$a_1, a_2, a_3, â¦$ is a non-constant arithmetic sequence, while both $a_1, a_2, a_6$ and $a_1, a_4$, an are finite geometric sequences. Find $n$
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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
sequences-and-series proof-explanation
add a comment |Â
up vote
1
down vote
favorite
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
sequences-and-series proof-explanation
It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
sequences-and-series proof-explanation
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
sequences-and-series proof-explanation
edited Aug 26 at 11:07
Jneven
641320
641320
asked Aug 26 at 7:46
Ernest Barzaga
133
133
It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12
add a comment |Â
It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12
It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12
It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
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 $n$.
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
add a comment |Â
up vote
2
down vote
$$a_2 = a_1 r$$
$$a_6 = a_2 r$$
$$r = fraca_2a_1=fraca_6a_2$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
add a comment |Â
up vote
0
down vote
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.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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 $n$.
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
add a comment |Â
up vote
0
down vote
accepted
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 $n$.
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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 $n$.
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 $n$.
answered Aug 26 at 7:56
user
527
527
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
add a comment |Â
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
Thank you friend. This is detailed.
â Ernest Barzaga
Aug 27 at 0:58
add a comment |Â
up vote
2
down vote
$$a_2 = a_1 r$$
$$a_6 = a_2 r$$
$$r = fraca_2a_1=fraca_6a_2$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
add a comment |Â
up vote
2
down vote
$$a_2 = a_1 r$$
$$a_6 = a_2 r$$
$$r = fraca_2a_1=fraca_6a_2$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$a_2 = a_1 r$$
$$a_6 = a_2 r$$
$$r = fraca_2a_1=fraca_6a_2$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
$$a_2 = a_1 r$$
$$a_6 = a_2 r$$
$$r = fraca_2a_1=fraca_6a_2$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
answered Aug 26 at 7:52
Siong Thye Goh
80.7k1453102
80.7k1453102
add a comment |Â
add a comment |Â
up vote
0
down vote
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.
add a comment |Â
up vote
0
down vote
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.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
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.
answered Aug 26 at 8:07
Euclid
122
122
add a comment |Â
add a comment |Â
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It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
â prog_SAHIL
Aug 26 at 8:12