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Laksh
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Asked: 2020-12-30T04:18:15+00:00 In:

Subba Rao started work in 1995 at an annual salary of rupay 5000 and received an increment of rupay 200 each year. In which year did his income reach ` 7000?

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NCERT Solutions for Class 10 Maths Chapter 5
Important NCERT Questions
Arithmetic Progression
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 5.2
Page No:107
Questions No:19

2020-2021arithmetic progressioncbsechapter 5class10mathematicsncert
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    2 Answers

    1. Get Hindi Medium and English Medium NCERT Solution for Class 10 Maths to download.
      Please follow the link to visit website for first and second term exams solutions.
      https://www.tiwariacademy.com/ncert-solutions/class-10/maths/chapter-5/

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    2. Best Answer

      Starting salary = a = Rs 5000 annual increment (common difference) = d = Rs 200
      Let, after n years, his salary become Rs 7000.
      Therefore, a_n = 7000
      ⇒ a + (n -1)d = 7000
      ⇒ 5000 + (n – 1)(200) = 7000
      ⇒ (n -1)(200) = 2000
      ⇒ n – 1 = 10 ⇒ n = 11
      Hence, in 11th year his salary become Rs 7000.

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