Question

Let $\{a_n\}_{n \ge 1}$ be a sequence of real numbers such that $a_1 = 1, a_2 = 7$, and $a_{n+1} = \dfrac{a_n + a_{n-1}}{2}$, $n \ge 2$. Assuming that $\lim_{n \to \infty} a_n$ exists, the value of $\lim_{n \to \infty} a_n$ is
 

• A. $\dfrac{19}{4}$
• B. $\dfrac{9}{2}$
• C. $5$
• D. $\dfrac{21}{4}$

Hint:

For linear recurrences with constant coefficients, express the solution in terms of characteristic roots to find the limit easily.

Answer 1

The Correct Option is C

Step 1: Given recursive relation. 
We have $a_{n+1} = \dfrac{a_n + a_{n-1}}{2}$ with $a_1 = 1$ and $a_2 = 7$. 
 

Step 2: Assume the limit exists. 
Let $\lim_{n \to \infty} a_n = L$. Taking the limit on both sides, \[ L = \frac{L + L}{2} \Rightarrow L = L. \] So this gives no direct value; we must analyze the pattern. 
 

Step 3: Compute next few terms to identify pattern. 
\[ a_3 = \frac{a_2 + a_1}{2} = \frac{7 + 1}{2} = 4, a_4 = \frac{a_3 + a_2}{2} = \frac{4 + 7}{2} = 5.5, a_5 = \frac{a_4 + a_3}{2} = \frac{5.5 + 4}{2} = 4.75. \] The sequence alternates around a point near 5. 
 

Step 4: Find exact limit using linear recurrence method. 
The characteristic equation is $2r^2 - r - 1 = 0$. \[ r = 1, -\frac{1}{2}. \] Hence, \[ a_n = A(1)^n + B\left(-\frac{1}{2}\right)^n = A + B\left(-\frac{1}{2}\right)^n. \] Using initial conditions: $a_1 = A - \dfrac{1}{2}B = 1$, $a_2 = A + \dfrac{1}{4}B = 7$. Solving, $B = 4$, $A = 5$.

Step 5: Take the limit. 
As $n \to \infty$, $\left(-\dfrac{1}{2}\right)^n \to 0$, so $\lim a_n = A = 5$. 
 

Step 6: Conclusion. 
Hence, $\boxed{\lim_{n \to \infty} a_n = 5}$.