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.

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.

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.