Basics Questions on Sequence & Series

Duration: 6 min

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Module outline

  1. Quantitative Aptitude: Number System, Divisibility Rules, HCF LCM, Arithmetic, Percentage, Ratio and Proportion (Ratios), Average, Powers and Exponents (Surds and Indices), Profit and Loss, Simple Interest and Compound Interest, Mixture and Alligation, Age Problems, Partnership, Algebra, Permutation and Combination, Probability, Logarithm, Mensuration and Geometry, Time and Work, Speed Time and Distance, Data Interpretation, Sequence and Series (Series), Gamified / Game Based Aptitude, Speed Mathematics and Basic Mathematics
  2. Problem Solving & Reasoning: Coding Decoding, Data Sufficiency, Seating Arrangements, Puzzle Solving (Puzzles), Data Arrangement, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Inequality, Venn Diagram, Ranking, Order and Sequence, TCS NQT Advanced Reasoning Ability, Crypt Arithmetic, Analytical Reasoning (Counting Figures Reasoning), Series (Number and Letter Series) (Numerical Relations and Reasoning), Logical Sequence Of Words
  3. Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Completion (Fill in the blanks), Sentence Correction (Error Correction), Subject Verb Agreement (Verb Noun Agreement), Tenses, Articles, Preposition, Modals, Vocabulary
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video is a lecture on sequences and series, specifically focusing on arithmetic progressions (AP). The instructor, Yash Jain, begins by introducing the topic with a title card and then presents a series of problems to demonstrate the application of key formulas. The first problem asks for the sum of an AP with 12 terms, a first term of -19, and a last term of 36. The instructor writes down the formula for the sum of an AP, S_n = n/2 (a + a_n), and substitutes the given values to calculate the sum as 66. The second problem involves an AP with 17 terms, a first term of -20, and a last term of 28, which is solved using the same formula to find a sum of 136. The final problem is a multiple-choice question where the relationship 7 times the 7th term equals 11 times the 11th term is given, and the task is to find the 18th term. The instructor uses the general term formula, a_n = a + (n-1)d, to set up the equation 7(a+6d) = 11(a+10d), which simplifies to -4a = 68d, or a = -17d. Substituting this into the formula for the 18th term, a_18 = a + 17d, results in a_18 = -17d + 17d = 0. The video concludes with a 'Thanks for Watching' screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with an animated title card for a lesson on 'SEQUENCE & SERIES'. The screen displays a cartoon boy at a desk with a chalkboard that shows the formula for the sum of a series, S_N = a1 + a2 + a3 + ... + aN. The instructor, Yash Jain, is visible in a small window in the bottom right corner. The title card also includes the text 'Basic To Advance' and 'KNOWLEDGE GATE EDUCATOR'. The instructor then introduces the first problem: 'In an AP, there are 12 terms, the first term is -19 and last term is 36. Find the sum of AP?'. He begins to solve it by writing down the given values: AP, n=12, a=-19, a_n=36.

  2. 2:00 – 5:00 02:00-05:00

    The instructor proceeds to solve the first problem. He writes the formula for the sum of an arithmetic progression: S_n = n/2 (a + a_n). He substitutes the values from the problem: S_12 = 12/2 (-19 + 36). He then performs the calculation: S_12 = 6 * 17, which equals 102. He then moves to the second problem: 'In an AP, there are 17 terms, the first term is -20 and last term is 28. Find the sum of AP?'. He writes down the values: AP, n=17, a=-20, a_n=28. He applies the same formula: S_17 = 17/2 (-20 + 28). He calculates: S_17 = 17/2 * 8 = 17 * 4 = 68. He then transitions to the third problem, a multiple-choice question: 'If 7 times of the seventh term of an AP is equal to 11 times its eleventh term, then the 18th term of the AP will be __'. He writes the given condition: 7a_7 = 11a_11.

  3. 5:00 – 6:23 05:00-06:23

    The instructor begins solving the third problem. He writes the general formula for the nth term of an AP: a_n = a + (n-1)d. He then expresses the 7th and 11th terms: a_7 = a + 6d and a_11 = a + 10d. He substitutes these into the given equation: 7(a + 6d) = 11(a + 10d). He expands the equation: 7a + 42d = 11a + 110d. He rearranges the terms to get: 7a - 11a = 110d - 42d, which simplifies to -4a = 68d. He then solves for a: a = -17d. The question asks for the 18th term, so he writes a_18 = a + 17d. He substitutes a = -17d into this formula: a_18 = -17d + 17d = 0. The video ends with a 'Thanks for Watching' screen.

The video presents a structured lesson on arithmetic progressions, progressing from basic sum calculations to a more complex problem involving the relationship between terms. The core teaching method is the step-by-step application of formulas. The first two problems demonstrate the direct use of the sum formula S_n = n/2 (a + a_n) with given values. The third problem requires a deeper understanding, using the general term formula a_n = a + (n-1)d to establish a relationship between the first term (a) and the common difference (d), which is then used to find a specific term. The key insight is that the 18th term is independent of the specific values of a and d, as the relationship forces it to be zero. This demonstrates a powerful problem-solving technique in algebra.

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