Chapter 5:Arithmetic Progressions

Class 10 Case based problem of Chapter 5 A.P. 1

Case Based :1

The students of a school decided to beautify the school on an Annual day by fixing colourful flags on the straight passage of the school. A total of 27 flags have to be fixed at intervals of every 3 m. All the flags are kept at the position of the middle most flag. Arzoo was given the responsibility of placing the flags on both sides of the middle most flag. Arzoo kept her bag wher the flags were stored. She could carry only one flag at a time . So she carries one flag from the middle most point. places the flag at its designated place and come back to the middle most point. She continues this for all flags on both sides.

(A) Find the distance covered by Arzoo in placing successive flags on one side of the middle most flag in form an A.P.

(B) Find the total distance covered by Arzoo in placing all the flags on both sides of the middle most flag.

Solution:

(A) All the flags are kept at the position of middle most flag $= \dfrac{27+1}{2} = 14^{th}$ flag

There are 13 flags on either side of the middle flag at a distance of 3 m from each other.

The first flag is ata distance of 3 m, the second at a 6 m, the third at a distance of 9 m and so on and this is on both side of the middle flag

As Arzoo has to come back to the position of the middle flag each time, The distance travelled for placing the first flag

and come back = 3 + 3= 6 m

The distance travelled for placing the first flag and come back = 6 + 6= 12 m

The distance covered by Arzoo in placing flags on one side of the middle post flag

= 6 m, 12 m, 18 m ……..

This forms an A.P. with a = 6 m and d = 6 m.

(B) Given A.P. = 6 m, 12 m, 18 m ……….

Total distance covered by Arzoo = 2 × (The distance covered by Arzoo in placing 13 flags on one side)

$= 2 ×S_n$

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

$= 2\times \frac{13}{2}\left[2(6) + (13-1)6\right]$

$= 13\times\left[ 12 + 72\right]$

$= 13\times 84$

$= 1092$ m

Case Based: 2

Excavators are important and widely used equipment in construction industry their general purpose is to excavate but other than that they are also used for many purposes like heavy lifting. demolition, river dredging, cutting of trees etc.

Class 10 Case based problem of Chapter 5 A.P.

ABC Machinery Pvt Ltd started making excavators about 15 years ago. The company increased its production uniformly by a fixed number every year. The company produces 700 excavators in the $5^{th}$ year and 1180 excavators in the $11^{th}$ year.

(A) What was the company’s production of excavators in the first year ?

(B) What was the total production of excavators in the first 8 years ?

Solution:

(A) The company increased its production uniformly by a fixed number every year, then the production will make an A.P.

Let the first term of A.P. = a

Common difference = d

Given that, $a_5 = 700 and a_{11} = 1180$

So, $a_5 = 700$

$\Rightarrow a + 4d = 700$ ……….(i)

and $a_{11} = 1180$

$\Rightarrow a + 10d = 1180$ ……….(ii)

Subtracting (ii) to (i)

$(a + 10d)-(a + 4d) = 1180 - 700$

$\Rightarrow a + 10d - a - 4d = 480$

$\Rightarrow 6d = 480$

$\Rightarrow d = 80$

Putting in (i)

$a + 4d = 700$

$\Rightarrow a + 4\times 80 = 700$

$\Rightarrow a + 320 = 700$

$\Rightarrow a = 700-320$

$\Rightarrow a = 380$

Therefore, the company production of excavators in the first year = 380.

(B) First term = a = 380

Common difference = d = 80

$S_8 = \frac{8}{2}\left[2(380) + (8-1)\times 80 \right]$

$= 4\left[760 + 560\right]$

$= 4 \times 1320$

$= 5280$

The total production of excavators in first 8 years = 5280

Case Based: 3

Seats of an auditorium are arranged such that there are a total of 6 seats in the first row, 8 seats in the second row, 10 seats in the third row and so on.

Class 10 Case based problem of Chapter 5 A.P. — Seats of an auditorium are arranged such that there are a total of 6

(A) Find the total number of seats in the auditorium if there are are total of 30 rows.

(B) The sum of the $5^{th}$ and the $7^{th}$ terms of an A.P. is 52 and the $10^{th}$ term is 46. Find the first term ‘a’ and common difference ‘d’ of the A.P.