Class 11 RD Sharma Solutions- Chapter 19 Arithmetic Progressions- Exercise 19.7 | Set 1

In Chapter 19 of RD Sharma's Class 11 Mathematics students delve into Arithmetic Progressions (AP). This chapter introduces the fundamental concepts of the sequences where each term after the first is obtained by adding a constant value to the preceding term. Exercise 19.7 focuses on applying these concepts to solve various problems involving arithmetic progressions such as finding the nth term the sum of n terms and identifying specific terms within the sequence.

Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms remains constant. This difference is known as the common difference and is denoted by the 'd'. The general form of an arithmetic progression can be written as a a+d,a+2d,a+3d,… where 'a' is the first term. Arithmetic progressions are widely used in various fields including finance, computer science, and physics due to their straightforward and predictable nature.

Question 1. A man saved Rs. 16500 in 10 years. In each year after the first, he saved Rs. 100 more than he did in the preceding year. How much did he save in the first year?²

Solution:

Let the money saved by man in the first year be Rs. x then,

as given in the problem, his saving increases by Rs. 100 per year so,

x + (x + 100)+ (x + 200)+ (x + 300)+ ..... + (x + 900) = 16500

⇒ 10x + 100 (1+2+3+....+9) = 16500

⇒ 10x + 100*(9 * 10)/2 = 16500 (Since, sum of n natural numbers = n(n+1)/2)

⇒ 10x + 4500 = 16500

⇒ 10x = 12000

⇒ x = 1200

Therefore, the man saved Rs. 1200 in the first month.

Question 2. A man saves Rs. 32 during the first year, Rs. 36 in the second year and in this way he increases his saving by Rs. 4 every year. Find in what time his saving will be Rs. 200.

Solution: 

Let the time taken by the man to save Rs. 200 be n years.

Here, 32 + 36 + 40 +.... = 200

We can notice, first term, a=32, common difference, d=4 and sum, S_n = 200.

We know, Sn = n [2a + (n-1)*d]/2 using this formula, we get

⇒ 200 = n [2*32 + (n-1)*4]/2

⇒ 400 = 64n + 4n² -4n

⇒ 4n² + 60n - 400 = 0

⇒ n² + 15n -100 = 0

⇒ n -5n + 20n -100 = 0

⇒ n(n-5) + 20(n-5) = 0

⇒ (n-5) (n+20) = 0

⇒ n=5 or n= -20

We consider, n =5 since years cannot be in negative.

Therefore, the man saves Rs. 200 in 5 years.

Question 3. A man arranges to pay of a debt of Rs. 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one third of the debt unpaid, find the value of the first installment.

Solution: 

According to the question, 40 annual installments are paid in form an arithmetic series, so

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

⇒ 3600 = 40[2a + (40-1)d]/2

⇒ 2a + 39d = 180 ---------------------------- (eqn.1)

Also, sum of first 30 installments is two-third of 3600 which is 2400

⇒ 30[2a + (30-1)d] = 2400

⇒ 2a + 29d = 160 ---------------------------- (eqn.2)

Subtracting eqn.2 from eqn.1, we get

10d = 20 ⇒d = 2

Putting the value of d in eqn.1 we get a= 51

Therefore, the value of first installment is Rs. 51

Question 4. A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) production in the first year, (ii) the total product in seven years and (iii) the product in the 10th year.

Solution: 

According to question, a₃ = a + 2d = 600 and a₇ = a + 6d = 700

Solving for a and d, we get a = 550 and d = 25.

Therefore, (i) total 550 radios were produced in the first year.

(ii) Total radios produced in first seven years

= 7 [550 + 700]/2 [since, S_n = n (a + l) /2]

= 2375

(iii) Product produced in 10th year = a + 9d = 550 + 9x25 = 775

Question 5. There are 25 trees at a distance of 5 meters in a line with a well. The distance of the well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Solution: 

As given in the question, the distance of the well from the nearest tree = 10 m. and in total there are 25 trees separated by a distance of 5 m. all in a line with the well.

