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 \(\dfrac{1}{4}\) 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.
(i) Yes. The total fare after each km is \(15, 23, 31, \ldots\), which forms an AP because each succeeding term is obtained by adding 8 to the preceding term.
(ii) No. The volumes of air are \(V, \dfrac{3V}{4}, \left(\dfrac{3}{4}\right)^2 V, \ldots\), which form a geometric progression, not an AP.
(iii) Yes. The costs are \(150, 200, 250, \ldots\), which form an AP with common difference 50.
(iv) No. The amounts are \(10000\left(1 + \dfrac{8}{100}\right), 10000\left(1 + \dfrac{8}{100}\right)^2, 10000\left(1 + \dfrac{8}{100}\right)^3, \ldots\), which form a geometric progression, not an AP.
Recall: A list of numbers forms an arithmetic progression (AP) if the difference between any two consecutive terms is constant (same throughout). This constant is called the common difference.
• Fare for the first km = ₹15.
• For every additional km, ₹8 is added to the previous fare.
So, the total fare after:
1 km: ₹15
2 km: ₹15 + 8 = ₹23
3 km: ₹23 + 8 = ₹31
4 km: ₹31 + 8 = ₹39, and so on.
This gives the sequence: \(15, 23, 31, 39, \ldots\).
Check differences:
\(23 - 15 = 8,\ 31 - 23 = 8,\ 39 - 31 = 8\).
The difference between consecutive terms is always 8 (constant), so the list of numbers is an AP.
• Let the initial amount of air be \(V\).
• Each time, the pump removes \(\dfrac{1}{4}\) of the remaining air, so \(\dfrac{3}{4}\) of the air remains.
After each operation, the air amounts are:
Initially: \(V\)
After first removal: \(\dfrac{3}{4}V\)
After second removal: \(\left(\dfrac{3}{4}\right)^2 V\)
After third removal: \(\left(\dfrac{3}{4}\right)^3 V\), and so on.
The sequence is: \(V, \dfrac{3V}{4}, \left(\dfrac{3}{4}\right)^2 V, \left(\dfrac{3}{4}\right)^3 V, \ldots\).
Here, each term is obtained by multiplying the previous term by \(\dfrac{3}{4}\), not by adding a fixed number.
So the ratio between successive terms is constant, not the difference. Hence this is a geometric progression (GP), not an AP.
• Cost for the first metre = ₹150.
• For each subsequent metre, cost increases by ₹50.
So, the cost after each metre:
1st metre: ₹150
2nd metre: ₹150 + 50 = ₹200
3rd metre: ₹200 + 50 = ₹250
4th metre: ₹250 + 50 = ₹300, and so on.
This gives the sequence: \(150, 200, 250, 300, \ldots\).
Check differences:
\(200 - 150 = 50,\ 250 - 200 = 50,\ 300 - 250 = 50\).
The difference between consecutive terms is constantly 50, so this is an arithmetic progression with common difference 50.
• Principal (initial deposit) = ₹10000.
• Rate of interest = 8% per annum, compounded yearly.
After 1 year, amount:
\[A_1 = 10000\left(1 + \dfrac{8}{100}\right)\]
After 2 years:
\[A_2 = 10000\left(1 + \dfrac{8}{100}\right)^2\]
After 3 years:
\[A_3 = 10000\left(1 + \dfrac{8}{100}\right)^3\]
and so on.
The sequence is: \(10000\left(1 + \dfrac{8}{100}\right), 10000\left(1 + \dfrac{8}{100}\right)^2, 10000\left(1 + \dfrac{8}{100}\right)^3, \ldots\).
Each term is obtained by multiplying the previous term by \(1 + \dfrac{8}{100}\), not by adding a fixed amount.
Therefore, the list of numbers forms a geometric progression, not an AP.