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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

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  • $\begingroup$ It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 \cdot a_6$ $\endgroup$ – prog_SAHIL Aug 26 '18 at 8:12
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We know $a_1=a, $ and $a_2=a+d$. Hence $a^2=(a+d)^2$

If you imagine an 'extract' from a geometric sequence $$\dots,ar^{n-1},ar^n,ar^{n+1},\dots$$ you can see that for any three terms in a geometric sequence, the square of the middle term is equal to the product of the two adjacent terms.

This is important in the context of your solution because it establishes the value of $d$ in relation to $a$, since $(a+d)^2=a(a+5d)\implies d=3a$. Thus we know that $(a_1,a_4,a_n)=(a,10a,a_n)$ and thus $a_n=100a$. From here, we can work out the value of $n$.

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  • $\begingroup$ Thank you friend. This is detailed. $\endgroup$ – Maths2020 Aug 27 '18 at 0:58
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$$a_2 = a_1 r$$ $$a_6 = a_2 r$$

$$r = \frac{a_2}{a_1}=\frac{a_6}{a_2}$$

Hence we have $$a_2^2=a_1a_6$$

It enables us to find a relationship between $a$ and $d$.

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Given three consecutive members of a geometric sequence, the square of the middle is equal to the product of the other two (Prove this by writing all three values in terms of the first member).

This fact was used to find a connection between a1 and d.

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