ML Aggarwal Arithmetic and Geometric Progression Exe-9.2 Class 10 ICSE Maths Solutions . We Provide Step by Step Answer of Exe-9.2 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 Exe-9.2 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-9 | Arithmetic and Geometric Progression |

Writer / Book | Understanding |

Topics | Solutions of Exe-9.2 |

Academic Session | 2024-2025 |

### Arithmetic and Geometric Progression Exe-9.2

(ML Aggarwal AP GP Class 10 ICSE Maths Solutions)

Page-173

**Question-1 ****Find the A.P. whose nth term is 7 – 3n Also find the 20th term.**

**Answer -1**

T_{n} = 7 – 3n

Giving values 1, 2, 3, 4, … to n, we get

T_{1} = 7 – 3 x 1 = 7 – 3 = 4

T_{2} = 7 – 3 x 2 = 7 – 6 = 1

T_{3} = 7 – 3 x 3 = 7 – 9 = -2

T_{4} = 7 – 3 x 4 = 7 – 12 = -5

T_{20} = 7 – 3 x 20 = 7 – 60 = -53

A.P. is 4, 1, -2, -5, …

20th term = -53

**Question- 2 ****Find the indicated terms in each of following A.P.s:**

**(i) 1, 6, 11, 16, …; a**_{20}

**(ii) – 4, – 7, – 10, – 13, …, a**_{25}, a_{n}

_{20}

_{25}, a

_{n}

**Answer -2**

**(i) 1, 6, 11, 16, …**

Here, a = 1, d = 6 – 1 – 5

a_{20} = a + (n – 1 )d

= 1 + (20 – 1) x 5

= 1 + 19 x 5

= 1 + 95

= 96

**(ii) – 4, – 7, – 10, – 13, …, a _{25}, a_{n}**

The first term a = -4

Then, difference d = -7 – (-4) = – 7 + 4 = -3

-10 – (-7) = -10 + 7 = -3

-13 – (-10) = -13 + 10 = -3

Therefore, common difference d = -3

From the formula, a_{n} = a + (n – 1)d

So, a_{25} = a + (25 – 1)d

= -4 + (25 – 1)(-3)

or -4 + (24)-3

= – 4 – 72

= -76

Therefore, a_{25} = -76

Now, a_{n} = a + (n – 1)d

a_{n} = -4 + (n – 1)-3

= -4 – 3n + 3

= -1 – 3n

**Question -3 ****Find the nth term and the 12th term of the list of numbers: 5, 2, – 1, – 4, …**

**Answer -3**

5, 2, -1, -4, …

Here, a = 5 d = 2 – 5 = -3

**nth term**

T_{n} = a + (n – 1)d

= 5 + (n – 1) (-3)

= 5 – 3n + 3

= 8 – 3n

**and 12th term**

T_{12} = a + 11d

= 5 + 11(-3)

= 5 – 33

= -28

**Question -4**

**(i) If the common difference of an A.P. is – 3 and the 18th term is – 5, then find its first term.**

**(ii) If the first term of an A.P. is – 18 and its 10th term is zero, then find its common difference.**

**Answer-4**

**(i) **Common difference (d) = -3

T_{18} = -5

T_{n} = a + (n – 1)d

So, T_{18} = a + (18 – 1)d

-5 = a + (18 – 1)(-3)

-5 = a + (17)(-3)

-5 = a – 51

a = 51 – 5

a = 46

Therefore, first term a = 46

**(ii) **The 10^{th} term = 0

Then, first term a = -18

T_{n} = a + (n – 1)d

So, T_{10} = a + (10 – 1)d

0 = -18 + (10 – 1)d

0 = -18 + 9d

9d = 18

d = 18/9

d = 2

Therefore, common difference d = 2

**Question -5 ****Which term of the A.P.**

**(i) 3, 8, 13, 18, … is 78?**

**(ii) 18, 15½, 13, … is – 47?**

**Answer -5**

**(i) 3, 8, 13, 18, … is 78**

Let 78 is nth term

Here, a = 3, d = 8 – 3 = 5

78 = a + (n-1) d

78 = 3 + (n-1) 5

78 – 3 = (n-1) 5

75/5 = n-1

15+1 = n

16 = n

**(ii) 18, 15½, 13, … is – 47?**

15½ = 31/2

Let -47 as n^{th} term.

