Hello fellow math lovers,
This question is a fairly trivial one from the New Jersey Undergraduate Mathematics competition. I'm having trouble understanding the solution and I'm fairly frustrated because it seems, on the surface, so simple.
Problem:
The sequence $a_1, a_2, a_3, \dotsc$ is a non-constant arithmetic sequence, while both $a_1, a_2, a_6$ and $a_1, a_4, a_n$ are finite geometric series.
Find $n$.Solution:
Let $a_k = a + (k-1)d$ for all $k$.
The fact that $a_1, a_2, a_6$ is a geometric sequence tells us that
$$
(a + d)^2 = a_2^2 = a_1 a_6 = a(a + 5d).
$$
From this it follows easily that $d = 3a$.
Hence, $a_4 = a + 3d = 10a$ and $a_n = a + (n-1)(3a) = (3n-2)a$.
Since $a_1, a_4, a_n$ is also a geometric sequence, we must have
$$
(10a)^2
= a \cdot (3n – 2)a
\longrightarrow
100
= 3n – 2
\longrightarrow
n
= 34.
$$(Original image here.)
My question is in regards to the squaring the second term of the geometric sequence step. How is that equal to $a_1 a_6$ and why was it important to do this to being with? The rest is very clear and easy to follow but the crux move is leaving me at a loss. If there is an easier way to understand and solve this problem please let me know.
-Ernie
Best Answer
$$a_2 = a_1 r$$ $$a_6 = a_2 r$$
$$r = \frac{a_2}{a_1}=\frac{a_6}{a_2}$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.