1. Let the geometric progression be

   $a,\, ar,\, ar^2,\, ar^3,\, ar^4,\, ar^5,\dots$

   Since it is an increasing GP of positive terms, we have:

   $a>0,\quad r>1.$

2. Use the product of the 3rd and 5th terms

   The 3rd term is $ar^2$ and the 5th term is $ar^4$.

   Their product is given as $49$:

   $(ar^2)(ar^4)=a^2r^6=49.$

   Taking positive square root (since terms are positive),

   $ar^3=7.$

   But $ar^3$ is the 4th term.

   So, the 4th term is:

   $T_4=7.$

3. Use the sum of the 2nd and 6th terms

   The 2nd term is $ar$ and the 6th term is $ar^5$.

   Given:

   $ar+ar^5=\frac{70}{3}.$

   Factor out $ar^3$:

   $ar^3\left(r^{-2}+r^2\right)=\frac{70}{3}.$

   Since $ar^3=7$,

   $7\left(r^2+\frac{1}{r^2}\right)=\frac{70}{3}.$

   Therefore,

   $r^2+\frac{1}{r^2}=\frac{10}{3}.$

4. Find $r+\frac{1}{r}$ indirectly

   We use:

   $\left(r-\frac{1}{r}\right)^2=r^2+\frac{1}{r^2}-2=\frac{10}{3}-2=\frac{4}{3}.$

   So,

   $r-\frac{1}{r}=\frac{2}{\sqrt{3}}$

   because $r>1$.

   Let $x=r+\frac{1}{r}$. Then

   $x^2=r^2+\frac{1}{r^2}+2=\frac{10}{3}+2=\frac{16}{3}.$

   Hence

   $r+\frac{1}{r}=\frac{4}{\sqrt{3}}.$

5. Required sum: 4th, 6th and 8th terms

   These are:

   $ar^3,\quad ar^5,\quad ar^7.$

   Their sum is:

   $ar^3+ar^5+ar^7=ar^5\left(\frac{1}{r^2}+1+r^2\right).$

   A simpler factorization is:

   $ar^3(1+r^2+r^4).$

   Since $ar^3=7$,

   $\text{Required sum}=7(1+r^2+r^4).$

   Now,

   $r^2+\frac{1}{r^2}=\frac{10}{3}.$

   Let $y=r^2$. Then

   $y+\frac{1}{y}=\frac{10}{3}.$

   Multiplying by $y$:

   $y^2+1=\frac{10}{3}y$

   $3y^2-10y+3=0.$

   Solving:

   $y=3 \quad \text{or} \quad y=\frac{1}{3}.$

   Since $r>1$, we need $r^2>1$, so

   $r^2=3.$

   Therefore,

   $1+r^2+r^4=1+3+9=13.$

   Hence,

   $\text{Required sum}=7\cdot 13=91.$

6. Check options

   • A: $78$ ❌
   • B: $96$ ❌
   • C: $91$ ✅
   • D: $84$ ❌

Therefore, the correct answer is Option C.