Key Concepts and Formulas

• A geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
• The $n^{\text {th }}$ term of a GP is given by $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.
• For a GP with positive terms and an increasing nature, the first term $a > 0$ and the common ratio $r > 1$.

Step-by-Step Solution

Let the first term of the increasing geometric progression be $a$ and the common ratio be $r$. Since the terms are positive and the progression is increasing, we have $a > 0$ and $r > 1$.

The terms of the GP are $a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, ar^7, ar^8, \dots$

We are given two conditions:

1. The sum of the second and sixth terms is $\frac{70}{3}$: $a_2 + a_6 = ar + ar^5 = \frac{70}{3}$ (Equation 1)

2. The product of the third and fifth terms is 49: $a_3 \cdot a_5 = (ar^2)(ar^4) = 49$ (Equation 2)

We need to find the sum of the fourth, sixth, and eighth terms: $S = a_4 + a_6 + a_8 = ar^3 + ar^5 + ar^7$.

Step 1: Simplify the product condition to find a key term. From Equation 2, we have: $(ar^2)(ar^4) = 49$ $a^2 r^{2+4} = 49$ $a^2 r^6 = 49$ Taking the square root of both sides, and since $a$ and $r$ are positive, we get: $\sqrt{a^2 r^6} = \sqrt{49}$ $ar^3 = 7$ Explanation: This step simplifies the product condition to find the value of the fourth term ($a_4$), which is a crucial term for the sum we need to calculate.

Step 2: Use the simplified product to rewrite the sum condition. From Equation 1, we have $ar + ar^5 = \frac{70}{3}$. We can factor out $ar$ from the left side: $ar(1 + r^4) = \frac{70}{3}$ We know that $ar^3 = 7$, which implies $a = \frac{7}{r^3}$. Substitute this expression for $a$ into the factored sum equation: $\left(\frac{7}{r^3}\right) r (1 + r^4) = \frac{70}{3}$ $\frac{7}{r^2} (1 + r^4) = \frac{70}{3}$ Divide both sides by 7: $\frac{1 + r^4}{r^2} = \frac{10}{3}$ Explanation: By substituting the value of $a$ derived from the product condition into the sum condition, we eliminate $a$ and create an equation solely in terms of $r$, making it solvable.

Step 3: Solve for the common ratio $r$. From the simplified sum condition, we have: $\frac{1}{r^2} + \frac{r^4}{r^2} = \frac{10}{3}$ $\frac{1}{r^2} + r^2 = \frac{10}{3}$ Let $x = r^2$. The equation becomes: $\frac{1}{x} + x = \frac{10}{3}$ Multiply the entire equation by $3x$ to clear the denominators: $3 + 3x^2 = 10x$ Rearrange into a quadratic equation: $3x^2 - 10x + 3 = 0$ Factor the quadratic equation: $(3x - 1)(x - 3) = 0$ This gives two possible values for $x$: $3x - 1 = 0 \implies x = \frac{1}{3}$ $x - 3 = 0 \implies x = 3$ Since $x = r^2$, we have $r^2 = \frac{1}{3}$ or $r^2 = 3$. This means $r = \pm \frac{1}{\sqrt{3}}$ or $r = \pm \sqrt{3}$. Given that the geometric progression is increasing and has positive terms, we must have $r > 1$. Therefore, we choose $r = \sqrt{3}$. Explanation: Solving the quadratic equation for $r^2$ allows us to find the possible values for the common ratio. The condition that the GP is increasing ($r>1$) is used to select the correct value of $r$.

Step 4: Calculate the required sum. We need to find $S = ar^3 + ar^5 + ar^7$. We already know $ar^3 = 7$. Now we can find $ar^5$ and $ar^7$ using $ar^3$ and $r = \sqrt{3}$: $ar^5 = ar^3 \cdot r^2 = 7 \cdot (\sqrt{3})^2 = 7 \cdot 3 = 21$. $ar^7 = ar^5 \cdot r^2 = 21 \cdot (\sqrt{3})^2 = 21 \cdot 3 = 63$.

Therefore, the sum is: $S = ar^3 + ar^5 + ar^7 = 7 + 21 + 63 = 91$ Explanation: Using the value of $ar^3$ and the common ratio $r$, we can efficiently calculate the required terms ($ar^5$ and $ar^7$) and then sum them up. This method avoids the need to explicitly calculate the first term $a$.

Common Mistakes & Tips

• Sign of the common ratio: Remember that for an increasing GP with positive terms, the common ratio $r$ must be greater than 1.
• Algebraic simplification: Look for opportunities to simplify expressions. In this problem, realizing that $ar^3=7$ was a significant shortcut.
• Avoid calculating $a$ if not necessary: The problem asks for a sum of terms, and often these can be expressed in terms of known terms like $ar^3$, thus avoiding the calculation of $a$ and potential arithmetic errors.

Summary

The problem involved an increasing geometric progression of positive terms. We used the given sum and product conditions to establish equations relating the first term ($a$) and the common ratio ($r$). By simplifying the product condition, we found that the fourth term, $ar^3$, is equal to 7. This value was then used to simplify the sum condition, leading to a quadratic equation in $r^2$. Solving this equation and applying the condition $r > 1$ yielded the common ratio $r = \sqrt{3}$. Finally, we calculated the sum of the fourth, sixth, and eighth terms using the value of $ar^3$ and $r$, resulting in a sum of 91.

The final answer is \boxed{91}, which corresponds to option (C).