RD Sharma Solutions for Class 11 Maths Chapter 19 Arithmetic Progressions

Mathematics is a subject that helps us understand the world around us using numbers, sequences, and mathematical patterns. One of the most fundamental topics in sequences and series is Arithmetic Progressions, and it plays a big role in our daily lives too—like when we calculate the total amount saved through monthly deposits, analyze the growth pattern of population over years, or determine the seating capacity in a stadium with uniform row arrangements. Chapter 19 in the RD Sharma Class 11 Maths book takes us deeper into advanced concepts of arithmetic progressions, building upon the foundation laid in earlier chapters. To make learning easier and more enjoyable, RD Sharma Solutions for this chapter give clear explanations and step-by-step answers to all the exercise questions.

In this chapter, students will explore advanced applications of arithmetic progressions, including complex word problems, relationship between multiple APs, and practical applications in various fields. Students will learn about the properties of arithmetic progressions when terms are manipulated, such as selecting alternate terms, inserting multiple arithmetic means, and finding relationships between different arithmetic sequences. These concepts go beyond basic AP formulas and help students understand how arithmetic progressions work in more complex mathematical scenarios. Then, we move on to advanced problem-solving techniques involving multiple conditions, constraints, and real-world applications that require deeper analytical thinking.

The chapter also covers advanced topics like arithmetic progression of higher order, relationship between arithmetic and geometric means in the context of APs, and solving systems of equations involving arithmetic progressions. These help us understand how sequences connect with other mathematical concepts and how they are applied in advanced mathematics. One of the most important parts of this chapter is mastering the art of setting up equations from complex word problems and solving them systematically using AP principles.

The RD Sharma Solutions for Class 11 Chapter 19 give students detailed answers to all textbook problems. Each solution is written in a simple way so that students can understand the logic and steps behind it. These solutions also include alternative methods, important shortcuts, and useful tips to help students solve similar questions easily in their board exams and competitive entrance tests like JEE.

By studying these solutions, students will not only do well in tests but also build a strong foundation in advanced sequence analysis. These concepts are essential for higher mathematics, calculus, and various entrance examinations too. So, whether you're preparing for your board exams or competitive entrance tests, RD Sharma Solutions for Class 11 Chapter 19 are a comprehensive guide to help you master advanced arithmetic progressions.

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RD Sharma Class 11 Chapter 19 PDF includes detailed solutions, examples, and extra questions to help you master advanced arithmetic progressions and other topics. Click here to download the RD Sharma Class 11 Chapter 19 PDF.

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In this chapter, students will learn about advanced arithmetic progressions and how to solve complex problems involving multiple sequences and real-world applications. The solutions provided here are detailed and easy to follow, helping students understand each concept thoroughly.

Q1. If the sum of first p terms of an AP is q and the sum of first q terms is p, find the sum of first (p+q) terms.

Let the first term be a and common difference be d.
Given: Sp = q and Sq = p

Using Sn = n/2[2a + (n-1)d]:
Sp = p/2[2a + (p-1)d] = q ... (1)
Sq = q/2[2a + (q-1)d] = p ... (2)

From (1): p[2a + (p-1)d] = 2q
From (2): q[2a + (q-1)d] = 2p

Expanding: 2ap + p(p-1)d = 2q ... (3)
2aq + q(q-1)d = 2p ... (4)

Subtracting (4) from (3):
2a(p-q) + d[p(p-1) - q(q-1)] = 2(q-p)
2a(p-q) + d[p² - p - q² + q] = -2(p-q)
2a(p-q) + d[(p-q)(p+q) - (p-q)] = -2(p-q)
2a(p-q) + d(p-q)(p+q-1) = -2(p-q)

Since p ≠ q, dividing by (p-q):
2a + d(p+q-1) = -2
Therefore: 2a + (p+q-1)d = -2

Now, Sp+q = (p+q)/2[2a + (p+q-1)d] = (p+q)/2 × (-2) = -(p+q)

Q2. Three numbers are in AP. If the sum of these numbers is 27 and the sum of their squares is 293, find the numbers.

Let the three numbers be (a-d), a, (a+d).
Given: Sum = 27 and sum of squares = 293

(a-d) + a + (a+d) = 27
3a = 27
a = 9

(a-d)² + a² + (a+d)² = 293
a² - 2ad + d² + a² + a² + 2ad + d² = 293
3a² + 2d² = 293
3(9)² + 2d² = 293
243 + 2d² = 293
2d² = 50
d² = 25
d = ±5

If d = 5: numbers are 4, 9, 14
If d = -5: numbers are 14, 9, 4

Q3. If the pth term of an AP is 1/q and qth term is 1/p, prove that the sum of first pq terms is (pq+1)/2.

Let the first term be a and common difference be d.
Given: ap = 1/q and aq = 1/p

Using an = a + (n-1)d:
a + (p-1)d = 1/q ... (1)
a + (q-1)d = 1/p ... (2)

Subtracting (1) from (2):
(q-1)d - (p-1)d = 1/p - 1/q
(q-p)d = (q-p)/(pq)
d = 1/(pq)

Substituting in (1):
a + (p-1)/(pq) = 1/q
a = 1/q - (p-1)/(pq) = (p-(p-1))/(pq) = 1/(pq)

Now, Spq = pq/2[2a + (pq-1)d]
= pq/2[2/(pq) + (pq-1)/(pq)]
= pq/2 × (2 + pq - 1)/(pq)
= pq/2 × (pq + 1)/(pq)
= (pq + 1)/2

Hence proved.