The first term a = 18

Then, difference d = 31/2 – 18 = (31 – 36)/2 = -5/2

13 – 31/2 = (26 – 31)/2 = -5/2

Therefore, common difference d = -5/2

T_{n} = a + (n – 1)d

So, -47 = a + (n – 1)d

-47 = 18 + (n – 1) (-5/2)

-47 = 18 – 5/2n + 5/2

-47 – 18 = -5/2n + 5/2

-65 = -5/2n + 5/2

-65 – 5/2 = – 5/2n

(-130 – 5)/2 = -5/2n

-135/2 = -5/2n

n = (-135/2) × (-2/5)

n = -135/-5

n = 27

### AP GP, Exe-9.2,

ML Aggarwal Arithmetic and Geometric Progression Class 10 ICSE Maths Solutions

**Question -6**

**(i) Check whether – 150 is a term of the A.P. 11, 8, 5, 2, …**

**(ii) Find whether 55 is a term of the A.P. 7, 10, 13, … or not. If yes, find which term is it.**

**Answer-6**

**(i) **A.P. is 11, 8, 5, 2, …

Here, a = 11, d = 8 – 11 = -3

Let -150 = nth term, then

T_{n} = a + (n – 1)d

So, -150 = 11 + (n – 1) (-3)

-150 = 11 – 3n + 3

-150 = 14 – 3n

3n = 150 + 14

3n = 164

n = 164/3

n is not a natural number Therefore, – 150 is not a term of the A.P. of 11, 8, 5, 2,

**(ii) **The first term a = 7

Then, difference d = 10 – 7 = 3

13 – 10 = 3

Then, common difference d = 3

Let us assume 55 as n^{th} term,

T_{n} = a + (n – 1)d

So, 55 = 7 + (n – 1)3

55 = 7 + 3n – 3

55 = 4 + 3n

3n = 55 – 4

3n = 51

n = 51/3

n = 17

Therefore, 55 is 17^{th} term of the A.P. 7, 10, 13

**Question-7 **

**(i) Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.**

**(ii) Find the 12th from the end of the A.P. – 2, – 4, – 6, …; – 100.**

**Answer-7**

(i) A.P. is 3, 8, 13, …, 253

12th term from the end

Last term = 253

Here, a = 3, d = 8 – 3 = 5

Therefore, common difference d = 5

T_{n} = a + (n – 1)d

So, 253 = a + (n – 1)d

253 = 3 + (n – 1)5

253 = 3 + 5n – 5

253 = -2 + 5n

5n = 253 + 2

5n = 255

n = 255/5

n = 51

Therefore, 253 is 51^{th} term.

Now, assume ‘P’ be the 20^{th} term from the last.

Then, P = L – (n – 1)d

= 253 – (20 – 1) 5

= 253 – (19) 5

= 253 – 95

P = 158

Therefore, 158 is the 20^{th} term from the last.

(ii) Let -100 as n^{th} term.

The first term a = -2

Then, difference d = – 4 – (-2) = – 4 + 2 = -2

-6 – (-4) = -6 + 4 = – 2

common difference d = -2

T_{n} = a + (n – 1)d

So, – 100 = a + (n – 1)d

– 100 = -2 + (n – 1)(-2)

– 100 = -2 – 2n + 2

– 100 = -2n

n = -100/-2

n = 50

Therefore, -100 is 50^{th} term.

let ‘P’ be the 12^{th} term from the last.

