Question 1 – All Parts
Part (i)
📋 Given Information
- Fare for first km = \(\text{₹}15\)
- Fare for each additional km = \(\text{₹}8\)
🎯 To Determine
Whether the taxi fare after each km forms an Arithmetic Progression (AP) or not, and explain why.
💡 Concept Used
A sequence forms an Arithmetic Progression (AP) if the difference between any two consecutive terms is constant. This constant difference is called the common difference (d).
Formula: \(d = a_{n+1} – a_n\) (must be constant for all consecutive terms)
📝 Step-by-Step Solution
Let’s find the total fare after traveling 1 km, 2 km, 3 km, 4 km, and so on.
- After 1 km: \(\text{₹}15\)
- After 2 km: \(\text{₹}15 + \text{₹}8 = \text{₹}23\)
- After 3 km: \(\text{₹}23 + \text{₹}8 = \text{₹}31\)
- After 4 km: \(\text{₹}31 + \text{₹}8 = \text{₹}39\)
- After 5 km: \(\text{₹}39 + \text{₹}8 = \text{₹}47\)
So the sequence of fares is: \(15, 23, 31, 39, 47, \ldots\)
Let’s calculate the differences:
\[ \begin{aligned} a_2 – a_1 &= 23 – 15 = 8 \\ a_3 – a_2 &= 31 – 23 = 8 \\ a_4 – a_3 &= 39 – 31 = 8 \\ a_5 – a_4 &= 47 – 39 = 8 \end{aligned} \]The difference between consecutive terms is constant and equals \(8\).
Since the common difference \(d = 8\) is constant throughout the sequence, this situation forms an Arithmetic Progression (AP).
The AP has:
- First term: \(a = 15\)
- Common difference: \(d = 8\)
Yes, the taxi fare forms an AP with first term \(a = 15\) and common difference \(d = 8\).
Part (ii)
📋 Given Information
- A vacuum pump removes \(\frac{1}{4}\) of the remaining air each time
- This means \(\frac{3}{4}\) of the air remains after each operation
🎯 To Determine
Whether the amount of air remaining in the cylinder forms an AP or not.
📝 Step-by-Step Solution
Let the initial amount of air be \(V\) units.
- Initially: \(V\)
- After 1st operation: \(V – \frac{V}{4} = \frac{3V}{4}\)
- After 2nd operation: \(\frac{3V}{4} – \frac{1}{4} \times \frac{3V}{4} = \frac{3V}{4} \times \frac{3}{4} = \frac{9V}{16}\)
- After 3rd operation: \(\frac{9V}{16} \times \frac{3}{4} = \frac{27V}{64}\)
The sequence is: \(V, \frac{3V}{4}, \frac{9V}{16}, \frac{27V}{64}, \ldots\)
Let’s assume \(V = 64\) units for easy calculation:
Sequence becomes: \(64, 48, 36, 27, \ldots\)
\[ \begin{aligned} a_2 – a_1 &= 48 – 64 = -16 \\ a_3 – a_2 &= 36 – 48 = -12 \\ a_4 – a_3 &= 27 – 36 = -9 \end{aligned} \]The differences are \(-16, -12, -9, \ldots\) which are NOT constant.
Notice that each term is obtained by multiplying the previous term by \(\frac{3}{4}\):
\[ \frac{a_{n+1}}{a_n} = \frac{3}{4} \text{ (constant ratio)} \]This is a Geometric Progression (GP), not an AP.
No, this does NOT form an AP. It forms a Geometric Progression with common ratio \(r = \frac{3}{4}\).
Part (iii)
📋 Given Information
- Cost for first metre = \(\text{₹}150\)
- Cost increases by \(\text{₹}50\) for each subsequent metre
🎯 To Determine
Whether the cost of digging after every metre forms an AP or not.
📝 Step-by-Step Solution
- 1st metre: \(\text{₹}150\)
- 2nd metre: \(\text{₹}150 + \text{₹}50 = \text{₹}200\)
- 3rd metre: \(\text{₹}200 + \text{₹}50 = \text{₹}250\)
- 4th metre: \(\text{₹}250 + \text{₹}50 = \text{₹}300\)
- 5th metre: \(\text{₹}300 + \text{₹}50 = \text{₹}350\)
The sequence is: \(150, 200, 250, 300, 350, \ldots\)
The common difference \(d = 50\) is constant.
Since the difference between consecutive terms is constant, this situation forms an AP with:
- First term: \(a = 150\)
- Common difference: \(d = 50\)
Yes, the cost forms an AP with first term \(a = 150\) and common difference \(d = 50\).
Part (iv)
📋 Given Information
- Principal amount (P) = \(\text{₹}10000\)
- Rate of interest (r) = \(8\%\) per annum
- Interest is compounded annually
🎯 To Determine
Whether the amount in the account every year forms an AP or not.
💡 Concept Used
Compound Interest Formula:
\[ A = P\left(1 + \frac{r}{100}\right)^n \]where \(A\) is the amount after \(n\) years.
📝 Step-by-Step Solution
Using the compound interest formula with \(P = 10000\) and \(r = 8\%\):
- After 0 years (Initially): \(\text{₹}10000\)
- After 1 year: \(10000 \times 1.08 = \text{₹}10800\)
- After 2 years: \(10000 \times (1.08)^2 = \text{₹}11664\)
- After 3 years: \(10000 \times (1.08)^3 = \text{₹}12597.12\)
- After 4 years: \(10000 \times (1.08)^4 = \text{₹}13604.89\)
The sequence is: \(10000, 10800, 11664, 12597.12, 13604.89, \ldots\)
The differences are \(800, 864, 933.12, 1007.77, \ldots\) which are NOT constant.
Each term is obtained by multiplying the previous term by \(1.08\):
\[ \frac{a_{n+1}}{a_n} = 1.08 \text{ (constant ratio)} \]This is a Geometric Progression (GP) with common ratio \(r = 1.08\), not an AP.
No, this does NOT form an AP. It forms a Geometric Progression with common ratio \(r = 1.08\).
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Confusing AP with GP when percentages are involved.
✅ Correct: When something increases/decreases by a fixed amount, it’s an AP. When it increases/decreases by a fixed percentage (or ratio), it’s a GP.
❌ Mistake 2: Not checking all consecutive differences before concluding it’s an AP.
✅ Correct: Always verify that the difference remains constant for at least 3-4 consecutive terms.
❌ Mistake 3: Thinking compound interest forms an AP because interest is added.
✅ Correct: In compound interest, interest is calculated on the accumulated amount, making it a GP, not an AP. Only simple interest would form an AP.
💡 Key Points to Remember
- An Arithmetic Progression (AP) has a constant difference between consecutive terms.
- To verify if a sequence is an AP, calculate \(a_{n+1} – a_n\) for multiple consecutive terms.
- Fixed addition/subtraction → AP (e.g., taxi fare, well digging cost)
- Fixed multiplication/division → GP (e.g., compound interest, vacuum pump)
- The formula for the \(n^{th}\) term of an AP is: \(a_n = a + (n-1)d\)
- Real-life situations involving uniform increase/decrease in cost, distance, or quantity often form APs.
