Cost of digging the well after 1 metre of digging of Rs. 150 = a4
Cost of digging the well after 2 metres of digging
= Rs. 150 + Rs. 50
= Rs. 200 = a2
Cost of digging the well after 3 metres of digging
= Rs. 150 + Rs. 50
= Rs. 2a = a3
Cost of digging the well after 4 metres of digging
= Rs. 200 + Rs. 50
= Rs. 250 = a4
and so on.
a2 – a4 = Rs. 200 – Rs. 150 = Rs. 50
a3 – a2 = Rs. 250 – Rs. 200 = Rs. 50
a4 – a3 = Rs. 350 – Rs. 250 = Rs. 50
i.e., ak +1 – ak is the same every time.
So this list of numbers forms an A .P. with the first term a Rs. 150 and the common difference d = Rs. 50.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
Taxi fare for 1 km = Rs. 15 = a1 Taxi fare for 2 kms
= Rs. 15 + Rs. 8 = Rs. 23 = a2 Taxi fare for 3 kms
= 23 + Rs. 8 = Rs. 31 = a3 Taxi fare for 4 kms
= Rs. 31 + Rs. 8 = Rs. 39 = a4
and so on.
a2 – a1 = Rs. 23 – Rs. 15 = Rs. 8
a3 – a2 = Rs. 31 – Rs. 23 = Rs. 8
a4 – a3 = Rs. 39 – Rs. 31 = Rs. 8
i.e., ak + 1 – ak is the same every time.