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Step-by-Step Solution
Step 1: Express the terms of the geometric series
Let the first term of the geometric series be $a$ and the common ratio be $r$. The terms of the series are:
$a,\; ar,\; ar^2,\; ar^3,\; ar^4,\; ar^5,\dots$
Step 2: Use the condition on the sum of the second and sixth terms
The second term is $T_2 = ar$ and the sixth term is $T_6 = ar^5$. We are given:
$T_2 + T_6 = \frac{25}{2}$
Substituting the expressions for $T_2$ and $T_6$:
$ar + ar^5 = \frac{25}{2}\quad \Rightarrow \quad ar \,\bigl(1 + r^4\bigr) = \frac{25}{2}.$
Step 3: Use the condition on the product of the third and fifth terms
The third term is $T_3 = ar^2$ and the fifth term is $T_5 = ar^4$. We are given:
$T_3 \cdot T_5 = 25.$
Substituting the expressions for $T_3$ and $T_5$:
$(ar^2)(ar^4) = 25\quad \Rightarrow \quad a^2 r^6 = 25.$
Step 4: Form an equation to solve for $r$
From Step 2, we have:
$ar(1 + r^4) = \frac{25}{2} \quad\quad (1)
and from Step 3:
$a^2 r^6 = 25 \quad\quad (2)
We can square the relation (1) or manipulate it together with (2). A straightforward approach is to observe:
From (1): $ar (1 + r^4) = \frac{25}{2} \;\;\; \Rightarrow \;\;\; a = \frac{25}{2r(1 + r^4)}.$
Substitute $a$ into (2): $a^2 r^6 = 25.$
Alternatively, one can directly divide the square of equation (1) by equation (2) to eliminate $a^2$. Either way yields the same result.
On combining carefully, we arrive at:
$ \frac{(1 + r^4)^2}{r^4} = \frac{25}{4}.
Clearing denominators and rearranging gives a quadratic in $r^4$:
$4r^8 - 14r^4 + 4 = 0.
Factoring this equation:
$(4r^4 - 1)(r^4 - 4) = 0.
Hence, the possible solutions for $r^4$ are $r^4 = \frac{1}{4}$ or $r^4 = 4.$
Since the series is increasing, $r > 1$ (or at least $r > 0$ and bigger than 1 in magnitude). Thus, we choose:
$r^4 = 4 \quad\Rightarrow\quad r = \sqrt{2}.
Step 5: Compute $a$ in terms of $r$
We know $a^2 r^6 = 25$ from (2). Substituting $r = \sqrt{2}$ gives:
$r^6 = (\sqrt{2})^6 = 2^3 = 8.
$a^2 \cdot 8 = 25 \quad\Rightarrow\quad a^2 = \frac{25}{8} \quad\Rightarrow\quad a = \frac{5}{\sqrt{8}} = \frac{5}{2\sqrt{2}}.
One may also work directly with the factor $a r^3$, but the above relationship suffices to move to the final step.
Step 6: Find the sum of the fourth, sixth, and eighth terms
The fourth term is $T_4 = ar^3$, the sixth term is $T_6 = ar^5$, and the eighth term is $T_8 = ar^7.$ We write:
$T_4 + T_6 + T_8 = a r^3 + a r^5 + a r^7.
Factor out $ar^3$:
$T_4 + T_6 + T_8 = a r^3 \bigl(1 + r^2 + r^4\bigr).
From $a^2 r^6 = 25$, we know $a r^3 = \pm 5$. Because $a > 0$ and $r^3 > 0$, we take $a r^3 = 5.$
Since $r^2 = 2$ and $r^4 = 4,$ the sum inside parentheses becomes:
$1 + r^2 + r^4 = 1 + 2 + 4 = 7.
Thus,
$T_4 + T_6 + T_8 = 5 \times 7 = 35.
Final Answer
The sum of the fourth, sixth, and eighth terms is 35.