Then, P = L – (n – 1)d

= -100 – (12 – 1) (-2)

= -100 – (11) (-2)

= -100 + 22

P = – 78

Hence, -78 is the 12^{th} term from the last of the A.P. – 2, – 4, – 6, …

**Question-8 ****Find the sum of the two middle most terms of the A.P.**

**Answer -8**

Last term (n^{th}) ** **= 13/3

First term a = -4/3

Then, difference d = -1 – (-4/3) = -1 + 4/3 = (-3 + 4)/3 = 1/3

= -2/3 – (-1) = -2/3 + 1 = (-2 + 3)/3 = 1/3

common difference d = 1/3

T_{n} = a + (n – 1)d

So, 13/3 = -4/3 + (n – 1)(1/3)

13/3 + 4/3 = 1/3n – 1/3

13/3 + 4/3 + 1/3 = 1/3n

(13 + 4 + 1)/3 = 1/3n

18/3 = 1/3n

6 = 1/3n

n = 6 × 3

n = 18

So, middle term is 18/2 and (18/2) + 1

= 9^{th} and 10^{th} term

Then, a_{9} + a_{10} = a + 8d + a + 9d

= 2a + 17d

Now substitute the value of ‘a’ and ‘d’.

= 2(-4/3) + 17(1/3)

= -8/3 + 17/3

= (-8 + 17)/3

= 9/3

= 3

Hence, the sum of the two middle most terms of the A.P is 3.

**Question-9 ****Which term of the A.P. 53, 48, 43,… is the first negative term ?**

**Answer -9**

Let nth term is the first negative term of the A.P. 53, 48, 43, …

Here, a = 53, d = 48 – 53 = -5

.’. T_{n} = a + (n – 1 )d

= 53 + (n – 1) x (-5)

= 53 – 5n + 5

= 58 – 5n

5n = 58

n= 58/5

n = 11.6 = 12th

Hence, 12^{th} term is the first negative term** **of the A.P. 53, 48, 43,…

**Question-10 ****Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12**

**Answer -10**

T_{3} = 16

T_{7} – T_{5} = 12

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

T_{n} = a + (n – 1)d

So, T_{3} = a + 2d = 16 … equation (i)

T_{7} – T_{5 }= (a + 6d) – (a + 4d) = 12 … equation (ii)

12 = a + 6d – a – 4d

12 = 2d

d = 12/2

d = 6

substitute value of d in equation (i)

Then, T_{3} = a + 2d

16 = a + 2(6)

a = 16 – 12

a = 4

So, A.P. is 4 + 6 = 10,

10 + 6 = 16,

16 + 6 = 22

Hence, the four term of A.P. are 4, 10, 16, 22, ….

### Arithmetic and Geometric Progression Exe-9.2

(ML Aggarwal AP-GP Class 10 ICSE Maths Solutions)

**Question -11 ****Find the 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.**

**Answer -11**

a= 12 given, and 7th term is 24 less than the 11th term = T_{11} – T_{7} = 24

T_{11} – T_{7 }= (a + 10d) – (a + 6d) = 24

24 = a + 10d – a – 6d

24 = 4d

d = 24/4

d = 6

Now, T_{20} = a + 19d

Substitute the values of a and d,

T_{20} = 12 + 19(6)

T_{20} = 12 + 114

T_{20} = 126

**Question-12 ****Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.**

**Answer-12**

T_{11} = 38, T_{6} = 73

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

a + 10d = 38..(i)

a + 5d = 73…(ii)

on Subtracting

(a + 10d) – (a + 5d) = 73 – 38

a + 10d – a – 5d = 35

5d = 35

d = 35/5

d = 7

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

a + 10d = 38

a + 10(7) = 38

a + 70 = 38

a = 38 – 70

a = -32

Therefore, T_{31} = a + 30d

= -32 + 30(7)

= -32 + 210

= 178

**Question-13 **** If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its 63**^{rd} term.

^{rd}term.

