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 $$a^2_2+a^2_3 =a^2_4+a^2_5$$

Question

Uniteasy Admin · Accepted Answer

Since $$a_n$$ is an arithmetic sequence, let $$d$$ be the common difference of the sequence 
$$a_n=a_1+(n-1)d$$
$$a_2=a_1+d$$
$$a_3=a_1+2d$$
$$a_4=a_1+3d$$
$$a_5=a_1+4d$$
$$a_7=a_1+6d$$

$$\because a_1 + a_3 + a_7 = 1$$
$$	herefore   3a_1+8d = 1$$n
$$\because  a^2_2+a^2_3 =a^2_4+a^2_5$$
$$	herefore   (a_1+d)^2+(a_1+2d)^2=(a_1+3d)^2+(a_1+4d)^2$$
$$2da_1+d^2+4da_1+4d^2 = 6da_1+9d^2+8da_1+16d^2$$
$$2a_1+d+4a_1+4d = 6a_1+9d+8a_1+16d$$
$$6a_1+5d = 14a_1+25d$$

$$d = -\dfrac{2}{5}a_1  $$n
Solve the equations (1) and (2) 
$$a_1=-5  $$
$$d = 2  $$
$$a_n = -5+2(n-1) = 2n-7$$n If $$\dfrac{a_ma_{m+1}}{a_{m+2}} $$ is a term of the sequence,

$$a_m = 2m-7$$
$$a_{m+1} =2(m+1)-7=2m-5$$
$$a_{m+2} = 2(m+2)-7 = 2m-3$$
$$\dfrac{a_ma_{m+1}}{a_{m+2}} $$
$$=\dfrac{(2m-7)(2m-5)}{2m-3}  $$n
$$=(1-\dfrac{4}{2m-3})(2m-5)  $$
$$\dfrac{4(2m-5)}{2m-3}  $$ 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

Question

Answer