Here is another excellent solution from Andrei Lazanu, age 12, School no. 205, Bucharest, ROMANIA. Well done Andrei.

I started with a1 = 9 and b1 = 3, and I used the recursive relations from the problem. I obtained the following values (to three decimal places):

a _{1} =9.000 |
b _{1} =3.000 |

arithmetic means | geometric means |

a _{2} =6.000 |
b _{2} =5.196 |

a _{3} =5.598 |
b _{3} =5.584 |

a _{4} =5.591 |
b _{4} =5.591 |

a _{5} =5.591 |
b _{5} =5.591 |

a _{6} =5.591 |
b _{6} =5.591 |

a _{7} =5.591 |
b _{7} =5.591 |

I represented the successive "a" terms and "b" terms on a number line and I saw that both of them go toward the same value. Using 3 decimals, practically, after the 3-rd iteration (a4 and b4) the same value is reached.

[Note that each time the geometric mean is less than the arithmetic mean. The arithmetic means are decreasing (can you explain why?) and the geometric means are increasing (why?) so they must converge.]

Now, I choose my numbers, and I looked for more distant numbers to see how fast they arrive at the same value: a1 = 2 and b1 = 15000.

I wrote the following programme in Matlab, to generate the terms of the sequences "a" and "b":

% arithmetic and geometric means

%a(1)=9;b(1)=3;

a(1)=15000;b(1)=2

for i=1:6

a(i+1)=(a(i)+b(i))/2;

b(i+1)=sqrt(a(i)*b(i));

end

I obtained the following values: for the "a" (left column) and "b" (right column) sequences respectively (all values must be multiplied by 104).

1.5000 | 0.0002 |

arithmetic means | geometric means |

0.7501 | 0.0173 |

0.3837 | 0.1140 |

0.2488 | 0.2091 |

0.2290 | 0.2281 |

0.2286 | 0.2286 |

0.2286 | 0.2286 |

I saw that after the 5-th iteration, to the precision used, the two sequences arrived at the same value.

[You may like to use a spreadsheet to do this investigation.]