**Answer -13**

T_{9} = 1/7

T_{7} = 1/9

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So, T_{9} = a + 8d = 1/7 equation (i)

T_{7} = a + 6d = 1/9 equation (ii)

Subtracting both equation (i) and equation (i),

(a + 6d) – (a + 8d) = 1/9 – 1/7

a + 6d – a – 8d = (7 – 9)/63

-2d = -2/63

d = (-2/63) × (-1/2)

d = 1/63

now, substitute the value of d in equation (ii) to find out a, we get

a + 6(1/63) = 1/9

a = 1/9 – 6/63

a = (7 – 6)/63

a = 1/63

Therefore, T_{63} = a + 62d

= 1/63 + 62(1/63)

= 1/63 + 62/63

= (1 + 62)/63

= 63/63 = 1

**Question-14**

**(i) The 15th term of an A.P. is 3 more than twice its 7th term. If the 10th term of the A.P. is 41, find its nth term.**

**(ii) The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.**

**Answer -14**

(i) T_{10} = 41

T_{10} = a + 9d = 41 … equation (i)

T_{15} = a + 14d = 2T_{7} + 3

= a + 14d = 2(a + 6d) + 3

= a + 14d = 2a + 12 d + 3

-3 = 2a – a + 12d – 14d

a – 2d = -3 … equation (ii)

, subtracting equation (ii) from (i), we get,

(a + 9d) – (a – 2d) = 41 – (-3)

a + 9d – a + 2d = 41 + 3

11d = 44, d = 44/11, d = 4

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

a + 9(4) = 41

a + 36 = 41

a = 41 – 36, a = 5

Hence, n^{th} term = T_{n} = a + (n – 1)d

= 5 + (n – 1)4

= 5 + 4n – 4

= 1 + 4n

(ii) a_{5} + a_{7} = 52

(a + 4d) + (a + 6d) = 52

a + 4d + a + 6d = 52

2a + 10d = 52

Divide both the side by 2

a + 5d = 26 … equation (i)

Given, a_{10} = a + 9d = 46

a + 9d = 46 … equation (ii)

Now subtracting equation (i) from equation (ii),

(a + 9d) – (a + 5d) = 46 – 26

a + 9d – a – 5d = 20

4d = 20 , d = 20/4 , d = 5

Substitute the value of d in equation (i) to find out a,

a + 5d = 26

a + 5(5) = 26

a + 25 = 26

a = 26 – 25, a = 1

Then, a_{2} = a + d

= 1 + 5 = 6

a_{3} = a_{2} + d

= 6 + 5

= 11

a_{4} = a_{3} + d

= 11 + 5

= 16

Therefore, 1, 6, 11, 16,… are A.P.

**Question -15 ****If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.**

**Answer -15**

T_{8} = 0

To prove that T_{38} = 3 x T_{18}

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

T_{8} = a + 7d = 0

T_{8} = a = -7d

T_{38} = a + 37d

= -7d + 37d

= 30d

Take, T_{18} = a + 17d

Substitute the value of a and d,

T_{18} = -7d + 17d

T_{18} = 10d

on comparing …… T_{38} and T_{18}, it is clear that 38^{th} term is triple of its 18^{th} term.

### Arithmetic and Geometric Progression (AP GP) Exe-9.2 ML Aggarwal Class 10 ICSE Maths Solutions

**Question -16 ****Which term of the A.P. 3, 10, 17,… will be 84 more than its 13th term?**

**Answer -16**

A.P. is 3, 10, 17, …

Here, a = 3, d – 10 – 3 = 7

T_{13} = a + 12d

= 3 + 12 x 7

= 3 + 84

= 87

Let , n^{th} term is 84 more than its 13^{th} term

So, T_{n} = 84 + 87

= 171

T_{n} = a + (n – 1)d = 171

3 + (n – 1)7 = 171

3 + 7n – 7 = 171

7n – 4 = 171

7n = 171 + 4

7n = 175

n = 175/7

n = 25

**Question -17**

**(i) How many two digit numbers are divisible by 3 ?**

**(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.**

**Answer -17**

(i) The two digits numbers which is divisible by 3 are

= 12, 15, 18, 21, 24,…..,99.

first number a = 12

Common difference d = 15 – 12 = 3

last number is 99

T_{n }(last number) = a + (n – 1)d

99 = 12 + (n – 1)3

99 = 12 + 3n – 3

99 = 9 + 3n

99 – 9 = 3n

3n = 90

n = 90/3 , n = 30

(ii) The natural numbers which are divisible by both 2 and 5 are

= 110, 120, 130, 140, ….,990

The above numbers are A.P.

