Monotonicity of a GP

Let the first term of the geometric progression be $$a$$ and the common ratio be $$r$$.
Since the sequence consists of increasing positive numbers, $$a > 0$$ and $$r > 1$$.

Given that the product of the third and fifth terms is $$\frac{1}{9}$$:
$$a_3 \cdot a_5 = \frac{1}{9}$$
$$(ar^2)(ar^4) = \frac{1}{9}$$
$$a^2r^6 = \frac{1}{9}$$

Taking the positive square root: $$ar^3 = \frac{1}{3}$$

Given that the sum of the sixth and eighth terms is 2:
$$a_6 + a_8 = 2$$
$$ar^5 + ar^7 = 2$$
$$ar^5(1 + r^2) = 2$$

We can rewrite $$ar^5$$ as $$(ar^3)r^2$$:

$$(ar^3)r^2(1 + r^2) = 2$$

Substitute $$ar^3 = \frac{1}{3}$$:
$$\frac{1}{3}r^2(1 + r^2) = 2$$
$$r^2(1 + r^2) = 6$$

Let $$r^2 = t$$.
The equation becomes:
$$t(t + 1) = 6$$
$$t^2 + t - 6 = 0$$
$$(t + 3)(t - 2) = 0$$

Since $$r > 1$$, $$r^2$$ must be positive, which gives $$t = 2$$.Thus $$r^2 = 2$$.

Substitute $$r^2 = 2$$ back into $$a^2r^6 = \frac{1}{9}$$:
$$a^2(2)^3 = \frac{1}{9}$$
$$8a^2 = \frac{1}{9}$$
$$a^2 = \frac{1}{72}$$

We need to evaluate the expression $$6(a_2 + a_4)(a_4 + a_6)$$.

Expressing this in terms of $$a$$ and $$r$$:
$$E = 6(ar + ar^3)(ar^3 + ar^5)$$
$$E = 6[ar(1 + r^2)][ar^3(1 + r^2)]$$
$$E = 6a^2r^4(1 + r^2)^2$$

Substitute $$a^2 = \frac{1}{72}$$ and $$r^2 = 2$$:

$$E = 6 \left(\frac{1}{72}\right) (2)^2 (1 + 2)^2$$
$$E = \frac{6}{72} \times 4 \times 9$$
$$E = \frac{1}{12} \times 36$$
$$E = 3$$

Hence the correct option is A.

Let the first term of the G.P. be $$a$$ and the common ratio be $$r\;(\,r\gt1\,)$$, because the terms are increasing and positive.

Then the terms are
$$a_1 = a,\; a_2 = ar,\; a_3 = ar^{2},\; a_4 = ar^{3},\; a_5 = ar^{4}$$

Step 1 : Use the product condition

Given $$a_3\,a_5 = 729$$, substitute the expressions for $$a_3$$ and $$a_5$$:
$$(ar^{2})(ar^{4}) = a^{2}r^{6} = 729$$
Rewrite it as $$\bigl(a\,r^{3}\bigr)^{2} = 729$$, so
$$a\,r^{3} = \sqrt{729} = 27 \quad -(1)$$

Step 2 : Use the sum condition

Given $$a_2 + a_4 = \dfrac{111}{4}$$, substitute $$a_2$$ and $$a_4$$:
$$ar + ar^{3} = ar\,(1 + r^{2}) = \dfrac{111}{4} \quad -(2)$$

Step 3 : Eliminate $$a$$

From $$(1)$$, express $$a$$:
$$a = \dfrac{27}{r^{3}} \quad -(3)$$

Insert $$(3)$$ in $$(2)$$:
$$\dfrac{27}{r^{3}}\;r\,(1 + r^{2}) = \dfrac{111}{4}$$
$$\dfrac{27}{r^{2}}\,(1 + r^{2}) = \dfrac{111}{4}$$

Multiply both sides by $$r^{2}$$:
$$27\,(1 + r^{2}) = \dfrac{111}{4}\,r^{2}$$

Clear the fraction by multiplying by $$4$$:
$$108\,(1 + r^{2}) = 111\,r^{2}$$

Expand and group $$r^{2}$$ terms:
$$108 + 108\,r^{2} = 111\,r^{2}$$
$$108 = 3\,r^{2}$$
$$r^{2} = 36 \;\Longrightarrow\; r = 6$$ (only the positive value suits an increasing G.P.)

Step 4 : Find $$a$$

Substitute $$r = 6$$ in $$(3)$$:
$$a = \dfrac{27}{6^{3}} = \dfrac{27}{216} = \dfrac{1}{8}$$

Step 5 : Compute the required expression

The sum of the first three terms is
$$a_1 + a_2 + a_3 = a\,(1 + r + r^{2})$$
$$= \dfrac{1}{8}\,(1 + 6 + 36) = \dfrac{1}{8}\times 43 = \dfrac{43}{8}$$

Finally,
$$24\,(a_1 + a_2 + a_3) = 24 \times \dfrac{43}{8} = 3 \times 43 = 129$$

Hence, the value of $$24(a_1 + a_2 + a_3)$$ is $$129$$, which corresponds to Option C.

The arithmetic progression is $$a_1,a_2,a_3,\ldots$$ with common difference $$d$$, so $$a_n = a_1 + (n-1)d$$.

The geometric progression is $$g_1,g_2,g_3,\ldots$$ with common ratio $$r$$, and we are told $$g_1 = a_1$$. Hence $$g_n = g_1\,r^{\,n-1} = r^{\,n-1}$$ because $$g_1 = a_1 = 1$$.

First condition: $$a_1 = a_2 + g_2 = 1$$.
  • $$a_2 = a_1 + d = 1 + d$$.
  • $$g_2 = r$$.
  • Therefore $$1 + d + r = 1 \;\Rightarrow\; d + r = 0 \;\Rightarrow\; r = -d$$. $$-(1)$$

Second condition: $$a_3 + g_3 = 4$$.
  • $$a_3 = a_1 + 2d = 1 + 2d$$.
  • $$g_3 = r^{\,2}$$.
  • Hence $$1 + 2d + r^{\,2} = 4 \;\Rightarrow\; 2d + r^{\,2} = 3$$. $$-(2)$$

Substitute $$r = -d$$ from $$(1)$$ into $$(2)$$:
$$2d + (-d)^{2} = 3 \;\Rightarrow\; 2d + d^{2} = 3 \;\Rightarrow\; d^{2} + 2d - 3 = 0.$$

Solving the quadratic,
$$d = 1 \quad \text{or} \quad d = -3.$$ But $$r = -d$$ must satisfy $$r \gt 1$$ because the G.P. is increasing.
  • If $$d = 1$$, then $$r = -1$$ (not > 1) - reject.
  • If $$d = -3$$, then $$r = 3$$ (acceptable).

Thus $$d = -3$$ and $$r = 3$$.

Compute the required terms.
  • Arithmetic term: $$a_{10} = a_1 + 9d = 1 + 9(-3) = 1 - 27 = -26.$$br/>   • Geometric term: $$g_5 = r^{\,4} = 3^{\,4} = 81.$$

Therefore, $$a_{10} + g_5 = -26 + 81 = 55.$$

Option D which is: 55