arithmetic progression is 42 and the ratio of 10th term and 30th term is 1: 3 calculate first and thirteen term .**

**Answer -13**

S_{6} = 42 and T_{10}/T_{30} = 1/3

We know that,

S_{n} = (n/2) (2a + (n – 1)d)

T_{n} = a + (n – 1)d

So, we have

S_{6} = (6/2) × [2a + (6 – 1)d] = 42

3 × (2a + 5d) = 42

2a + 5d = 14 … (i) Dividing by 3

And,

T_{10}/T_{30} = [a + (10 – 1)d]/ [a + (30 – 1)d] = 1/3

(a + 9d)/ (a + 29d) = 1/3

On cross-multiplication,

3(a + 9d) = a + 29d

3a + 27d = a + 29d

3a – a + 27d – 29d = 0

2a – 2d = 0

a – d = 0

a = d

Using the above the relation in (i),

2a + 5a = 14

7a = 14

a = 14/7, a = 2

Hence, the first term of the A.P is 2.

the thirteenth term

T_{13} = 2 + (13 – 1)(2)

= 2 + 12 × 2

= 2 + 24, = 26

**Question-14 **In an AP the sum of first n term is 6n – n². Find is 25th term

**Answer-14**

S_{n} = 6n – n²

T_{25} = ?

S_{(n–1)} = 6(n – 1) – (n – 1)^{2}

= 6n – 6 – (n^{2} – 2n + 1)

= 6n – 6 – n^{2} + 2n –1

= 8n – n^{2} – 7

a_{n} = S_{n} – S_{n} – 1

= 6n – n^{2} – 8n + n^{2} + 7

= –2n + 7

a_{25 }= –2(25) + 7

= –50 + 7

= –43.

**Question-15 **If S_{n} donate the sum of first n term of an AP prove that S_{n} = 3( S_{20} – S_{10} )

**Answer-15**

S_{n} = n/2 × [2a + (n – 1)d]

S_{10} = 10/2 × [2a + (10 – 1)d]

= 5 × (2a + 9d)

= 10a + 45d

S_{20} = 20/2 × [2a + (20 – 1)d]

= 10 × (2a + 19d)

= 20a + 190d

S_{30} = 30/2 × [2a + (30 – 1)d]

= 15 × (2a + 29d)

= 30a + 435d

taking L.H.S we have

3 (S_{20} – S_{10}) = 3 [20a + 190d – (10a + 45d)]

= 3 (20a + 190d – 10a – 45d)

= 3 (10a + 145d)

= 30a + 435d

= S_{30}

= R.H.S

– Hence proved

**Arithmetic and Geometric Progression (AP GP )Exe-9.3**

**Question-16 **

#### (i) find the sum of first 1000 positive integers.

#### (ii) find the sum of first 15 multiple of 8.

**Answer-16**

**(i) the sum of first 1000 positive integers.**

Let 1 + 2 + 3 + …….. + 1000

This is an A.P with first term a = 1 and common difference d = 1

We know that,

S_{n} = n/2 × [2a + (n – 1)d]

S_{1000} = 1000/2 × [2(1) + (1000 – 1)(1)]

= 500 × (2 + 99)

= 500 × 101

= 50500

the** **sum of first 1000 positive integers is 50500.

**(ii) The first 15 multiples of 8 are:**

8, 16, 24, ….. where n = 15

This forms an A.P with first term a = 8 and common difference d = 16 – 8 = 8

S_{n} = n/2 × [2a + (n – 1)d]

S_{15} = 15/2 × [2(8) + (14 – 1)(8)]

= 15/2 × (16 + 13 × 8)

= 15/2 × (16 + 104)

= 15/2 × 120

= 15 × 60

= 900

Hence the sum of first 15 multiples of 8 is 900

Question-17

**(i) Find the sum of two digit natural numbers which are divisible by 4**

**(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4**

**(iii) Find the sum of all multiple of 9 lying between 300 and 700**

**(iv) Find the sum of all natural numbers less than 100 which are divisible by 6**

**Answer-17**

**(i) The two-digit natural numbers which are divisible by 4 are:**

4, 8, 12, 16, …..

This form an A.P.

The last term in this series is found out by dividing 100 by 4

100/4 = 25 and remainder is zero

So, the last two-digit number which is divisible by 4 is 96

And, it is the 24^{th} term

So, we have

a = 4, d = 4, last term l = 96 and n = 24

Thus, sum of these numbers is

S_{24} = 24/2 (4 + 96)

= 12 × 100

= 1200

**(ii) The natural numbers between 100 and 200 which are divisible by 4 are:**

104, 108, 112, …..

This form an A.P.

The last term in this series is found out by dividing 200 by 4

200/4 = 50 and remainder is zero

So, the last natural number between 100 and 200 which is divisible by 4 is 196

So, we have

a = 104, d = 4 and last term l = 96

The number of terms is found out by

T_{n} = 196 = 104 + (n – 1)(4)

196 = 104 + 4n – 4

4n = 196 – 104 + 4

4n = 96

n = 96/4

n = 24

sum of these numbers is

S_{24} = 24/2 (104 + 196)

= 12 × 200

= 2400

**(iii) The multiples of 9 lying between 300 and 700 are:**

306, 315, 324, ……

This form an A.P. where a = 306 and d = 9

The last term in this series is found out by dividing 700 by 9

700/9 gives 77 as quotient and 7 as remainder.

the last number between 300 and 700 which is a multiple is 700 – 7 = 693

a = 306, d = 9 and l = 693

The number of terms

T_{n} = 693 = 306 + (n – 1)(9)

693 = 306 + 9n – 9

9n = 693 – 306 + 9

9n = 396

n = 396/9

n = 44

, sum of these numbers is

S_{44} = 44/2 (306 + 693)

= 22 × 999

= 21978

**(iv) The natural numbers less than 100 which are divisible by 4 are:**

4, 8, 12, …., 96

This forms an A.P. where a = 4, d = 4 and l = 96

T_{n} = 96 = 4 + (n – 1)4

96 = 4 + 4n – 4

4n = 96

n = 96/4

n = 24

sum of these numbers is

S_{24} = 24/2 (4 + 96)

= 12 × 100

= 1200

— : End of ML Aggarwal **Arithmetic and Geometric Progression Exe-9.3** Class 10 ICSE Maths Solutions (AP GP) : –

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