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: \[ n = 163 \]