Question :
In which of the following situations ,does the list of numbers involved make an arithmetic progression,and why?
(i) The taxi fare each km when the fare is Rs.15 for the first km and Rs.8 for each additional km.
Solution :
From the above given information we come to know that,
Taxi fare for 1st km = 15
Taxi fare for 2nd km = 15 + 8 = 23
Taxi fare for 3rd km = 23 + 8 = 31
15, 23, 31,.................
Every term of this sequence is 8 more than the previous term. This sequence clearly form an A.P
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Solution :
Let "x" be the amount of air present in a cylinder
In each stroke vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
In other words, after every stroke, only 1-(1/4) = (3/4)th part of air will remain.
The volumes will be x,(3x/4),(3x/4)²,,,,,,,,,
Since the common difference are not same, the given sequence will be arithmetic progression.
(iii) The cost of digging a well after every meter of digging,when it costs Rs.150 for first meter and rises by Rs.50 for each subsequent meters.
Solution :
Cost of digging for the first meter = Rs.150
Cos t of digging for the second meter = 150 + 50
= Rs. 200
Cost of digging for the third meter = 200 + 50
= Rs. 250
150, 200, 250,..............
Every term is 50 more than the previous term, since the common difference is same. It is A.P
(iv) The amount of money in the account every year,when Rs.10000 is deposited at compound interest at 8% per annum.
Solution :
We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be p[1+(r/100)]^n after n years.
Therefore, after every year, our money will be
10000(1.08),10000(1.08)²,10000(1.08)³,..............
Since the common difference is not same. The given statement is not Arithmetic progression.
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