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 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.

Arithmetic Progressions | Exercise 5.1

Question 1

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.

Solution

Understand the Question

A sequence of numbers forms an Arithmetic Progression (AP) if the difference between any two consecutive terms ( $a_{n+1} - a_n$ ) is constant throughout. This fixed difference is called the common difference ( $d$ ).
To determine if each situation forms an AP, we find the first few terms of the sequence ( $a_1, a_2, a_3, a_4, \dots$ ) and check whether the consecutive differences ( $a_2 - a_1, a_3 - a_2, \dots$ ) are equal.

(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Step 1 · Calculate Fares and Check Common Difference

Given:

Fare for the first $\text{km}$ ( $a_1$ ) $= ₹ 15$
Fare for each additional $\text{km}$ $= ₹ 8$

Fares for subsequent kilometers:

\begin{aligned} a_1 &= 15 \\ a_2 &= 15 + 8 = 23 \\ a_3 &= 23 + 8 = 31 \\ a_4 &= 31 + 8 = 39 \end{aligned}

Differences between consecutive terms:

\begin{aligned} a_2 - a_1 &= 23 - 15 = 8 \\ a_3 - a_2 &= 31 - 23 = 8 \\ a_4 - a_3 &= 39 - 31 = 8 \end{aligned}

Since the difference between consecutive terms is constant ( $d = 8$ ), the terms form an AP.

Answer

(i) Yes, it forms an AP because the common difference is constant ( $d = 8$ ).

(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.

Step 1 · Calculate Remaining Air and Check Common Difference

Let the initial volume of air in the cylinder be $V$ .

Each stroke removes $\dfrac{1}{4}$ of the remaining air, leaving $1 - \dfrac{1}{4} = \dfrac{3}{4}$ of the previous volume.

Volumes after successive strokes:

\begin{aligned} a_1 &= V \\[0.6em] a_2 &= V \times \dfrac{3}{4} = \dfrac{3}{4}V \\[0.6em] a_3 &= \dfrac{3}{4}V \times \dfrac{3}{4} = \dfrac{9}{16}V \\[0.6em] a_4 &= \dfrac{9}{16}V \times \dfrac{3}{4} = \dfrac{27}{64}V \end{aligned}

Differences between consecutive terms:

\begin{aligned} a_2 - a_1 &= \dfrac{3}{4}V - V = -\dfrac{1}{4}V \\[0.6em] a_3 - a_2 &= \dfrac{9}{16}V - \dfrac{3}{4}V = -\dfrac{3}{16}V \end{aligned}

Since $a_2 - a_1 \neq a_3 - a_2$ , the difference is not constant.

Answer

(ii) No, it does not form an AP because the difference between consecutive terms is not constant.

(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.

Step 1 · Calculate Digging Costs and Check Common Difference

Given:

Cost for the first metre ( $a_1$ ) $= ₹ 150$
Cost increase for each subsequent metre $= ₹ 50$

Costs for subsequent metres:

\begin{aligned} a_1 &= 150 \\ a_2 &= 150 + 50 = 200 \\ a_3 &= 200 + 50 = 250 \\ a_4 &= 250 + 50 = 300 \end{aligned}

Differences between consecutive terms:

\begin{aligned} a_2 - a_1 &= 200 - 150 = 50 \\ a_3 - a_2 &= 250 - 200 = 50 \\ a_4 - a_3 &= 300 - 250 = 50 \end{aligned}

Since the difference between consecutive terms is constant ( $d = 50$ ), the terms form an AP.

Answer

(iii) Yes, it forms an AP because the common difference is constant ( $d = 50$ ).

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.

Step 1 · Calculate Annual Amounts and Check Common Difference

Given:

Principal ( $P$ ) $= ₹ 10000$
Rate of interest ( $r$ ) $= 8\%\text{ per annum}$

The amount after $n$ years is given by: $A = P\left(1 + \dfrac{r}{100}\right)^n$

Amounts at the end of each year:

\begin{aligned} a_1 &= 10000 \\[0.6em] a_2 &= 10000 \left(1 + \dfrac{8}{100}\right) = 10800 \\[0.6em] a_3 &= 10000 \left(1 + \dfrac{8}{100}\right)^2 = 11664 \\[0.6em] a_4 &= 10000 \left(1 + \dfrac{8}{100}\right)^3 = 12597.12 \end{aligned}

Differences between consecutive terms:

\begin{aligned} a_2 - a_1 &= 10800 - 10000 = 800 \\ a_3 - a_2 &= 11664 - 10800 = 864 \end{aligned}

Since $a_2 - a_1 \neq a_3 - a_2$ , the difference is not constant.

Answer

(iv) No, it does not form an AP because the difference between consecutive terms is not constant.

Common Mistakes

Compound vs. Simple Interest: In simple interest, a constant interest amount is added every year, forming an AP. In compound interest, the interest is calculated on the growing principal, so the differences increase each year and do not form an AP.
Fraction of Remaining Air: Do not mistakenly subtract a fixed $\dfrac{1}{4}V$ at each step. The vacuum pump removes $\dfrac{1}{4}$ of the remaining volume, meaning each step reduces the volume by a decreasing amount ( $-\dfrac{1}{4}V, -\dfrac{3}{16}V, \dots$ ).

Question 1

More questions in Exercise 5.1