Let $ b_1 = 3, b_2, b_3, \ldots $ be a geometric progression of increasing positive numbers. Let \[ \sum_{n=1}^{20} b_{3n} = 4\sum_{n=1}^{20} b_{3n-2}. \] Then, the sum of the first ten terms of the G.P. is:

Question

If $a + b = 5$ and $ab = 6$, find $a^2 + b^2$:

Answer 1

To solve the given problem, we start by understanding the formula and properties of a geometric progression (G.P.). For a G.P., if the first term is a and the common ratio is r, the n^{th} term can be expressed as:

b_n = a \cdot r^{n-1}

Here, we know that the first term b_1 = 3, so a = 3. The common ratio, r, needs to be determined. We are given:

\sum_{n=1}^{20} b_{3n} = 4 \sum_{n=1}^{20} b_{3n-2}

First, compute expressions for the series:

The terms b_{3n} in the geometric progression are: b_3, b_6, b_9, \ldots, and generally, b_{3n} = a \cdot r^{3n-1}.
The terms b_{3n-2} are: b_1, b_4, b_7, \ldots, and generally, b_{3n-2} = a \cdot r^{3n-3}.

Substitute these general terms into the equation:

\sum_{n=1}^{20} (a \cdot r^{3n-1}) = 4 \sum_{n=1}^{20} (a \cdot r^{3n-3})

Canceling a (since a = 3 \neq 0), the equation becomes:

\sum_{n=1}^{20} r^{3n-1} = 4 \sum_{n=1}^{20} r^{3n-3}

Rearrange terms:

r^2 \sum_{n=1}^{20} r^{3(n-1)} = 4 \sum_{n=1}^{20} r^{3(n-1)}

Further simplify (the sums are identical):

r^2 = 4 \Rightarrow r = 2

So, the common ratio r = 2. The first ten terms of the G.P. are:

b_1 = 3, \, b_2 = 6, \, b_3 = 12, \ldots, \, b_{10} = 3 \cdot 2^9

The sum of the first ten terms S_{10} of a G.P. is given by:

S_{10} = a \frac{r^{10} - 1}{r - 1}

Substituting a = 3 and r = 2:

S_{10} = 3 \frac{2^{10} - 1}{2 - 1} = 3(1024 - 1) = 3 \times 1023 = 3069

The correct answer is 3069.

Answer 2