In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ` 15 for the first km and ` 8 for each additional km. (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. (iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre. (iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.
Step-by-Step Solution
Key Concept: An arithmetic progression (AP) is a sequence of numbers in which the difference between any two successive terms is constant. If the successive terms are $a, a+d, a+2d, \dots$, the constant $d$ is called the common difference. To decide whether a given situation forms an AP, we examine whether the successive terms differ by the same amount.
1. Recall the definition: A sequence $\{T_n\}$ is an AP if $T_{n+1}-T_n = d$ (constant) for all $n$.
2. Examine each situation:
- (i) Taxi fare:
*First km*: $\text{Fare}_1 = 15$ rupees.
*Each additional km*: an extra $8$ rupees is added.
Hence the fare after $n$ km is $\text{Fare}_n = 15 + (n-1)\times 8$.
The difference between successive terms is $\text{Fare}_{n+1}-\text{Fare}_n = 8$, a constant. Therefore the list of fares forms an AP with first term $a=15$ and common difference $d=8$.
- (ii) Air in a cylinder:
Let $A_n$ be the amount of air remaining after the $n^{\text{th}}$ pump. The pump removes $\frac{1}{4}$ of the air present, so $A_{n+1}=\frac{3}{4}A_n$.
The ratio $\frac{A_{n+1}}{A_n}=\frac{3}{4}$ is constant, but the *difference* $A_{n+1}-A_n$ is not constant (it changes with $A_n$). Hence the sequence is geometric, not arithmetic.
- (iii) Cost of digging a well:
Cost for the first metre = $150$ rupees.
Cost for each subsequent metre rises by $50$ rupees.
Thus the cost for the $n^{\text{th}}$ metre is $C_n = 150 + (n-1)\times 50$.
The successive difference $C_{n+1}-C_n = 50$ is constant. Hence the list of costs per metre is an AP with $a=150$, $d=50$.
- (iv) Money in a bank account with compound interest:
Initial amount $M_0 = 10000$ rupees.
Each year the amount grows by $8\%$, i.e., $M_{n+1}=1.08\,M_n$.
This gives a constant *ratio* $1.08$, not a constant *difference*. Therefore the yearly amounts form a geometric progression, not an AP.
3. Conclusion: Situations (i) and (iii) produce arithmetic progressions because the successive terms differ by a fixed amount (₹8 and ₹50 respectively). Situations (ii) and (iv) produce geometric progressions because the successive terms are obtained by multiplying by a constant factor.
4. Answer in short: (i) and (iii) are AP; (ii) and (iv) are not.
Correct Answer: (i) and (iii) form arithmetic progressions because the successive terms differ by a constant amount (₹8 and ₹50 respectively). (ii) and (iv) do not form APs; they are geometric progressions.