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A machine bought for ₹60000 loses ₹4500 of its value every year. What is its value at the end of 8 years?

Class 10Arithmetic ProgressionsAlgebra — Sequences
  • A₹28500
  • B₹27000
  • C₹19500
  • D₹24000

Step-by-Step Solution

The values 60000, 55500, 51000, … form an AP with d = -4500, where ₹60000 is the value at the start. After 8 years the value is 60000 + 8(-4500) = 60000 - 36000 = ₹24000. Treating ₹60000 as the value after one year and subtracting only 7 lots of ₹4500 gives ₹28500.
Answer: D) ₹24000

Key idea: Common AP word problems

**Salaries**: 'Person earns ₹X in year 1 with annual increase of ₹Y. Find salary in year n / total over n years.' → AP with a = X, d = Y.

**Stadium seating**: 'Front row has 30 seats; each row 4 more. 25 rows. Total seats?' → AP with a = 30, d = 4. S₂₅ = ?

**Loans/installments**: equal monthly payments form an AP.

**Number patterns**: 'Multiples of 5 between 100 and 500: how many? Sum?' → AP with a = 100, d = 5, last = 500.

Always identify a, d, and what's asked (term aₙ or sum Sₙ). Check whether n is the COUNT or the POSITION.

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