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Arithmetic Progression | Exercise 5.1 | Chapter 5 | "लक्ष्य" 2025

Class 10 Learn With Mansi・2 minutes read

The completion of Exercise 5.1 in the video involves calculating different scenarios involving arithmetic progression and identifying common differences, with an emphasis on practice for scoring well in the chapter. The exercise covers various scenarios such as calculating taxi fare, compound interest, identifying AP sequences, and determining subsequent terms based on common differences.

Insights

**Identifying Common Differences:** The exercise focuses on recognizing common differences between terms in arithmetic sequences, crucial for understanding patterns and calculations in the sequence.
**Practice for Mastery:** The completion of Exercise 5.1 underscores the significance of practice in mastering arithmetic progression concepts, emphasizing the need for consistent practice to score well and build proficiency in simple calculations.

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Recent questions

What is the importance of Exercise 5.1 introduction video?
Understanding AP concepts for scoring well.
How are common differences calculated in AP sequences?
By subtracting consecutive terms in the sequence.
What is the significance of practicing AP exercises?
Enhancing understanding and proficiency in AP concepts.
How are subsequent terms calculated in AP sequences?
By adding the common difference to each term.
How does the evolution of a list of numbers in AP occur?
By incrementing or decrementing with a common difference.

Related videos

Summary

00:00

Understanding Arithmetic Progressions in Exercise 5.1

Exercise 5.1 introduction video is crucial to understand AP.
Question 1 asks about the evolution of a list of numbers with a common difference.
The first part involves calculating taxi fare with Rs 15 for the first km and Rs 8 for each additional km.
The common difference is confirmed by subtracting consecutive terms.
The second part deals with the amount of air in a cylinder removed by a vacuum pump.
The common difference is checked by subtracting consecutive terms.
The third part involves the cost of Digi, increasing by Rs 50 for each meter.
The common difference is evident in the increase of Rs 50 for each meter.
The fourth part calculates the compound interest on Rs 10,000 deposited annually.
The common difference is determined by subtracting consecutive terms.
Question 2 requires finding the first four terms of an AP with given first term and common difference.
The method involves adding the common difference to each term or using the formula a + nd.

12:30

Identifying A Sequences and Calculating Terms

A is 10 + 2, resulting in d being 30.
The second part involves a = -2 and d = 0, where all terms are the same due to a zero difference.
Moving to the third part, a = 4 and d = -3, leading to the calculation of subsequent terms using the formula.
The fourth term is found by applying the formula a + 3d, resulting in -5.
In the fourth part, a = -1 and d = 1/2, leading to the calculation of subsequent terms.
The third term is found to be 1/2, followed by the fourth term at 0.
The fifth part involves a = -1.25 and d = -0.25, with calculations done using the formula.
The fourth term is found to be -2, followed by -1.75 for the next term.
The fourth question involves identifying which set forms an A sequence, calculating the common difference, and finding three more terms.
The set with a common difference of 1/2 forms an A sequence, with the next three terms calculated as 4, 4.5, and 5.
The third part includes terms -1.2, -3.2, -5.2, and -7.2, with the common difference and subsequent terms calculated accordingly.

24:34

"Calculating Common Differences in Arithmetic Sequences"

The common difference between the terms is calculated by subtracting the first term from the second term, resulting in a difference of -2.
Further, the common difference is confirmed by subtracting the second term from the third term, yielding the same result of -2.
The process of finding the next three terms (a5, a6, a7) is initiated by calculating a + 4d for the fifth term, resulting in 9.2.
Moving on to the sixth term, a + 5d is computed, leading to a value of 11.2.
The seventh term, a + 6d, is then determined, resulting in 13.2.
Shifting focus to a new set of terms, the first term is -10, the second term is -6, the third term is -2, and the fourth term is 2.
The common difference is calculated by subtracting the first term from the second term, resulting in a difference of 4.
Subsequently, the common difference is confirmed by subtracting the second term from the third term, again yielding a difference of 4.
The process of finding the next three terms (a5, a6, a7) is initiated by calculating a + 4d for the fifth term, resulting in 6.
Moving on to the sixth term, a + 5d is computed, leading to a value of 10.

37:36

Determining Common Difference in Sequences

To find the common difference in a sequence, subtract 2a from 3a, resulting in a, which is the common difference denoted as 'a'.
The task involves determining the fifth, sixth, and seventh terms in a sequence with the given first four terms.
The fifth term is calculated as a + 4d, which simplifies to 5a, indicating a common difference of 5a.
Moving forward, the sixth term is found as 6a, and the seventh term as 7a, showcasing a common difference of a.
In a different sequence involving square roots, the common difference is identified by subtracting the first term from the second term.
The sequence involving square roots is analyzed to determine if the common difference remains consistent, indicating the presence of 'a'.
The subsequent sequence involves squares of numbers, with the common difference being assessed to determine if it aligns, indicating the presence of 'a'.
Following the confirmation of a consistent common difference, the next three terms are calculated in the sequence.
The fifth term is found to be 97, the sixth term as 121, and the seventh term as 145 in the sequence.
The completion of the exercise involves successfully determining the next three terms in the sequence based on the established common difference.

49:52

"Practice for Scoring Well in Math"

Practice chapter with many parts and questions
Emphasizes importance of practice for scoring well
Simple calculations involved
Completion of Exercise 5.1
Next exercise to start in the next session

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