Question

Let $a_1, a_2, a_3, \dots$ be a geometric progression of positive integers such that $a_1 = 3$ and $a_{n+2} - 2a_n = a_{n+1}$ for all positive integers $n$. What is the value of $a_1 + a_2 + a_3 + a_4 + a_5$?

• A. 93
• B. 120
• C. 255
• D. 99

Hint:

For any GP relation $a_{n+2} + p a_{n+1} + q a_n = 0$, you can directly write the characteristic equation $r^2 + p r + q = 0$ to find the common ratio $r$.
This completely bypasses the need to write out the individual terms.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a geometric progression (GP) that also satisfies a linear recurrence relation.
Key Formula or Approach:
For a GP, the general term is \( a_{n} = a_{1}r^{n-1} \).
The sum of the first \( n \) terms is \( S_{n} = \frac{a_{1}(r^{n}-1)}{r-1} \).

Step 2: Detailed Explanation:
1. Find the common ratio \( r \):
Substitute the GP terms into the recurrence relation \( a_{n+2} - a_{n+1} - 2a_{n} = 0 \):
\[ a_{1}r^{n+1} - a_{1}r^{n} - 2a_{1}r^{n-1} = 0 \]
Divide by \( a_{1}r^{n-1} \) (as \( a_{1}, r > 0 \)):
\[ r^{2} - r - 2 = 0 \implies (r-2)(r+1) = 0 \]
Since the terms are positive integers, \( r \) must be positive, so \( r = 2 \).
2. Calculate the sum of the first 5 terms:
Using \( a_{1} = 3 \) and \( r = 2 \):
\[ S_{5} = \frac{3(2^{5}-1)}{2-1} = 3(32-1) = 3 \times 31 = 93 \]

Step 3: Final Answer:
The value of the sum is 93.