$a_1, a_2, a_3, …$ 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







share|cite|improve this question






















  • It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
    – prog_SAHIL
    Aug 26 at 8:12














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







share|cite|improve this question






















  • It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
    – prog_SAHIL
    Aug 26 at 8:12












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







share|cite|improve this question














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









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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 $a_2^2=a_1 cdot a_6$
    – prog_SAHIL
    Aug 26 at 8:12
















  • It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
    – prog_SAHIL
    Aug 26 at 8:12















It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
– prog_SAHIL
Aug 26 at 8:12




It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 cdot a_6$
– prog_SAHIL
Aug 26 at 8:12










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 $n$.






share|cite|improve this answer




















  • Thank you friend. This is detailed.
    – Ernest Barzaga
    Aug 27 at 0:58

















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 $d$.






share|cite|improve this answer



























    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.






    share|cite|improve this answer




















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      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 $n$.






      share|cite|improve this answer




















      • Thank you friend. This is detailed.
        – Ernest Barzaga
        Aug 27 at 0:58














      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 $n$.






      share|cite|improve this answer




















      • Thank you friend. This is detailed.
        – Ernest Barzaga
        Aug 27 at 0:58












      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 $n$.






      share|cite|improve this answer












      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 $n$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Aug 26 at 7:56









      user

      527




      527











      • Thank you friend. This is detailed.
        – Ernest Barzaga
        Aug 27 at 0:58
















      • Thank you friend. This is detailed.
        – Ernest Barzaga
        Aug 27 at 0:58















      Thank you friend. This is detailed.
      – Ernest Barzaga
      Aug 27 at 0:58




      Thank you friend. This is detailed.
      – Ernest Barzaga
      Aug 27 at 0:58










      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 $d$.






      share|cite|improve this answer
























        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 $d$.






        share|cite|improve this answer






















          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 $d$.






          share|cite|improve this answer












          $$a_2 = a_1 r$$
          $$a_6 = a_2 r$$



          $$r = fraca_2a_1=fraca_6a_2$$



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



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







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 26 at 7:52









          Siong Thye Goh

          80.7k1453102




          80.7k1453102




















              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.






              share|cite|improve this answer
























                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.






                share|cite|improve this answer






















                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 26 at 8:07









                  Euclid

                  122




                  122



























                       

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