Here a = 110, d = 120 – 110 = 0

a_{n} = 990

⇒ a + (n – 1)d = 990

⇒ 110 + (n – 1)(10) = 990

⇒ (n – 1)(10)

= 990 – 110

= 880

⇒ (n – 1) = 880/10 = 88

∴ n = 88 + 1

= 89

Hence, number between 101 and 999 which are divisible by both 2 and 5 are 89.

**Question -18 ****If the numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n.**

**Answer-18**

given that,

n – 2, 4n – 1 and 5n + 2 are in A.P.

If AP then common deference should be same

(4n – 1 )-(n-2) =(5n + 2)- (4n – 1 )

on solving

4n -1 -n+2 = 5n+2 -4n+1

3n+1 = n+3

3n-n= 3-1

2n = 2, n= 2/2 , n= 1

**Question-19 ****The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.**

**Answer -19**

Sum of three numbers which are in A.P. = 3

Their product = -35

Let three numbers which are in A.P.

a – d, a, a + d

a – d + a + a + d = 3

⇒ 3a = 3 , a = 1

product of 3 numbers is – 35

So, (a – d) × (a) × (a + d) = – 35

(1 – d) × (1) × (1 + d) = – 35

1^{2} – d^{2} = – 35

d^{2} = 35 + 1

d^{2} = 36

d = √36

d = ±6

the numbers are (a – d) = 1 – 6 = – 5

a = 1

(a + d) = 1 + 6 = 7

If d = – 6

The numbers are (a – d) = 1 – (-6) = 1 + 6 = 7

a = 1

(a + d) = 1 + (-6) = 1 – 6 = -5

Hence, the numbers -5, 1, 7,… and 7, 1, -5,… are in A.P.

** AP GP Exe-9.2 ML Aggarwal Solutions**

Page-174

**Question-20 ****The sum of three numbers in A.P. is 30 and the ratio of first number to the third number is 3 : 7. Find the numbers.**

**Answer -20**

Sum of three numbers in A.P. = 30

Ratio between first and the third number = 3 : 7

Let numbers be

a – d, a, a + d, then

a – d + a + a + d = 30

3a = 30

a = 30/3

a = 10

Given ratio 3 : 7 = a – d : a + d

3/7 = (a – d)/(a + d)

(a – d)7 = 3(a + d)

7a – 7d = 3a + 3d

7a – 3a = 7d + 3d

4a = 10d

4(10) = 10d

40 = 10d

d = 40/10

d = 4

the numbers are a – d = 10 – 4 = 6

a = 10

a + d = 10 + 4 = 14

So, 6, 10, 14, … are in A.P.

**Question -21 ****The sum of the first three terms of an A.P.is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.**

**Answer -21**

Let the three numbers in A.P. are

a – d, a, a + d

Now, a – d + a + a + d = 33

⇒ 3a = 33

a = 11

It is given that

the product of the first and the third terms exceeds the second term by 29.

(a – d) (a + d) = a + 29

a^{2} – d^{2} = 11 + 29

11^{2} – d^{2} = 40

121 – 40 = d^{2}

d^{2} = 81

d =√81

d = ±9

If d = 9

Therefore, the numbers are (a – d) = 11 – 9 = 2

a = 11

(a + d) = 11 + 9 = 20

If d = – 9

The numbers are (a – d) = 1 – (-9) = 11 + 9 = 20

a = 11

(a + d) = 11 + (-9) = 11 – 9 = 2

Therefore, the numbers 2, 11, 20,… and 20, 11, 2,… are in A.P.

— : End of ML Aggarwal Arithmetic and Geometric Progression Exe-9.2 Class 10 ICSE Maths Solutions : –

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