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.

=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.

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. =Problem 3 = (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 ($$\sqrt{9b X b}$$ + 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