Arithmetic Progression Class 10 in One Shot πŸ”₯ | Class 10 Maths Chapter 5 AP | Shobhit Nirwan

Shobhit Nirwan・2 minutes read

The video explains arithmetic progression, highlighting steps to understand and solve problems related to AP, with detailed explanations and practical examples provided. Viewers are encouraged to study thoroughly, apply the formula for finding the nth term, and focus on practicing to tackle questions from NCRT and previous papers.

Insights

  • Understanding the concept of Arithmetic Progression (AP) is crucial for solving questions from NCRT, previous years, and sample papers related to the topic.
  • The formula for calculating the common difference in AP, as well as the importance of identifying patterns and sequences, plays a significant role in solving mathematical problems.
  • The application of formulas for determining specific terms in a sequence, along with the emphasis on practical examples and real-life scenarios, aids in comprehending and utilizing the concepts effectively.

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Recent questions

  • What is Arithmetic Progression (AP)?

    A list of numbers with a fixed difference.

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Summary

00:00

Understanding Arithmetic Progression in Class 10

  • The video discusses the Arithmetic Progression Chapter of Class 10th, also known as AP.
  • The chapter receives many comments from children and is filled with sections.
  • The video emphasizes studying AP to understand the concepts thoroughly.
  • The concept of AP is explained step by step, ensuring clarity.
  • After watching the video, viewers will be able to tackle questions from NCRT, previous years, and sample papers related to AP.
  • The definition of arithmetic progression is detailed, focusing on a list of numbers obtained by adding a fixed number to the previous term.
  • The common difference in AP is explained as the difference between consecutive terms.
  • The formula for finding the common difference is illustrated as the next term minus the previous term.
  • The general form of AP is presented, showing how each term is obtained by adding the common difference.
  • Practical examples are provided to demonstrate how to calculate terms in AP using the first term and common difference.

11:43

"Identifying Patterns in Number Sequences"

  • Adding zero to term two results in two.
  • The sequence continues with twos.
  • The process involves checking for a specific pattern.
  • Verification is done by comparing differences.
  • The focus is on identifying a specific sequence.
  • The common difference is crucial in determining the sequence.
  • The process involves subtracting consecutive terms.
  • The goal is to establish if the sequence follows a specific pattern.
  • The sequence is analyzed for consistency in differences.
  • The final step is to confirm if the sequence fits the pattern.

23:07

"Vacuum Pump Reduces Air, Compound Interest Analysis"

  • To indicate the amount of air present in a cylinder, a Vacuum Pump removes 1/4 of the air remaining in the cylinder at a time.
  • The initial air in the cylinder is progressively reduced by 1/4 each time the vacuum pump operates.
  • The air present in the cylinder is calculated as 9v/16 after multiple removals of 1/4 of the air.
  • The process involves removing 1/4 of the air present each time until no air remains in the cylinder.
  • The final amount of air left in the cylinder after these removals is 9v/16.
  • In a compound interest scenario, an amount of β‚Ή10000 is deposited at a certain rate, increasing annually by 8%.
  • The formula for compound interest is explained, detailing the amount left at the end of each year.
  • The calculation involves determining the amount left at the end of each year based on the compound interest formula.
  • The progression of the amount left each year is analyzed to determine if it follows an Arithmetic Progression (AP).
  • The process involves calculating the difference between consecutive amounts to ascertain if they form an AP sequence.

35:09

"Financial Formula for Savings and Investments"

  • Formula for calculating savings in FD at Rs. 7 with 7% interest rate after one year.
  • Encouragement to remember the formula for future financial planning.
  • Discussion on creating one's own FD and investing in mutual funds.
  • Explanation of the formula for calculating the nth term in a sequence.
  • Detailed steps on forming a formula for nth term using common differences.
  • Practical examples provided for understanding the formula application.
  • Clarification on the importance and versatility of the formula.
  • Application of the formula to determine the 10th, 20th, and 100th terms in a sequence.
  • Detailed calculations for finding specific terms in a sequence using the formula.
  • Practical demonstration of determining the number of terms in a sequence using the formula.

