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 $a_1 + a_2 + a_3 + a_4 + a_5$?
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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.
Step 1: Understanding the Question:
We are given a geometric progression (GP) of positive integers starting with $a_1 = 3$.
The terms of the progression satisfy the recurrence relation $a_{n+2} - 2a_n = a_{n+1}$ for all $n \ge 1$.
We need to find the common ratio of this progression and calculate the sum of the first 5 terms. Step 2: Key Formula or Approach:
For a geometric progression, the general term is given by:
\[ a_n = a_1 r^{n-1} \]
where $r$ is the common ratio. We substitute this into the given recurrence relation to find the value of $r$. Step 3: Detailed Explanation:
• The given recurrence relation is:
\[ a_{n+2} - a_{n+1} - 2a_n = 0 \]
• Let us substitute $a_n = a_1 r^{n-1}$ into the relation:
\[ a_1 r^{n+1} - a_1 r^n - 2 a_1 r^{n-1} = 0 \]
• Since $a_1 = 3 \neq 0$ and the terms are positive (so $r \neq 0$), we can divide the entire equation by $a_1 r^{n-1}$:
\[ r^2 - r - 2 = 0 \]
• We factor the quadratic equation:
\[ (r - 2)(r + 1) = 0 \]
This gives two possible values for the common ratio: $r = 2$ or $r = -1$.
• Since we are given that the progression consists of positive integers, the common ratio $r$ must be positive.
Therefore, we choose $r = 2$.
• Now, we can find the first 5 terms of the geometric progression:
\[ a_1 = 3 \]
\[ a_2 = 3(2) = 6 \]
\[ a_3 = 6(2) = 12 \]
\[ a_4 = 12(2) = 24 \]
\[ a_5 = 24(2) = 48 \]
• The sum of the first 5 terms is:
\[ S_5 = a_1 + a_2 + a_3 + a_4 + a_5 \]
\[ S_5 = 3 + 6 + 12 + 24 + 48 = 93 \]
Step 4: Final Answer:
The sum of the first 5 terms of the geometric progression is 93.
