Class 11 RD Sharma Solutions - Chapter 19 Arithmetic Progressions- Exercise 19.7 | Set 2

In this article, we will explore the solutions to the problems in Exercise 19.7 Set 2 from Chapter 19 of the Class 11 RD Sharma textbook which focuses on Arithmetic Progressions (AP). This exercise provides practice in solving problems related to the sum of the first n terms of an AP finding specific terms in an AP and more.

Arithmetic Progressions (AP)

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference denoted by the d. The general form of an AP is given by the a,a+d,a+2d,… where a is the first term and d is the common difference.

Question 11. A man starts repaying a loan as the first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment?

Solution:

In the first installment, the man pays Rs. 100, so a₁ = 100 and in next installment the man pays Rs. 105, so a₂ = 105.

So we can conclude, common difference, d = 105 - 100 = 5

Thus, the money he will pay in 30th installment, a₃₀ = a₁ + 29d = 100 + 29 x 5 = 100 + 145 = 245

Hence, he will pay Rs. 245 in the 30th installment.

Question 12. A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many did it take him to finish the job?

Solution:

On the first day, the carpenter makes 5 frames, so a = 5 and thereafter each day he makes 2 extra frames, so d = 2.

Total frames to be made, Sn = 192

We know the formula,

We will neglect n= -16 and consider n = 12

Therefore, the carpenter took 12 days to make 192 frames.

Question 13. We know that the sum of interior angled of a triangle is 180°. Show that the sums of the interior angles of polygons with 3,4,5,6.... form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.

Solution:

The sum of interior angles of a polygon having n-sides is = (n-2) x 180°

Thus, for a polygon with 3 sides, we have a₃ = (3-2) x 180° = 180°

similarly, for a polygon with 4 sides, we have a₄ = (4-2) x 180° = 360°

for a triangle with 5 sides, we have a₅ = (5-2) x 180° = 540°

Now, a₄ - a₃ = 360° - 180° = 180°

also, a₅ - a₄ = 540° - 360° = 180°, we are getting the same common difference for every succeeding and preceding term, hence we can say that it forms an AP.

Now,  the sum of the interior angles for a 21 sided polygon = (21-2) x 180° = 3420°

Question 14. In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Solution: 

Since, there are in total 20 potatoes and they are placed in a line at intervals of 4 m, therefore. n = 20 and d=4  

First potato is placed at a distance of 24 m, so a₁ = 24, now a₂ = 28, similarly a₂₀ = 24+ 19 x 4 = 100.

We know the formula, 

therefore, 

The contestant has to bring back the potatoes hence, he will run back and forth, so total distance covered = 1240 x 2  =2480 m.

Question 15. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter. 

(i) Find his salary for the tenth month.

(ii) What is total earnings during the first year?

Solution:

Given: initial salary, a = Rs. 5200 and d = Rs. 320.

(i) his salary in 10th month = a + 9d = 5200 + 9 x 320 = 8080

(ii)  his total earnings during the first year

Question 16. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?

Solution:

Given: Sn = 66000, n=20, d=200.

We know the formula,

The man saved Rs. 1400 in the first year.

Question 17. In a cricket team tournament 16 teams participated. A  sum of Rs 8000 is to be awarded among themselves as prize money. If the last place team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

Solution:

Total number of teams participating in the tournament, n = 16

Total prize money, Sn = 8000

The last placed team received prize money, a = 275

Suppose, for every successive team the prize money increases by d, then by the sum formula

therefore, the amount received by the first placed team = a+15d = 275 + 15 x 30 = 725

Hence, the first placed team received Rs 725 as the prize money.

Read More:

• Arithmetic Progressions Class 10- NCERT Notes
• Arithmetic Progression

Conclusion

Understanding Arithmetic Progressions is crucial as it forms the basis for the many mathematical concepts and problems. This exercise helps reinforce the concepts by providing the various problems that test a student's ability to the apply formulas and derive results accurately.