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: \[ n = 163 \]

Answer 2

Step 1: Set up AP and GP relations.
AP terms: 7th = 2 ⇒ A + 6d = 2; 8th = p ⇒ A + 7d = p; 13th = q ⇒ A + 12d = q.
GP terms: 1st = 2, 2nd = p, 3rd = q ⇒ common ratio r = p/2 and q = 2r^2 = 2(p/2)^2 = p^2/2.

Step 2: Relate p and q via AP.
From AP: d = (A+7d) − (A+6d) = p − 2; also q − 2 = 6d ⇒ d = (q − 2)/6.
Equate d: p − 2 = (q − 2)/6 ⇒ q = 6p − 10.

Step 3: Solve for p, q.
From GP and AP: q = p^2/2 and q = 6p − 10 ⇒ p^2/2 = 6p − 10 ⇒ p^2 − 12p + 20 = 0.
Discriminant = 144 − 80 = 64 ⇒ p = (12 ± 8)/2 ∈ {10, 2}. Given q ≠ 2 ⇒ p ≠ 2 ⇒ p = 10.
Then q = p^2/2 = 100/2 = 50 (also 6p − 10 = 60 − 10 = 50).

Step 4: Find A, d.
d = p − 2 = 8; A + 6d = 2 ⇒ A = 2 − 48 = −46.

Step 5: Match 5th GP term to n-th AP term.
5th GP term = 2r^4 = 2(10/2)^4 = 2·5^4 = 1250.
A + (n − 1)d = 1250 ⇒ −46 + 8(n − 1) = 1250 ⇒ 8(n − 1) = 1296 ⇒ n − 1 = 162 ⇒ n = 163.

Final Answer:
\[ \boxed{163} \]