5.1-Arithmetic Progressions

5.1-Arithmetic Progressions Important Formulae

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Grade 10 → Math → Arithmetic Progressions → 5.1-Arithmetic Progressions

After successful completion of this topic, you should be able to:

  • Produce patterns in order to observe that succeeding terms are obtained by adding a fixed number to the preceding terms.
  • Distinguish between finite and infinite AP in order to determine the nature and write the last term of the given AP.

An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference, denoted by $d$. The general form of an arithmetic progression can be expressed as:

$a, a + d, a + 2d, a + 3d, \ldots$

Here, $a$ is the first term of the sequence, and $d$ is the common difference. The $n^{th}$ term of an arithmetic progression can be calculated using the formula:

$a_n = a + (n - 1)d$

Where:

  • $a_n$ = $n^{th}$ term
  • $a$ = First term
  • $d$ = Common difference

For example, in the arithmetic progression $2, 5, 8, 11, \ldots$, the first term $a = 2$ and the common difference $d = 3$.

To find a specific term in the sequence, we can use the formula mentioned above. For instance, to find the 10th term of the sequence:

$a_{10} = 2 + (10 - 1) \cdot 3 = 2 + 27 = 29$

The sum of the first $n$ terms of an arithmetic progression, denoted as $S_n$, can be calculated using the formula:

$S_n = \frac{n}{2} \cdot (2a + (n - 1)d)$

Alternatively, the sum can also be expressed as:

$S_n = \frac{n}{2} \cdot (a + a_n)$

Where:

  • $S_n$ = Sum of the first $n$ terms
  • $a_n$ = $n^{th}$ term

For example, if we want to find the sum of the first 10 terms of the AP $2, 5, 8, 11, \ldots$:

We already know:

  • First term $a = 2$
  • Common difference $d = 3$
  • 10th term $a_{10} = 29$

Using the formula for the sum:

$S_{10} = \frac{10}{2} \cdot (2 \cdot 2 + (10 - 1) \cdot 3)$

$S_{10} = 5 \cdot (4 + 27) = 5 \cdot 31 = 155

Thus, the sum of the first 10 terms of the arithmetic progression is $155$.

Arithmetic progressions have various applications in real-life scenarios, including finance (calculating interest), physics (uniform motion), and computer science (algorithms). Recognizing the structure of APs allows for efficient calculations and predictions.

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 Rs. 15 for the first km and Rs. 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 14 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 Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.

Solution:
Let's evaluate each situation to determine if it forms an arithmetic progression (AP).

(i) Taxi fare after each km:
- The fare is Rs. 15 for the first km and Rs. 8 for each additional km.
- The sequence of fares would be: 15, 23, 31, 39, ...
- This is an AP because the difference between the first term and the second is \(8\), but the difference between subsequent terms is always \(8\). Hence, it is an AP, as the difference is constant.

(ii) Amount of air in a cylinder with a vacuum pump:
- If 14 units of air are removed each time, the sequence would be: initial volume, initial volume - 14, initial volume - 28, ...
- This does not form an AP because the amount removed is constant, but the amount remaining does not have a constant difference in the context of what is left after each operation. Hence, it is not an AP.

(iii) Cost of digging a well:
- The cost for the first metre is Rs. 150, and it rises by Rs. 50 for each subsequent metre.
- The sequence would be: 150, 200, 250, 300, ...
- This is an AP because the difference between consecutive terms is constant at \(50\).

(iv) Amount of money in the account with compound interest:
- When Rs. 10,000 is deposited at compound interest, the amount grows exponentially.
- The sequence of amounts would not have a constant difference (it would be 10,800, 11,664, ...).
- This does not form an AP because the differences between consecutive terms are not constant.

Summary:
- (i) AP (constant difference of Rs. 8).
- (ii) Not AP.
- (iii) AP (constant difference of Rs. 50).
- (iv) Not AP.

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a=10, d=10
(ii) a=–2, d=0
(iii) a=4, d=–3
(iv) a=–1, d= 12
(v) a=–1.25, d=–0.25

Solution:

First Four Terms of the AP
(i) a=10, d=10
10, 20, 30, 40
(ii) a=–2, d=0
–2, –2, –2, –2
(iii) a=4, d=–3
4, 1, –2, –5
(iv) a=–1, d=12
–1, 11, 23, 35
(v) a=–1.25, d=–0.25
–1.25, –1.50, –1.75, –2.00

For the following APs, write the first term and the common difference:

(i) 3, 1, –1, –3,...
(ii) –5, –1, 3, 7, ...
(iii) 1 , 5 , 9 , 13 , ...
(iv) 0.6, 1.7, 2.8, 3.9, ... 3333

Solution:

(i) 3, 1, –1, –3, ...
First term = 3, Common difference = -2

(ii) –5, –1, 3, 7, ...
First term = -5, Common difference = 4

(iii) 1, 5, 9, 13, ...
First term = 1, Common difference = 4

(iv) 0.6, 1.7, 2.8, 3.9, ...
First term = 0.6, Common difference = 1.1