Question:

To solve this problem, let's denote the first term of the geometric progression as $a$ and the common ratio as $r$. Step 1: Given that the sum of the second and sixth terms is $\dfrac{70}{3}$, we can express this as an equation: $$ar + ar^5 = \dfrac{70}{3}$$ This can be rewritten as: $$ar(1 + r^4) = \dfrac{70}{3}$$ Step 2: The product of the third and fifth terms is given as 49, which translates to: $$ar^2 \cdot ar^4 = 49$$ Simplifying, we get: $$a^2r^6 = 49$$ Step 3: From the information given, we also have the relationship: $$ar^3 = 7$$ And from the options and the nature of geometric progressions, we find: $$r^2 = 3$$ Step 4: To find the sum of the 4th, 6th, and 8th terms, we use the formula for the nth term of a geometric progression, $ar^{n-1}$, and sum these terms: $$T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7$$ Substituting $ar^3 = 7$ and $r^2 = 3$, we get: $$T_4 + T_6 + T_8 = 7 + 7 \cdot 3 + 7 \cdot 3^2$$ $$T_4 + T_6 + T_8 = 7 + 21 + 63$$ $$T_4 + T_6 + T_8 = 7(1 + 3 + 9)$$ $$T_4 + T_6 + T_8 = 7 \cdot 13$$ $$T_4 + T_6 + T_8 = 91$$ Therefore: $91$ <div class="key-concept"><strong>Key Concept:</strong> Let GP have first term $a$ and ratio $r$. $ar+ar^5=70/3$ and $ar^2\cdot ar^4=49\Rightarrow a^2r^6=49\Rightarrow ar^3=7$ (positive terms). So $r+r^5=70/(3\cdot7/r^2)$... use $ar^3=7$.</div> <div class="trap-box"><strong>Trap:</strong> From $ar+ar^5=70/3$: $7(1/r^2+r^2)=70/3\Rightarrow r^2+1/r^2=10/3$. Increasing GP: $r>1$. $r^4-10r^2/3+1=0\Rightarrow r^2=3$ (taking larger root). $r=\sqrt{3}$, $a=7/r^3=7/(3\sqrt{3})$. Sum $=ar^3+ar^5+ar^7=7(1+r^2+r^4)=7(1+3+9)=91$.</div>