The speed of a car increases by 2 km/h after every one hour. If the distance…

Concept: when a quantity increases by the same fixed amount in every successive interval, its successive values form an arithmetic progression (AP) with first term a and common difference d. The sum of the first n terms of such a progression is given by the formula below, and it is also equal to n times the average of the first and the last term.

Sn = n/2 × [2a + (n − 1)d]

Applying this to the given data:

1. The car covers 35 km in the first hour, so its speed during that hour is 35 km/h.

2. The speed rises by 2 km/h after every hour and each interval lasts exactly one hour, so the distance covered in any hour is 2 km more than the distance covered in the previous hour: 35 km, 37 km, 39 km and so on.

3. These twelve hourly distances therefore form an AP with a = 35, d = 2 and n = 12.

4. S12 = 12/2 × [2(35) + (12 − 1)(2)] = 6 × [70 + 22] = 6 × 92 = 552

5. So the car covers 552 km in 12 hours.

Cross-check: the distance covered in the twelfth hour is a + 11d = 35 + 22 = 57 km. The average of the first and the last term is (35 + 57)/2 = 46 km, and 46 × 12 = 552 km, which confirms the total.

Common ways this total is mis-computed:

• Treating the speed as constant gives 35 × 12 = 420 km and ignores every hourly increase.

• Adding the 2 km/h increase only once gives 37 × 12 = 444 km instead of letting it accumulate hour after hour.

• The increases actually contribute 2 × (0 + 1 + 2 + … + 11) = 2 × 66 = 132 km, so the total is 420 + 132 = 552 km.