Arithmetic Progression Class 10 RS Aggarwal Exe-10B Goyal Brothers ICSE Maths Solutions Ch-10. Step by step solutions of AP questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

## Arithmetic Progression Class 10 RS Aggarwal Exe-10B Goyal Brothers ICSE Maths Solutions Ch-10

Board | ICSE |

Publications | Goyal Brothers Prakashan |

Subject | Maths |

Class | 10th |

Chapter-10 | Arithmetic Progression |

Writer | RS Aggarwal |

Book Name | Foundation |

Exe-10B | Sum of n Term of AP |

Edition | 2024-2025 |

### Sum of n Term of AP

##### The formula to find the sum of n terms in AP is **S**_{n} = n/2 (2a+(n−1)d)

_{n}= n/2 (2a+(n−1)d)

Consider the general form of AP with first term as a, common difference as d and last term i.e. the n^{th} term as l. The sum of n terms of AP will be** = n/2 (a+l)**

**Exercise- 10B**

Arithmetic Progression Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-10

Find the sum :

**Que-1: 2+7+12+17+……… to 19 terms.**

**Sol: **Sum of n terms Sn = (n/2) (2a+(n−1)d)

a = 2 , d = 7 – 2 = 5 and n = 19

S19 = (19/2) (2×2+(19−1)5)

S19 = (19/2) (4+90)

S19 = (19/2) × 94

S19 = 893

**Que-2: 9+7+5+3……… to 14 terms.**

**Sol: **Sum of n terms Sn = (n/2) (2a+(n−1)d)

a = 9, d = -2, n = 14

S14 = (14/2){2 × 9+(14-1)×-2}

S14 = 7{18-26}

S14 = 7×-8

S14 = -56

**Que-3: (-11)+(-7)+(-3)+1+………. to 12th terms.**

**Sol: **Sum of n terms Sn = (n/2) (2a+(n−1)d)

a = -11, d = 4, n = 12

S12 = (12/2){2×-11+(12-1)4}

S12 = 6{-22+(11×4)}

S12 = 6(-22+44)

S12 = 6×22

S12 = 132.

**Que-4: (1/15),(1/12),(1/10), ………….. to 11th terms.**

**Sol: **Sum of n terms Sn = (n/2) (2a+(n−1)d)

a = 1/15, n = 11

d = a2−a1 = 1/12 − 1/15 = (5−4)/60 = 160

S11 = (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) = 33/20

**Que-5: 0.6+17+2.8+…………. to 100 terms.**

**Sol: **For this A.P.,

a = 0.6, d = a2−a1 = 1.7−0.6 = 1.1

n = 100

We know that,

Sn = (n/2)[2a+(n−1)d]

S100 = (100/2)[1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

**Que-6: 7+[10*(1/2)]+14+……….. + 84.**

**Sol: **Given,

First-term, a = 7

Common Difference, d = 10 ½ – 7 = 21/2 – 7 = 7/2

Last term, l = 84

l = aₙ = a + (n – 1) d

84 = 7 + (n – 1) 7/2

77 = (n – 1) 7/2

22 = n – 1

n = 23

Hence, n = 23

Sₙ = n/2 [a + l]

S₂₃ = 23/2 [7 + 84]

= (23/2) × 91

= 2093/2

= 1046 ½

**Que-7: 32+30+28+………. +10.**

**Sol: **Given,

First-term, a = 32

Common Difference, d = 30-32 = -2

Last term, l = 10

l = aₙ = a + (n – 1) d

10 = 32 + (n – 1) -2

-22 = (n – 1) -2

11 = n – 1

n = 12

Hence, n = 12

Sₙ = n/2 [a + l]

S₂₃ = 12/2 [32 + 10]

= 6 × 42

= 252.

**Que-8: (-5)+(-8)+(-11)+……… +(-62).**

**Sol: **Given,

First-term, a = -5

Common Difference, d = -8-(-5) = -3

Last term, l = -62

l = aₙ = a + (n – 1) d

-62 = -5 + (n – 1) -3

-57 = (n – 1) -3

19 = n – 1

n = 20

Hence, n = 20

Sₙ = n/2 [a + l]

S₂₃ = 20/2 [-5 + (-62)]

= 10 × (-67)

= -670.