The total distance covered by the gardener in watering 25 trees will be given as:

2 [10 + 15 + 20 + 25 + ..... + 135] (Multiplying it by 2 because he moves back also)

⇒ 2 * [25 * [2*10 + (25-1)5]/2]

⇒ total distance = 3500 meters.

Hence, the total distance the gardener will cover in order to water all the trees is 3500 meters.

Question 6. A man is employed to count Rs. 10710 at the rate of Rs. 180 per minute for half an hour. After this he counts at the rate of Rs. 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

Solution: 

As given in the question. for half an hour he counts at the rate of Rs. 180 every minute, so amount = 180 * 30 = 5400. Now the amount left = 10710 - 5400 = 5310

Then, he counts Rs. 3 less every minute than the preceding minute, so

(180 -3) + (180 - 2x3) + (180 - 3x3) + .... = 5310

Let the time taken to count this amount be n years, so

5310 = n [(180-3) + (n-1) (-3)]/2

⇒ 5310 = n [200 - 3n]

⇒ 3n² -200n +5310 = 0

⇒ n = 59

Therefore, total time taken to count the entire amount of Rs. 10710 is 59+30 = 89 minutes.

Question 7. A piece of equipment cost a certain factory Rs. 600,000. If it deprecates in value, 15% the first, 13.5% the next year, 12% the third year and so on, what will be its value at the end of 10 years, all percentages applying to the original cost.

Solution: 

The piece deprecates by 15% of its value in first year which is 15% of 600,000 = 90,000

Therefore, value of the piece after first year = 600,000 - 90,000 = 510,000.

Value deprecates by 13.5% of its value in second year i.e, 13.5% of 600,000 = 81,000

Value deprecates by 12% of its value in second year i.e, 12% of 600,000 = 72,0006--

Therefore, total deprecation in 10 years, can be calculated as:

S = 10 *[2* 81000 + (10-1) (-9000)] /2

= 5 * 81000 = 405000

Therefore, the cost of piece after 10 years will be = 600,000 - 405,000 = 1050000

Question 8. A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual installments of Rs. 500 plus 12% interest on the unpaid amount. How much the tractor cost him.

Solution: 

Total cost of the tractor

= 6000 + [(500 + 12% of 6000) + (500 + 12% of 5500) + (500 + 12% of 5000) ..... to 12 times]

= 6000 +6000 +12% of (6000 + 5500 + 5000 + .... + 12 times)

= 12000 + 12% of [12 (6000 + 500)/2]

= 12000 + 12% of [6 * 6500]

= 12000 + 4680

= 16680

Therefore, in total cost of the tractor to the farmer is Rs. 16680

Question 9. Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installments of Rs. 1000 plus 10% interest on the unpaid amount. How much the scooter cost him.

Solution: 

Total cost of the scooter

= 4000 + [(1000 + 10% of 18000) + (1000 + 10% of 17000) + (1000 + 10% of 16000) ..... to 18 times]

= 4000 +18000 +10% of (1800+ 1700 + 1600 + .... + 18 times)

= 22000 + 10% of [18 (1800 + 100)/2]

= 22000 + 10% of [9 * 19000]

= 22000 + 17100

= 39100

Therefore, in total cost of the tractor to the farmer is Rs. 39100

Question 10. The income of a person is Rs. 300,000 in the first year, and he receives an increase of Rs. 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Solution: 

First year income of person, a = 300,000

here, common difference, d = 10,000

Since, Sn = n [2a + (n-1)*d]/2

⇒ Sn = 20 [2*300000 + (20 - 1) * 10000]/2

⇒ Sn =10 * [600000 + 190000]

⇒ Sn = 7900000

Therefore, total amount, he received in 20 years is Rs. 7900000

Read More:

• Arithmetic Progression
• Arithmetic Progressions Class 10- NCERT Notes

Conclusion

Chapter 19 of RD Sharma’s Class 11 Mathematics book provides students with the comprehensive understanding of the arithmetic progressions. Mastery of these concepts is essential for the solving more advanced mathematical problems and is also applicable in the real-life situations. Exercise 19.7 challenges students to apply their knowledge of the APs reinforcing their learning through the practice.