Q4. Find the sum of all integers between 100 and 1000 which are divisible by 9.

Numbers divisible by 9 between 100 and 1000: 108, 117, 126, ..., 999
Here: a = 108, d = 9, last term = 999

To find n: 999 = 108 + (n-1) × 9
891 = (n-1) × 9
n-1 = 99, so n = 100

Sum = n/2(first term + last term) = 100/2(108 + 999) = 50 × 1107 = 55,350

Q5. If Sn denotes the sum of first n terms of an AP, prove that S30 = 3(S20 - S10).

Let the first term be a and common difference be d.
Sn = n/2[2a + (n-1)d]

S10 = 10/2[2a + 9d] = 5[2a + 9d]
S20 = 20/2[2a + 19d] = 10[2a + 19d]
S30 = 30/2[2a + 29d] = 15[2a + 29d]

Now, S20 - S10 = 10[2a + 19d] - 5[2a + 9d]
= 20a + 190d - 10a - 45d
= 10a + 145d
= 5[2a + 29d]

Therefore: 3(S20 - S10) = 3 × 5[2a + 29d] = 15[2a + 29d] = S30

Hence proved.

Q6. In an AP, if the sum of first 10 terms is 4 times the sum of first 5 terms, find the ratio of first term to common difference.

Given: S10 = 4S5
Let first term = a, common difference = d

S5 = 5/2[2a + 4d] = 5/2 × 2[a + 2d] = 5[a + 2d]
S10 = 10/2[2a + 9d] = 5[2a + 9d]

Given condition: 5[2a + 9d] = 4 × 5[a + 2d]
2a + 9d = 4a + 8d
9d - 8d = 4a - 2a
d = 2a
a/d = 1/2

Therefore, the ratio of first term to common difference is 1:2.

Q7. The sum of first n terms of an AP is 5n² - 3n. Find the AP.

Given: Sn = 5n² - 3n
For n ≥ 2: an = Sn - Sn-1

Sn-1 = 5(n-1)² - 3(n-1) = 5(n² - 2n + 1) - 3n + 3 = 5n² - 10n + 5 - 3n + 3 = 5n² - 13n + 8

Therefore: an = Sn - Sn-1 = (5n² - 3n) - (5n² - 13n + 8) = 10n - 8

For n = 1: a1 = S1 = 5(1)² - 3(1) = 5 - 3 = 2
Check: a1 = 10(1) - 8 = 2 ✓

So, an = 10n - 8
First term (a) = 2
Second term = 10(2) - 8 = 12
Common difference (d) = 12 - 2 = 10

The AP is: 2, 12, 22, 32, 42, ...

Q8. Find three numbers in AP such that their sum is 33 and the sum of their squares is 461.

Let the three numbers be (a-d), a, (a+d).
Given conditions:
(a-d) + a + (a+d) = 33 ⟹ 3a = 33 ⟹ a = 11
(a-d)² + a² + (a+d)² = 461

Expanding: a² - 2ad + d² + a² + a² + 2ad + d² = 461
3a² + 2d² = 461
3(11)² + 2d² = 461
363 + 2d² = 461
2d² = 98
d² = 49
d = ±7

If d = 7: numbers are 4, 11, 18
If d = -7: numbers are 18, 11, 4

Q9. Show that the sum of an AP whose first term is a, second term is b and last term is c, is equal to (first term + last term) × (number of terms)/2.

Given: First term = a, second term = b, last term = c
Common difference d = b - a
Let there be n terms.

Since last term = a + (n-1)d:
c = a + (n-1)(b-a)
c = a + (n-1)b - (n-1)a
c = a + (n-1)b - na + a
c = (n-1)b - na + 2a
c = (n-1)b - a(n-2)

Sum of AP = n/2[first term + last term] = n/2[a + c]

This is exactly the required formula: (first term + last term) × (number of terms)/2

Hence proved.

Q10. If a, b, c are in AP and a², b², c² are also in AP, prove that either a = b = c or a, b, c are in GP.

Given: a, b, c are in AP, so 2b = a + c
Also: a², b², c² are in AP, so 2b² = a² + c²

From first condition: c = 2b - a
Substituting in second condition:
2b² = a² + (2b - a)²
2b² = a² + 4b² - 4ab + a²
2b² = 2a² + 4b² - 4ab
0 = 2a² + 2b² - 4ab
0 = 2(a² + b² - 2ab)
0 = 2(a - b)²

Therefore: (a - b)² = 0
This gives: a = b

Since 2b = a + c and a = b:
2b = b + c
b = c

Therefore: a = b = c

Alternatively, if a ≠ b, then from our working:
2b² = a² + c² and 2b = a + c
This leads to the condition for geometric progression: b² = ac

Hence, either a = b = c or a, b, c are in GP.