**Que-9: Find the sum of all 2-digits natural number divisible by 5.**

**Sol: **AP = 10 , 15 , 20…….95

Here,

First term ( A) = 10

Common Difference ( D) = 5

Tn = 95

A + ( N-1) × D = 95

10 + ( N-1) × 5 = 95

10 + 5N -5 = 95

10 + 5N = 95+5

10 + 5N = 100

5N = 100-10

5N = 90

N = 90/5 = 18

Therefore,

Sn = N/2 × [ 2A + ( N-1) × D ]

S18 = 18/2 × [ 2 × 10 + ( 18-1) × 5 ]

=> 9 × ( 20 + 85)

=> 9 × 105

=> 945.

**Que-10: Find the sum of all even numbers between 10 and 100.**

**Sol: **Even numbers between 10 to 100 = 12, 14, 16, 18,….98, 100.

l = 100, a = 12, n = ?

According to the question,

l = a + (n − 1)d

⇒ 100 = 12 + (n – 1)2

⇒ n = 45

Now,

Sn = n/2(a + l)

⇒ Sn = (45/2)(12 + 100)

= (45/2) × 112

= 45×56

= 2520.

**Que-11: Find the sum of first fifteen multiples of 8.**

**Sol: **The first 8 multiples of 8 are

8, 16, 24, 32, 40, 48, 56,64…….120

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

S15 = ?

Sn = (n/2)[2a + (n – 1)d]

⇒ S15 = (15/2)[2(8) + (15 – 1)8]

⇒ S15 = 7.5[6 + (14) (8)]

⇒ S15 = 7.5[16 + 112]

⇒ S15 = 7.5(128)

⇒ S15 = 960

**Que-12: Find the sum of all odd numbers between 100 and 150.**

**Sol: **Odd number between 100 and 150 are :101,103,105,……..,149

a = 101, d = 2, l = 149

l = a + (n − 1)d

149 = 101+(n-1)2

149-101 = 2n-2

48+2 = 2n

50 = 2n

n = 25

Sn = n/2(a + l)

= (25/2)(101+149)

= (25/2) 250

= 25 × 125

= 3125.

**Que-13: The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and common difference. Hence, find the sum of the series to 8 terms.**

**Sol: **Let a be the first term and d be the common difference of given A.P.

Now,

4th term = 22

⇒ a + 3d = 22 …(i)

15th term = 66

⇒ a + 14d = 66

Subtracting (i) from (ii), we have

11d = 44

⇒ d = 4

Substituting the value of d in (1) we get

a = 22 – 3 x 4 = 22 – 12 =10

⇒ First term = 10

Now

Sum of 8 terms = (8/2)[(2×10)+(7×4)]

= 4[20+28]

= 4×48

= 192.

**Que-14: In an Arithmetic Progression (A.P.), the fourth and sixth term are 8 and 14 respectively. Find the : (i) first term (ii) common difference (iii) the sum of the first 20 terms**

**Sol: **Let the first term of the sequence is a and the common difference is d.

a_{4} = a +3d = 8 …(1)

a_{6} = a + 5d = 14 …(2)

On solving equation (1) and (2) we get

-2d = -6

d = 3

Put d = 3 in equation (1)

a + 3 × 3 = 8

a = – 1

∴ (i) First term (a) = –1

(ii) Common difference (d) = 3

(iii) Sum of the first 20 terms = Sn = (n/2)[2a+(n-1)d]

= (20/2)[2×(-1)+19×3]

= 10[-2+57]

= 10 × 55

= 550

**Que-15: The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and common difference.**

**Sol: **Let the three times terms of an A.P. be (a – d ),a, (a + d)

sum = 42 = (a – d) + a + (a + d)

42 = 3a

⇒ a = 42/3

⇒ a = 14

Also , (a – d ) (a + d) = 52

⇒ a²-d² = 52

d² = a²-52

d² = 196-52

d² = 144

⇒ d = ±12

∴ First term = a-d = 14-12 = 2

a+d = 14+12 = 26.

a = 2, d = 12 or

a = 26, d = -12.

