Activities - progressions problems

Problem 1 from Exercise 3.2 ( Q.N.11 - page No. 37)

A company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785? Solution

Pre-requisites

Pupil know:-

1. Student know meaning of progression and arithmatic progression.
2. Student know the terms used in arithmatic progression problem.
3. Student know the formula under arithmatic progression .
4. Student know the basic operation to calculate the values.

Interpretation of the problem

1. Student should understand there are 400 persons in a company in 2001.
2. Student also know that every year they taken equal number of employees - this means that it is related to arithmatic progression

What is asked of student

In which year the number of employees is equal to 785. This means they know the first term is 400 and last term is 785. So they have to calculate the number of terms in this sequence.

Method 1:Solution by formula method

Assumption

1. He know the general form of A.P.
2. He know the formula of nth term of A.

Algorithm

Given:Arithmatic progression
400, 435, 470,......,785.
First term = a = 400
common differnce = d = 35
Last term = Tn = 785
Number of terms = n = ?
We know the n th term of A.P. is
Tn = a + (n-1) d
785 = 400 + ( n-1) 35
785 – 400 = 35n – 35
385 + 35 = 35n
420 = 35n
n = 12
So there are 12 terms in that progression . So In 2012 the number of employees equal to 785.

Method 2 - By Verification

Assumption

Student know the writing progression when he know first term and common differnce.

Algarithm

400, 435, 470, 505, 540, 575, 610, 645, 680, 715, 750, 785.
In 2001 -------> 400
In 2002 --------> 435
In 2003---------> 470
In 2004 --------> 505
In 2005 --------> 540
In 2006 --------> 575
In 2007 ----------> 610
In 2008 ----------> 645
In 2009 -----------> 680
In 2010 ----------->715
In 2011 -----------> 750
In 2012 -----------> 785
so in 2012 the number employees became 785.

Problem 2 from Exercise 3.3 ( Q.N.12 - page No. 43)

The sum of 6 terms which form an A.P is 345. The difference between the first and last terms is 55. Find the terms.

Pre-requisites

Pupil know:-

1. What is progression ?
2. What is arithmatic progression?
3. Student know the general form of A.P.
4. Student know the knowledge of simultaniouse equation and substitution method.

Premisis for problem solving

1. Student know to use the general form A.P.
2. Solve simultaniouse equation by substitution method

Interpretation of the problem

1. There are six terms in an A.P. and their sum is given.
2. Differnce between first and last terms is given.
3. We find the remaining terms of A.P.

What is asked of student

To find the remaining terms in A.P.

Method 1 :- Solution by formula method

Assumption

1. He know the general form of A.P.
2. He know the simultaniouse linear eq.

Algorithm

The general form A.P. is a, a+d, a+2d, a+3d, ......, a+(n-1)d.
But there are only six terms are there so
a, a+d, a+2d, a+3d, a+4d, a+5d.
Sum of six terms is 345.
so a+a+d+a+2d+a+3d+a+4d+a+5d = 345
6a + 15d = 345------------>1
Differnce between last term and first term is 55.
so (a+ 5d) – (a) = 55
5d = 55
d = 11 -------------->2
Now we find the first term of A.P. by substitute in eq 1.
6a + 15d = 345
6a + 15 (11) = 345
6a + 165 = 345
6a = 345 – 165
6a = 180
a = 30
then, a = 30, d = 11,
so A.P. is 30, 41,52,63,74,85.

Exercise 3.7 , Problem number 8, Page number 62

Find two numbers whose arithmetic mean exceeds their geometric mean by 2, and whose harmonic mean is one-fifth of the larger number.

Pre-requisites

Pupil know

1. The meaning of arithmetic mean , geometric mean and harmonic mean
2. The formula of A.M , G.M and H.M
3. Student know the basic operation to calculate the values.

Interpretation of the problem

1. Student should take a>b
2. Student know that arithmetic mean is exceeds the geometric mean by 2.
3. Harmonic mean is one – fifth of the larger number.

What is asked of student

Find the two numbers 'a' and 'b' such that whose arithmetic mean is exceeds their geometric mean by 2 and harmonic mean is one – fifth of the larger number.

Solution of the problem

Assumption :-

1. He know the formula for A.M , G.M and H.M
2. He know the basic operation.

Algorithm

Let 'a' and 'b' two numbers such that a>b
Arithmetic mean formula is A.M =${\frac {a+b}{2}}$
Geometric mean formula is G.M = ${\sqrt {ab}}$
A.M exceeds their G.M by 2
A.M = G.M +2
${\frac {a+b}{2}}$=${\sqrt {ab}}$ + 2
a + b = 2 ( ${\sqrt {ab}}$ + 2 )-------> 1
H.M is one – fifth larger number . Here 'a' is larger number so,
H.M = ${\frac {1}{5}}$ a
${\frac {2ab}{a+b}}$ = ${\frac {1}{5}}$ a
Here we cancel 'a' because it is in both side of the equal sign and after cancel 'a' we get,
${\frac {2b}{a+b}}$ = ${\frac {1}{5}}$
Cross multiply we get,
a + b = 10b ----------------->2
Take '10b' to left hand side, we get
a + b – 10b = 0
a – 9b = 0

 a = 9b

Consider equation 1
a + b = 2 (${\sqrt {ab}}$ + 2 )
Put a = 9b in the above equation,

9b + b = 2 (${\sqrt {9bXb}}$ + 2 )
10b = 2 (${\sqrt {9b^{2}}}$
10b = 2 ( 3b + 2 )
10b = 6b + 4
10b – 6b = 4
4b = 4
b = 1
Consider equation 2

a + b = 10b
Put b = 1
a + (1) = 10 (1)
a + 1 = 10
a = 10-1
a = 9