RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11C Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11C Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 11
Chapter NameArithmetic Progressions
Exercise11 C
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C PDF

Question 1.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.3
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.4

Question 2.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.3
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.4
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.5

Question 3.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 3.1

Question 4.
Solution:
Sn = 3n² + 6n
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 4.1

Question 5.
Solution:
Sn = 3n² – n
S1 = 3(1)² – 1 = 3 – 1 = 2
S2 = 3(2)² – 2 = 12 – 2 = 10
T2 = 10 – 2 = 8 and
T1 = 2
(i) First term = 2
(ii) Common difference = 8 – 2 = 6
Tn = a + (n – 1) d = 2 + (n – 1) x 6
= 2 + 6n – 6 = 6n – 4

Question 6.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.3

Question 7.
Solution:
Let a be first term and d be the common difference of an AP.
Since, we have,
am = a + (m – 1) d = 1n …(i)
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 7.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 7.2

Question 8.
Solution:
AP is 21, 18, 15,…
Here, a = 21,
d = 18 – 21 = -3,
sum = 0
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 8.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 8.2

Question 9.
Solution:
AP is 9, 17, 25,…
Here, a = 9, d = 17 – 9 = 8
Sum of terms = 636
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 9.1
Which is not possible being negative and fraction.
n = 12
Number of terms = 12

Question 10.
Solution:
AP is 63, 60, 57, 54,…
Here, a = 63, d = 60 – 63 = -3 and sum = 693
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 10.1
22th term is zero.
There will be no effect on the sum.
Number of terms are 21 or 22.

Question 11.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.3

Question 12.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49
Here, a = 1, d = 3 – 1 = 2, l = 49
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 12.1
= 252 x 50 = 25 x 25 = 625

Question 13.
Solution:
Numbers between 200 and 400 which are divisible by 7 will be 203, 210, 217,…, 399
Here, a = 203, d = 7, l = 399
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 13.1

Question 14.
Solution:
First forty positive integers are 0, 1, 2, 3, 4, …
and numbers divisible by 6 will be 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 12 – 6 = 6, n = 40 .
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 14.1

Question 15.
Solution:
First 15 multiples of 8 are 8, 16, 24, 32, … to 15 terms
Here, a = 8, d = 16 – 8 = 8 , n = 15
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 15.1

Question 16.
Solution:
Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, …, 693
Here, a = 306, d = 9, l = 693
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 16.1

Question 17.
Solution:
Three digit numbers are 100, 101, …, 999
and numbers divisible by 13, will be 104, 117, 130, …, 988
Here, a = 104, d = 13, l = 988
Tn (l) = a + (n – 1) d
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 17.1

Question 18.
Solution:
Even natural numbers are 2, 4, 6, 8, 10, …
Even natural numbers which are divisible by 5 will be
10, 20, 30, 40, … to 100 terms
Here, a = 10, d = 20 – 10 = 10, n = 100
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 18.1

Question 19.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 19.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 19.2

Question 20.
Solution:
Let a be the first term and d be the common difference of the AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 20.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 20.2

Question 21.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 21.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 21.2

Question 22.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 22.1

Question 23.
Solution:
In an AP
a = 17, d = 9, l = 350
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 23.1

Question 24.
Solution:
Let a be the first term, d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 24.1

Question 25.
Solution:
In an AP, let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 25.1

Question 26.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 26.1

Question 27.
Solution:
Let a be first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 27.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 27.2

Question 28.
Solution:
Let first term be a and d be the common difference in an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 28.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 28.2

Question 29.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 29.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 29.2

Question 30.
Solution:
Let a1 and a2 be the first terms of the two APs and d be the common difference, then
a1 = 3, a2 = 8
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 30.1

Question 31.
Solution:
Let a be the first term and d be the common difference, then
S10 = -150
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 31.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 31.2

Question 32.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 32.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 32.2

Question 33.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 33.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 33.2