**Que-16: If the 6th term of an A.P. is equal to four times its first term, and the sum of first six terms is 75, find the first term and common difference**

**Sol: **an = a1 + (n – 1) d

Taking n = 6,

⇒ a6 = a1 + (6 – 1) d

⇒ a6 = a1 + 5d —————– (1)

Given that, a6 = 4a1 —————— (2)

Substituting (2) in (1) we get,

⇒ 4a1 = a1 + 5d

⇒ 3a1 = 5d

⇒ a1 = 5d/3 ————– (3)

Sn = n / 2 [2a1 + ( n – 1) d ]

Given that, S6 = 75 and using a1 = 5d/3

⇒ 75 = 6 / 2 [2 (5d/3) + ( 6 – 1) d ]

⇒ 75 = 3 [ (10d/3) + 5d ]

⇒ 75 = 3 (10d + 15d ) / 3

⇒ 75 = 10d + 15d

⇒ 75 = 25d

⇒ d = 75 / 25

⇒ d = 3

Putting the value of d in (3) we get,

a1 = (5 × 3) /3

a1 = 5

**Que-17: The 5th term and the 9th term of an Arithmetic Progression are 4 and -12 respectively. Find: (i) the first term (ii) common difference (iii) sum of first 16 terms of an A.P.**

**Sol: **Let a and d be the first term and common difference of A.P.

Then by T_{n} = a + (n – 1)d

Given T_{5} = 4

⇒ a + 4d = 4 …(1)

And T_{9} = – 12

⇒ a + 8d = – 12 …(2)

Solving equations (1) and (2), we get

– 4d = 16

⇒ d = – 4

Put this value in equation (1)

a + 4 × (– 4) = 4

a – 16 = 4

a = 20

∴ (i) First term a = 20

(ii) Common difference d = – 4

(iii) Sum of n terms = (n/2)[2a+(n-1)d]

∴ Sum of 16 terms = (16/2)[2×20+15×(-4)]

= 8 [40 – 60]

= 8 × (– 20)

= – 160

**Que-18: Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60 ….. is 300? Hence find the sum of all the terms of the Arithmetic Progression (A.P.).**

**Sol: **Given A.P. is 15, 30, 45, 60………

Here a = 15, d = 30 – 15 = 15

Let T_{n} = 300

T_{n} = a + (n – 1)d

300 = 15 + (n – 1) × 15

300 = 15 + 15n – 15

15n = 300

∴ n = 20

Hence, 300 is the 20^{th} term

Also by S_{n} = (n/2)[2a+(n-1)d]

S_{20} = (20/2)[2×15+(20-1)×15]

= 10[30 + 285]

= 10 × 315

∴ S_{20} = 3150

**Que-19: A sum of Rs2800 is to be used to award four prizes. If each prices after the first is Rs200 less than the preceding prize, find the value of each of these prizes.**

**Sol: **Let the amount of the first prize be ₹a

Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.

Amount of the second prize = ₹ (a – 200)

Amount of the third prize = ₹ (a – 2 × 200) = (a-400)

Amount of the fourth prize = ₹(a -3 × 200) =(a-600)

Now,

Total sum of the four prizes = 2,800

∴ ₹a + ₹(a -200) + ₹ (a – 400) + ₹(a – 600) = ₹2,800

⇒ 4a – 1200 = 2800

⇒ 4a = 2800 + 1200 = 4000

⇒ a= 1000

Amount of the first prize = ₹1,000

Amount of the second prize = ₹ ( 1000 – 200) =₹800

Amount of the third prize = ₹( 1000 = 400) = ₹600

Amount of the fourth prize = ₹(1000-600) = ₹400

Hence, the value of each of the prizes is ₹1,000, ₹800, ₹600 and ₹400.

