1. Set up the G.P.

The first three terms of the G.P. are $2,\, p,\, q$ with common ratio say $r$.

So, $p=2r, \qquad q=2r^2$ Since $q\ne 2$, we have $r^2\ne 1$.

Also, the $5^{\text{th}}$ term of the G.P. is $T_5^{(GP)}=2r^4$

2. Set up the A.P.

Let the A.P. have first term $a$ and common difference $d$. Then its general term is $T_n^{(AP)}=a+(n-1)d$

Given:

• $7^{\text{th}}$ term of A.P. = first term of G.P. = $2$
• $8^{\text{th}}$ term of A.P. = second term of G.P. = $p=2r$
• $13^{\text{th}}$ term of A.P. = third term of G.P. = $q=2r^2$

So, $a+6d=2 \quad ...(1)$ $a+7d=2r \quad ...(2)$ $a+12d=2r^2 \quad ...(3)$

3. Use the A.P. property

From (2) $-$ (1): $d=2r-2=2(r-1)$

From (3) $-$ (1): $6d=2r^2-2=2(r^2-1)$ $3d=r^2-1$

Now substitute $d=2(r-1)$: $3\cdot 2(r-1)=r^2-1$ $6r-6=r^2-1$ $r^2-6r+5=0$ $(r-1)(r-5)=0$

Since $q\ne 2$, we cannot have $r=1$ (because then $q=2r^2=2$). Hence, $r=5$

4. Find the A.P. explicitly

Now, $d=2(r-1)=2(5-1)=8$

Using (1): $a+6d=2$ $a+48=2$ $a=-46$

So the A.P. is $-46, -38, -30, \dots$ with $T_n^{(AP)}=a+(n-1)d=-46+8(n-1)=8n-54$

5. Find the 5th term of the G.P.

The G.P. is $2, 10, 50, 250, 1250, \dots$

Thus, $T_5^{(GP)}=2\cdot 5^4=2\cdot 625=1250$

This is the $n^{\text{th}}$ term of the A.P., so $8n-54=1250$ $8n=1304$ $n=163$

6. Compare with options

Thus, $\boxed{n=163}$ So the correct option is C.

7. Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.