Q) In which of the following situations does the list of numbers involved make an arithmetic progression and why?

(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.

Ans: Let’s first start making a sequence and then check if the sequence makes an AP or not..

It is given that the cost of digging for the 1st metre is Rs 15.0 clearly this becomes our 1st term,

Next, it is given that the cost increase by Rs. 50 for each additional metre.

It means if we dig 2 metres, then the cost for 1st metre (Rs. 150) will be counted PLUS cost for 2nd  metre (Rs. 50) will be counted.

Hence, cost for 2 metres will become 150 + 50 = 200

This becomes 2nd term of sequence

Next, if we go 3 metres, then cost for 1st metre (Rs. 150) will be counted PLUS cost for 2nd metre (Rs. 50) will be counted + PLUS cost  for 3rd metre (Rs. 50) will be counted

Hence, cost for 3 metres will become = 150 + 50 + 50

or we can write it as: 15 + 2 x 50 = 150 + 100 = 250

This becomes 3rd term of sequence

Similarly, if we go 3 metres, then cost for 1st metre (Rs. 150) will be counted PLUS cost for 2nd metre (Rs. 50) will be counted + PLUS cost  for 3rd metre (Rs. 50) will be counted + PLUS cost  for 4th metre (Rs. 50) will be counted

Hence, cost for 4 metres will become 150 + 3 x 50 = 300

This becomes 4th term of sequence

Thus, the sequence starts to emerge as 150, 200, 250, 300,……

Here the difference between any two consecutive terms will be 50 because we have added Rs. 50 for each additional metre.

Therefore, the difference will be common across the sequence

and since the common difference is same across any two consecutive terms, the sequence formed is an AP.

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