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

Answer: (D)

163

Answer 2

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: $n = 163$