© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Represent the terms of the G.P.
Let the first term be $a$ and the common ratio be $r$.
The $n$th term of this geometric progression is given by
$a_n = a\,r^{n-1}$.
Step 2: Express the given conditions using $a$ and $r$
1) Sum of the 6th and 8th terms is 2:
$a_6 + a_8 = a\,r^{5} + a\,r^{7} = 2.$
2) Product of the 3rd and 5th terms is $\tfrac{1}{9}$:
$a_3 \cdot a_5 = (a\,r^2)\,(a\,r^4) = a^2\,r^6 = \tfrac{1}{9}.$
From the second condition,
$$
a^2 \, r^6 = \frac{1}{9}
\quad \Longrightarrow \quad
a \, r^3 = \frac{1}{3}.
$$
Step 3: Rewrite the sum condition using $a \, r^3 = \tfrac{1}{3}$
From $a \, r^3 = \tfrac{1}{3}$, we have $a = \tfrac{1}{3r^3}.$
Substitute $a = \tfrac{1}{3r^3}$ into $a\,r^{5} + a\,r^{7} = 2$:
$$
\frac{r^5}{3r^3} + \frac{r^7}{3r^3} = 2
\quad \Longrightarrow \quad
\frac{r^2}{3} + \frac{r^4}{3} = 2
\quad \Longrightarrow \quad
r^2 + r^4 = 6.
$$
Step 4: Solve for the common ratio $r$
Let $x = r^2.$ The equation becomes
$x + x^2 = 6,$
or
$x^2 + x - 6 = 0.$
Factorize this quadratic:
$$
(x - 2)(x + 3) = 0.
$$
Hence, $x = 2$ or $x = -3.$ Since $r^2$ must be positive, we choose
$r^2 = 2 \quad\Longrightarrow\quad r = \sqrt{2}$ (positive, as terms are increasing and positive).
Step 5: Find the first term $a$
We have $a \, r^3 = \tfrac{1}{3}.$ Substituting $r = \sqrt{2}$:
$$
a \cdot (\sqrt{2})^3 = a \cdot 2\sqrt{2} = \frac{1}{3}
\quad\Longrightarrow\quad
a = \frac{1}{3 \,\cdot\, 2\sqrt{2}}
= \frac{1}{6\sqrt{2}}.
$$
Step 6: Calculate $a_2 + a_4$ and $a_4 + a_6$
Note:
$a_2 = a\,r,\quad a_4 = a\,r^3,\quad a_6 = a\,r^5.$
• Compute $a_2 + a_4 = a(r + r^3).$
Substituting $a = \tfrac{1}{6\sqrt{2}}$ and $r = \sqrt{2}$:
$$
a_2 + a_4
= \frac{1}{6\sqrt{2}} \bigl(\sqrt{2} + (\sqrt{2})^3 \bigr)
= \frac{1}{6\sqrt{2}} \,\bigl(\sqrt{2} + 2\sqrt{2}\bigr)
= \frac{1}{6\sqrt{2}} \,\cdot 3\sqrt{2}
= \frac{3\sqrt{2}}{6\sqrt{2}}
= \frac{1}{2}.
$$
• Compute $a_4 + a_6 = a(r^3 + r^5).$
Substituting $a = \tfrac{1}{6\sqrt{2}}$ and $r = \sqrt{2}$:
$$
a_4 + a_6
= \frac{1}{6\sqrt{2}} \,\bigl((\sqrt{2})^3 + (\sqrt{2})^5 \bigr)
= \frac{1}{6\sqrt{2}} \,\bigl(2\sqrt{2} + 4\sqrt{2}\bigr)
= \frac{1}{6\sqrt{2}} \,\cdot 6\sqrt{2}
= 1.
$$
Step 7: Compute the required expression
We want to find
$6 \,(a_2 + a_4)\,(a_4 + a_6).$
From above,
$$
(a_2 + a_4) = \frac{1}{2},
\quad
(a_4 + a_6) = 1.
$$
Hence,
$$
6 \,\times \frac{1}{2} \,\times 1 = 3.
$$
Final Answer
3
