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Activities - progressions problems (view source)
Revision as of 04:00, 30 October 2019
, 04:00, 30 October 2019added Category:Progressions using HotCat
then, a = 30, d = 11,<br>
then, a = 30, d = 11,<br>
so A.P. is 30, 41,52,63,74,85.
so A.P. is 30, 41,52,63,74,85.
=Exercise 3.7 , Problem number 8, Page number 62=
=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.
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==
==Pre-requisites==
A.M exceeds their G.M by 2<br>
A.M exceeds their G.M by 2<br>
A.M = G.M +2<br>
A.M = G.M +2<br>
<math>\frac{ a + b } {2}</math>=<math>\sqrt{ab}</math>
<math>\frac{ a + b } {2}</math>=<math>\sqrt{ab}</math> + 2<br>
a + b = 2 ( <math>\sqrt{ab}</math> + 2 )-------> 1<br>
H.M is one – fifth larger number . Here 'a' is larger number so,<br>
H.M = <math>\frac{1} {5}</math> a<br>
<math>\frac{2ab} {a + b}</math> = <math>\frac{1} {5}</math> a<br>
Here we cancel 'a' because it is in both side of the equal sign and after cancel 'a' we get,<br>
<math>\frac{2b} {a + b}</math> = <math>\frac{1} {5}</math><br>
Cross multiply we get,<br>
a + b = 10b ----------------->2<br>
Take '10b' to left hand side, we get<br>
a + b – 10b = 0<br>
a – 9b = 0<br>
a = 9b<br>
Consider equation 1<br>
a + b = 2 (<math>\sqrt{ab}</math> + 2 )<br>
Put a = 9b in the above equation,
9b + b = 2 (<math>\sqrt{9b X b}</math> + 2 )<br>
10b = 2 (<math>\sqrt{9b^2}</math><br>
10b = 2 ( 3b + 2 )<br>
10b = 6b + 4<br>
10b – 6b = 4<br>
4b = 4<br>
b = 1<br>
Consider equation 2<br>
a + b = 10b<br>
Put b = 1<br>
a + (1) = 10 (1)<br>
a + 1 = 10<br>
a = 10-1<br>
a = 9<br>
=Problem 4=
''' Exercise 3.7 , Problem number 10, Page number 62 '''
If 'a' be the arithmetic mean between 'b' and 'c', and 'b' the geometric mean between 'a' and 'c', then prove that 'c' will be the harmonic mean between 'a' and 'b'.
==Pre-requisites==
Pupil know
#The meaning of arithmetic mean , geometric mean and harmonic mean
#The formula of A.M , G.M and H.M
#Student know the basic operation to calculate the values.
==Interpretation of the problem==
#In this problem 'a' is the arithmetic mean of 'b' and 'c'
#They also given that 'b' is the geometric mean of 'a' and 'c'
==What is asked of student==
#In this problem we prove that 'c' will be the harmonic mean between 'a' and 'b'
#That is we just show that c = <math>\frac{2ab} {a + b}</math>
==Solution of the problem==
Assumption :-
#He know the formula for A.M , G.M and H.M
#He know the basic operation.
#He know substitute the values
==Algorithm==
The formula for arithmetic mean between 'a' and 'b' is A.M = <math>\frac{a + b} {2}</math><br>
The formula for geometric mean between 'a' and 'b' is G.M = <math>\sqrt{ab}</math><br>
Then formula for harmonic mean between 'a' and 'c' is H.M = <math>\frac{2ab} {a + b}</math><br>
Let 'a' be the arithmetic mean of 'b' and 'c'<br>
That is a = <math>\frac{b + c} {2}</math><br>
Re-arrange the formula,<br>
2 = <math>\frac{b + c} {a}</math><br>
Multiply both side by 'ab' we get,<br>
2ab = <math>\frac{ab(b + c)} {a}</math><br>
Cancel 'a' in Right hand side and multiply 'b' inside to the bracket we get<br>
2ab = <math>b^{2}</math> + bc ---------->1<br>
Also 'b' is the geometric mean between 'a' and 'c'<br>
That is b = <math>\sqrt{ac}</math><br>
We also write this as <math>b^{2}</math> = ac.-------->2<br>
Now substitute thia value In equation 1,<br>
2ab = ac + bc<br>
Take common in right hand side ( c is common )<br>
2ab = c(a + b)<br>
Divide both side by (a + b),<br>
<math>\frac{2ab} {a + b}</math>= c<br>
Hence 'c' is the harmonic between 'a' and 'b'.
[[Category:Progressions]]
