2. Geometric Progressions
by M. Bourne
A Geometric Progression (GP) is formed by multiplying a starting number (a1) by a number r, called the common ratio.
The progression `5, 10, 20, 40, 80, 160`, has first term `a_1= 5`, and common ratio `r = 2`.
In this example, we started with `5` and multiplied by `2` each time to get the next number in the progression.
Formula for the `n`-th term of a GP
The n-th term of a geometric progression is given by:
an = a1rn−1
The first term is
The second term is obtained by multiplying the first by r
The third term is obtained by multiplying the second by r
The fourth term is obtained by multiplying the third by r
We continue this pattern and can see that in general, the n-th term is
Find the 50th term of the geometric progression 5, 10, 20, 40, 80, ...
Since `a_1= 5`, `r = 2`, and using
an = a1rn-1,
a50 = (5)(250−1)
≈ 2.81 × 1015
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The Sum of a Geometric Progression
The sum to n terms of a GP means:
a1 + a1r + a1r2 + a1r3 + ... + a1rn-1
We can show (using Proof by Induction) that this sum is equivalent to:
(We first saw this story in the Chapter Introduction).
A king once promised a prince anything he wanted because he saved the princess's life. The prince requested one grain of rice on the first square of a chess board, `2` on the second, `4` on the third, `8` on the fourth square, etc.
How much rice is there if one grain of rice weighs `20\ "mg"`?
We need `1 + 2 + 4 + 8 + ... + 2^63`
Now `a_1= 1`, `r = 2`, `n = 64`.
Our formula for the sum to n terms says:
Substituting our values:
Each grain weighs `20\ "mg" = 2 × 10^-5\ "kg" ` `= 2 × 10^-8\ "tonnes"`.
So the weight is
`(1.84467 × 10^19) × ` `(2 × 10^-8)\ "tonnes" ` `= 369\ "billion tonnes"`, so of course, the king cannot grant the Prince's wish.
NOTE 1: There are `1000\ "kg"` in one tonne.
NOTE 2: The world annual output of rice today is only `600` million (not billion) tonnes!
NOTE 3: We are using the US/French 'billion' (`10^9`) and not the British 'billion' (`10^12`). [See Short and Long Scales.]