Let the first term of the Geometric Progression (G.P.) be $a$ and the common ratio be $r$. The terms are $a_1, a_2, ..., a_n$. We are given:

• $a_1 + a_n = 66$
• $a_2 \cdot a_{n-1} = 128$
• $\sum_{r=1}^n a_r = 126$
• $n > m$ and $a_n > a_m$, implying the sequence is increasing.

G.P. Term Relations

Using the properties of a G.P.:

• $a_1 = a$
• $a_n = a r^{n-1}$
• $a_2 = a r$
• $a_{n-1} = a r^{n-2}$

The condition $a_2 \cdot a_{n-1} = 128$ becomes:

$ (ar) \cdot (a r^{n-2}) = a^2 r^{n-1} = 128 $

For a G.P., the product of terms equidistant from the beginning and end is constant, i.e., $a_1 \cdot a_n = a_2 \cdot a_{n-1}$. Therefore:

$ a_1 \cdot a_n = 128 $

Finding First and Nth Term

We have the sum $a_1 + a_n = 66$ and the product $a_1 \cdot a_n = 128$. These two terms are the roots of the quadratic equation $x^2 - (\text{sum})x + (\text{product}) = 0$.

$ x^2 - 66x + 128 = 0 $

Solving this quadratic equation:

$ (x - 2)(x - 64) = 0 $

The roots are $x=2$ and $x=64$. So, $\{a_1, a_n\} = \{2, 64\}$.

Since $n > m$ and $a_n > a_m$, the sequence is increasing. Given $a_1$ and $a_n$ are positive ($66>0, 128>0$), the common ratio $r$ must be positive. For an increasing sequence, $r > 1$. Therefore, the first term must be smaller than the nth term:

$ a_1 = 2 \quad \text{and} \quad a_n = 64 $

Calculating Common Ratio and Number of Terms

From $a_1 = 2$, we know the first term $a = 2$. Using $a_n = 64$:

$ a r^{n-1} = 64 $ $ 2 \cdot r^{n-1} = 64 $ $ r^{n-1} = 32 \quad (*) $

Now, using the sum formula for a G.P.:

$ \sum_{r=1}^n a_r = \frac{a(r^n - 1)}{r - 1} = 126 $

Substitute $a=2$:

$ \frac{2(r^n - 1)}{r - 1} = 126 $ $ \frac{r^n - 1}{r - 1} = 63 $

We can write $r^n$ as $r \cdot r^{n-1}$. Substitute $r^{n-1}=32$ from equation (*):

$ r^n = r \cdot 32 $

Substitute this into the sum equation:

$ \frac{32r - 1}{r - 1} = 63 $

Solve for $r$:

$ 32r - 1 = 63(r - 1) $ $ 32r - 1 = 63r - 63 $ $ 63 - 1 = 63r - 32r $ $ 62 = 31r $ $ r = \frac{62}{31} = 2 $

Now substitute $r=2$ back into equation (*):

$ 2^{n-1} = 32 $

Since $32 = 2^5$:

$ 2^{n-1} = 2^5 $

Equating the exponents:

$ n-1 = 5 $ $ n = 6 $