Question 34.
Solution:
(i) AP is 5, 12, 19,… to 50 terms
Here, a = 5, d = 12 – 5 = 7, n = 50
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 34.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 34.2

Question 35.
Solution:
Let a1, a2 be the first term and d1, d2 be common difference of the two AP’s respectively.
Given, ratio of sum of first n terms =
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 35.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 35.2

Question 36.
Solution:
Let a be the first term and d be the common difference of an AP.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 36.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 36.2

Question 37.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 37.1

Question 38.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 38.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 38.2

Question 39.
Solution:
AP is -12, -9, -6,…, 21
Here, a = -12, d = -9 – (-12) = -9 + 12 = 3, l = 21
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 39.1

Question 40.
Solution:
S14 = 1505
Let a be the first term and d be the common difference, then
a = 10
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 40.1

Question 41.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 41.1

Question 42.
Solution:
In the school, there are 12 classes, class 1 to 12 and each class has two sections.
Each class plants double of the class
i.e. class 1 plants two plant, class 2 plants 4 plants, class 3 plants 6 plants, and so on.
So, total plants will be for each class each sections = 2 + 4 + 6 + 8 … + 24
Here, a = 2, d = 2, l = 24, n = 12
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 42.1
Each class has. two sections.
Plants will also be doubt
i.e. Total plants = 156 x 2 = 312

Question 43.
Solution:
In a potato race,
Bucket is at 5 m from first potato and then the distance between the two potatoes is 3m.
There are 10 potatoes.
The player, pick potato and put it in the bucket one by one.
Total distance in m to be covered, for 1st, 2nd, 3rd, … potato.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 43.1

Question 44.
Solution:
Total number of trees = 25
Distance between them = 5 m in a line.
There is a water tank which is 10 m from the first tree.
A gardener waters these plants separately.
Total distance for going and coming back
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 44.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 44.2

Question 45.
Solution:
Total sum = ₹ 700
Number of cash prizes = 7
Each prize in ₹ 20 less than its preceding prize.
Let first prize = ₹ x
Then second prize = ₹ (x – 20)
Third prize = ₹ (x – 40) and so on.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 45.1

Question 46.
Solution:
Total savings = ₹ 33000
Total period = 10 months
Each month, a man saved ₹ 100 more than its preceding of month.
Let he saves ₹ x in the first month.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 46.1

Question 47.
Solution:
Total debt to be paid = ₹ 36000
No. of monthly installments = 40
After paying 30 installments, 13 of his debt left
i.e. ₹ 36000 x 13 = ₹ 12000
and amount paid = ₹ 36000 – ₹ 12000 = ₹ 24000
Monthly installments are in AP.
Let first installment = x
and common difference = d
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 47.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 47.2

Question 48.
Solution:
A contractor will pay the penalty for not doing the work in time.
For the first day = ₹ 200
For second day = ₹ 250
For third day = ₹ 300 and so on
The work was delayed for 30 days
Total penalty, he paid
200 + 250 + 300 + ….. to 30 terms
Here, a = 200, d = 50, n = 30
Total sum = n2 [2a + (n – 1) d]
= 302 [2 x 200 + (30 – 1) x 50]
= 15 [400 + 29 x 50]
= 15 [400 + 1450]
= 15 x 1850 = ₹ 27750

Question 49.
Solution:
Child will put 5 rupee on 1st day, 10 rupee (2 x 5 rupee)
on 2nd day, 15 rupee (3 x 5 rupee) on 3rd day etc.
Total saving = 190 coins = 190 x 5 = 950 rupee
The above problem can be written as Arithmetic Progression series
5, 10, 15, 20, ……
with a = 5, d = 5, Sn = 950
Let n be the last day when piggy bank become full.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 49.1
⇒ n (n + 20) – 19 (n + 20) = 0
⇒ (n + 20) (n – 19) = 0
⇒ n + 20 = 0 or n – 19 = 0
⇒ n = -20 or n = 19
cannot be negative, hence n = 19
She can put money for 19 days.
Total saving is 950 rupees.

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