Question

Let the first three terms $2$, $p$, and $q$, with $q \neq 2$, of a G.P. be respectively the $7^\text{th}$, $8^\text{th}$, and $13^\text{th}$ terms of an A.P. If the $5^\text{th}$ term of the G.P. is the $n^\text{th}$ term of the A.P., then $n$ is equal to:

• A. 151
• B. 169
• C. 177
• D. 163

Answer 1

The Correct Option is D

Define the Terms of the G.P.: 
Let the first term of the G.P. be $a = 2$ and the common ratio be $r$. Then the terms of the G.P. are:
$$2, \, 2r, \, 2r^2, \ldots$$ 
Thus: $2$ is the $7$th term of the A.P., $2r$ is the $8$th term of the A.P., $2r^2$ is the $13$th term of the A.P. 

Set Up the A.P. Terms: 
Let the first term of the A.P. be $A$ and the common difference of the A.P. be $d$. 

Then: $$A + 6d = 2, \quad A + 7d = 2r, \quad A + 12d = 2r^2$$ 
Solving the Equations: 
Using the equations, subtract the first equation from the second and the second from the third:
$$A + 7d - (A + 6d) = 2r - 2 \implies d = 2r - 2$$
$$A + 12d - (A + 7d) = 2r^2 - 2r \implies 5d = 2r^2 - 2r$$ 
Substitute $d = 2r - 2$ into $5d = 2r^2 - 2r$:
$$5(2r - 2) = 2r^2 - 2r \implies 10r - 10 = 2r^2 - 2r$$ 

Rearranging gives: $$2r^2 - 12r + 10 = 0 \implies r^2 - 6r + 5 = 0$$ 
Solving this quadratic equation for $r$: 
$$r = 5 \quad \text{or} \quad r = 1$$ Since $q \neq 2$, we discard $r = 1$, so $r = 5$. 

Calculate $n$: 
The $5$th term of the G.P. is $2r^4 = 2 \cdot 5^4 = 2 \cdot 625 = 1250$. 
We are given that the $n$th term of the A.P. is $1250$: $$A + (n - 1)d = 1250$$
Substitute $A = 2 - 6d$ and $d = 8$ (from previous calculations): $$2 - 6 \times 8 + (n - 1) \times 8 = 1250$$ 

Simplifying, we get: $$n = 163$$