47:32

"Finding Arithmetic Progression in Sequence"

  • The value of 'a' in the sequence is 3.
  • To find the value of 'n - 1 * d', it is given that 3 * 3 = 111.
  • Out of 111, three went and became 108, leading to the equation 'n - 1 = 108 / 3'.
  • Solving the equation, 'n' is equal to 36.
  • There are 37 terms in the sequence.
  • To find 'a' and 'd', two equations are given: 'a + 2d = 5' and 'a + 6d = 9'.
  • By solving the equations, 'a' is 3 and 'd' is 1.
  • The sequence is checked for being an arithmetic progression (AP).
  • The common difference is calculated by checking the differences between terms.
  • The sequence is confirmed to be an AP, and the formula for 'a' is applied to find the 301st term, resulting in 'n' being 151.

59:40

"Two-Digit Numbers Divisible by Three"

  • The first three two-digit numbers divisible by three are 12, 15, and 18.
  • The common difference between these numbers is 3.
  • The total number of two-digit numbers divisible by three between 12 and 99 is 30.
  • To find the number of terms in the sequence, the formula n-1 * 3 = 99 - 12 is used.
  • By solving the equation, the value of n is found to be 29.
  • Therefore, there are 30 two-digit numbers divisible by three in the sequence.
  • To find the 11th term from the last in the sequence, the common difference is calculated as 3.
  • The 11th term from the last is determined to be -32.
  • Subba Rao started working in 1995 with an annual salary of Rs 5000.
  • By calculating the increment, it is found that in the eighth year, his salary will be Rs 7000.
  • The sum of terms in an arithmetic progression can be calculated using the formula sn = n/2 * (2a + (n-1) * d).

01:12:25

"Sum, Terms, Formulas: Calculations Simplified"

  • The task involved finding the sum of seven terms, equating it to 7/2, and determining the value of n as 7 * 2 * a.
  • The expression 3 p n - 1 translates to 7 - 1 * d, resulting in 5 - 3 becoming 2.
  • The calculation of 7/2 * 2a leads to 3 * 2, which simplifies to 6 p 6.
  • Subsequently, 12 transforms into 7/2 * 18, then 2 * 9, resulting in 18 and 63, respectively.
  • The removal of a specific term from the formula directly is emphasized as crucial.
  • The formula for the sum of the first a terms of a series is highlighted as significant.
  • The importance of understanding the first and last terms to utilize specific formulas is stressed.
  • The formula for finding a specific term in a series is detailed, involving the subtraction of sums.
  • The significance of remembering key formulas, such as n / 2 * a p a and n / 2 * a + a, is underscored.
  • The process of determining the number of terms needed for a sum of 78 is explained, showcasing the application of formulas and solving equations.

01:25:21

Uniform Production Increase and Spiral Length Calculation

  • The manufacturer of TV sets produces 600 sets in the third year and 700 sets in the seventh year.
  • Production increases uniformly by a fixed number each year.
  • The common difference in production is fixed, indicating a uniform increase.
  • The value of the first term (a) is found to be 550 sets of TVs in the first year.
  • The total production in the tenth year is calculated to be 775 TV sets.
  • The sum of production in the first seven years is determined to be 5575 TV sets.
  • The total length of a spiral made up of 13 concentric semi-circles is calculated to be 91Ο€ units.
  • The length of each semi-circle is found by adding the circumferences of each semi-circle.
  • The common difference in the lengths of the semi-circles is calculated to be 2Ο€.
  • The sum of the lengths of the 13 semi-circles is determined to be 91Ο€ units.

01:38:18

"NCRT questions answered, focus on Trigonometry"

  • Answered all important and conceptual questions from NCRT RD and RS, advised to solve each question individually using NCRT, suggested focusing on Trigonometry chapter, and promised to bring the next chapter in three to four days, emphasizing the need to study for one and a half to two hours daily.
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