The service cost of a vehicle is (1200) rupees in the first year and increases by (300) rupees each next year. What is the cost in the (6)th year?

Correct answer: ₹2,700

The costs form an arithmetic progression (AP) with first term \(a=1200\) and common difference \(d=300\). The cost in the sixth year is \(a_6=a+(6-1)d=1200+5\times300=2700\) rupees. Choosing \(3000\) would mean adding the increase 6 times, but there are only 5 increases after the first year. Exam tip: Use \(a_n=a+(n-1)d\) for the \(n\)th term.