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Arithmetic progressions · 3 marks

Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Answer: After 10 years (10 increments of ₹20,000). His salary in the 11th year of work is ₹7,00,000.

Step-by-step solution

Given: First-year salary ₹5,00,000; Increase of ₹20,000 every year
To find: When the annual salary becomes ₹7,00,000

Idea: The yearly salaries form an AP with a = 5,00,000 and d = 20,000. Find which term is 7,00,000; the number of years that have passed is one less than that term number.

  1. Salaries: 5,00,000, 5,20,000, 5,40,000, … an AP with a = 500000, d = 20000.½ mark
  2. 500000 + (n − 1) × 20000 = 700000 ⇒ (n − 1) × 20000 = 200000.1 mark
  3. n − 1 = 10 ⇒ n = 11. So the 11th year’s salary is ₹7,00,000.1 mark
  4. He started on ₹5,00,000 and got 10 yearly increments, so his income reached ₹7,00,000 after 10 years.½ mark
His income reached ₹7,00,000 after 10 years (it is his salary in the 11th year).

Check: 5,00,000 + 10 × 20,000 = 5,00,000 + 2,00,000 = 7,00,000 ✓.

Answer to write in the exam

Salaries: 500000, 520000, 540000, … (AP, a = 500000, d = 20000)

500000 + (n − 1) × 20000 = 700000

(n − 1) × 20000 = 200000 ⇒ n − 1 = 10 ⇒ n = 11

∴ Income reached ₹7,00,000 after 10 years (in the 11th year).

Common mistakes that cost marks

  • Answering 11 years without explaining: 11 is the term number (11th year); the time that has passed is 10 years.
  • Dividing 7,00,000 by 20,000 = 35 and ignoring the starting salary.
  • Mixing up lakh commas: ₹5,00,000 is five lakh (500000), ₹20,000 is twenty thousand.

How this can come in the exam

Case-based (4 marks)

Meera joins a company at a monthly salary of ₹30,000. Each year her monthly salary rises by ₹1,500.
(i) Write her monthly salary in the first three years. (ii) Write the monthly salary in year n. (iii) In which year will it be ₹45,000? (iv) What is her monthly salary in the 8th year?

Show answer(i) ₹30,000, ₹31,500, ₹33,000 (1 mark). (ii) 30000 + 1500(n − 1) = 28500 + 1500n (1 mark). (iii) 28500 + 1500n = 45000 ⇒ n = 11: the 11th year (1 mark). (iv) 28500 + 12000 = ₹40,500 (1 mark).

Try one yourself

A plant is 15 cm tall when planted and grows 4 cm every week. After how many weeks will it be 63 cm tall?

Show answer

15 + 4w = 63 ⇒ w = 12 weeks.

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