Mathematics Lab Activity-08 Class X

Mathematics Laboratory Activities on Arithmatic Progression for class X students with complete observation tables strictly according to the CBSE syllabus.

Chapter - 05

arithmatic progression

Activity - 08

Objective

To establish a formula for the sum of first n terms of an Arithmetic Progression.

Material Required

Cardboard, coloured drawing sheets, white paper, cutter, adhesive.

Procedure

1. Take a rectangular cardboard of a convenient size and paste a white paper on it. Draw a rectangle ABCD of length (2a+9d) units and breadth 10 units.

2. Make some rectangular strips of equal length a units and breadth one unit and some strips of length d units and breadth 1 unit, using coloured drawing sheets.

3. Arrange/paste these strips on the rectangle ABCD as shown in Fig. 1.

Figure 1

4. The strips so arranged look like a stair case.

5. The first stair is of length a units, the second stair is of length a + d (units), third of a + 2d units and so on and each is of breadth 1 unit. So, the areas (in sq. units) of these strips are a, a + d, a + 2d, ....., a + 9d, respectively

6. This arrangement of strips gives a pattern a, a + d, a + 2d, a + 3d, ...

which is an AP with first term a and the common difference d.

Observations & calculations

1. The sum of the areas (in square units) of these strips

= a + (a + d) + (a + 2d) + ... + (a + 9d) = 10a + 45d ....... (1)

2. Area of the designe formed by the stair case

$=\frac{1}{2}\left(Area\;of\;rectangle\;ABCD\right)$

$=\frac{1}{2}\times 10\left[2a+(9)d\right]$

= (10a + 45d),

It is the same as obtained in equation (1) above.

3. This shows that the sum of first 10 terms of the AP

$=\frac{1}{2}\times 10\left[2a+(9)d\right]$

$S_{10}=\frac{10}{2}\left[2a+(10-1)d\right]$

4. This can be further generalised to find the sum of first n terms of an AP as

$S_{n}=\frac{n}{2}\left[2a+(n-1)d\right]$

5. On actual measurement we can calculate sum of n terms in an arithmatic progression as follows.

a = 3, d = 2, n = 5

$S_{5}=\frac{5}{2}\left[2\times 3+(5-1)2\right]=35$

a = 6, d = 4, n = 10

$S_{10}=\frac{10}{2}\left[2\times 6+(10-1)4\right]=240$

a = 12, d = 8, n = 40

$S_{40}=\frac{40}{2}\left[2\times 12+(40-1)8\right]=6720$

Result

Sum of first n terms of an Arithmetic Progression is

$S_{n}=\frac{n}{2}\left[2a+(n-1)d\right]$

Applications

This result may be used to find the sum of first n terms of the list of numbers :

1. 1², 2², 3², ...

2. 1³, 2³, 3³, ...

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