Question

Let a_n be the general term of an arithmetic sequence, where the common difference is a non-zero constant. It is given that a_1 + a_3 + a_7 = 1 and a^2_2+a^2_3 =a^2_4+a^2_5


Find the general term an of the sequence.

If \dfrac{a_ma_{m+1}}{a_{m+2}} is a term of the sequence, how many possible values of m are there? Explain your answer.

Collected in the board: Arithmetic sequence

Steven Zheng posted 2 months ago


Answer

Since a_n is an arithmetic sequence, let d be the common difference of the sequence

a_n=a_1+(n-1)d

a_2=a_1+d

a_3=a_1+2d

a_4=a_1+3d

a_5=a_1+4d

a_7=a_1+6d

\because a_1 + a_3 + a_7 = 1

\therefore 3a_1+8d = 1
(1)
\because a^2_2+a^2_3 =a^2_4+a^2_5

\therefore (a_1+d)^2+(a_1+2d)^2=(a_1+3d)^2+(a_1+4d)^2

2da_1+d^2+4da_1+4d^2 = 6da_1+9d^2+8da_1+16d^2

2a_1+d+4a_1+4d = 6a_1+9d+8a_1+16d

6a_1+5d = 14a_1+25d

d = -\dfrac{2}{5}a_1
(2)

Solve the equations (1) and (2)

a_1=-5

d = 2

a_n = -5+2(n-1) = 2n-7
(3)

If \dfrac{a_ma_{m+1}}{a_{m+2}} is a term of the sequence,


a_m = 2m-7

a_{m+1} =2(m+1)-7=2m-5

a_{m+2} = 2(m+2)-7 = 2m-3

\dfrac{a_ma_{m+1}}{a_{m+2}}

=\dfrac{(2m-7)(2m-5)}{2m-3}
(1)
=(1-\dfrac{4}{2m-3})(2m-5)

\dfrac{4(2m-5)}{2m-3} must be integers

\dfrac{4(2m-5)}{2m-3}

=4(1-\dfrac{2}{2m-3})

In order for \dfrac{8}{2m-3} to be be integers, m = 1, 2,

when m= 1

\dfrac{a_ma_{m+1}}{a_{m+2}}

=\dfrac{(2-7)(2-5)}{2-3}

=-15 Substitute to a_n = 7-2n, n = 11

When m= 2

\dfrac{a_ma_{m+1}}{a_{m+2}}

=\dfrac{(2m-7)(2m-5)}{2m-3}

=\dfrac{(4-7)(4-5)}{4-3}

=3

Substitute to a_n = 7-2n, n = 2

Therefore, there are 2 possibilities for \dfrac{a_ma_{m+1}}{a_{m+2}} to be a term of the sequence

Steven Zheng posted 1 month ago

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