**Que-20: The production of TV sets in a factory increase uniformly by a fixed number every year. It produced 8000 sets in 6th year, and 11300 in 9th year. Find the production in (i) first year (ii) 8th year (iii) 6 year.**

**Sol: **Let a and b be the production in the first year and the common difference of the A.P.

Given that 8000 TV’s are produced in** **6th years and** **11300 in 9th** **year.

a6 = 8000

⇒ a+5d = 8000 …1

a9 = 11300

⇒ a+8d = 11300 …2

Subtracting (1) from (2).

8d-5d = 11300-8000

⇒ 3d = 3300

⇒ d = 1100

Computing the production of 1st year

(i) Computing the value of a:

a+(5×1100) = 8000

⇒ a = 8000-5500

⇒ a = 2500

Computing the production of 8th year:

(ii) a8 = a+7d

= 2500+(7×1100)

= 2500+7700

= 10200

(iii) For 6 years n = 6

S_{n} = (n/2)[2a+(n-1)d]

= (6/2)[(2×2500)+(6-1)1100]

= 3[5000+5500]

= 3 × 10500

= 31500.

**Que-21: 200 logs are stacked in a such way that there are 20 logs in the bottom row, 19 logs in the next row, 18 logs in the next year and so on. How many rows are formed and how many logs are there in the top row?**

**Sol: **For this A.P.,

First terms, a = 20

Common difference, d = 19 – 20 = – 1

Sum of the n terms, Sₙ = 200

We know that of AP is given by the formula Sₙ = n/2 [2a + (n – 1) d]

200 = n/2 [2 × 20 + (n – 1)(- 1)]

400 = n [40 – n + 1]

400 = n [41 – n]

400 = 41n – n²

n² – 41n + 400 = 0

n² – 16n – 25n + 400 = 0

n(n – 16) – 25(n – 16) = 0

(n – 16)(n – 25) = 0

Either (n -16) = 0 or (n – 25) = 0

∴ n = 16 or n = 25

aₙ = a + (n – 1) d

a₁₆ = 20 + (16 -1) × (- 1)

a₁₆ = 20 – 15

a₁₆ = 5

Therefore, 200 logs can be placed in 16 rows. The number of logs in the top (16^{th}) row is 5.

**Que-22: In a flower bed there are 43 rose plants in the first row, 41 in the second row, 39 in the third row and so on. There are 11 rose plants in the last row. How many rows are are there in the flower bed? How many rose plants are there in the flower bed?**

**Sol: **The number of rose plants in consecutive rows are 43,41,39…11.

Difference of rose plant between two consecutive rows = (41-43) = (39-41) = -2

So, the given progression is an AP

Here, first term = 43

Common difference = -2

Last term 11

Let n be the last term, then we have

Tn = a+(n-1) d

11 = 43+(n-1) (-2)

11 = 45-2n

34 = 2n

n = 17

S_{n} = (n/2)[2a+(n-1)d]

= (17/2)[(2×43)+(17-1)(-2)]

= (17/2)[86-32]

= (17/2) (54)

= 17 × 27

= 459.

**Que-23: A man saved Rs33000 in 10 months. In each month after the first, he saves Rs100 more than he did in the preceding month. How much did he save in the first month?**

**Sol: **Let the money saved by the man in the first month be ₹a

It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.

∴ d = Rs100

Number of months, n = 10

Sum of money saved in 10 months, s_{10} = ₹ 33,000

Using the formula ,Using the formula ,Sn = (n/2)[2a+(n-1)d],we get

S10 = (10/2)[2a+(10-1)×100] = 33000

⇒ 5 (2a +900 ) = 33000

⇒ 2a +900 = 6600

⇒ 2a = 6600-900 = 5700

⇒ a = 2850

–: End of Arithmetic Progression Class 10 RS Aggarwal Exe-10B Goyal Brothers ICSE Maths Solutions :–

Return to :– ** RS Aggarwal ICSE Class 10 Solutions Goyal Brothers**

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