Let $$a_n$$ be the n-th term of an arithmetic sequence, where the common difference is a non-zero constant. It is given that $$a_1 + a_3 + a_5 =\dfrac{72}{5} $$ and $$a^2_1= a^2_{11}$$

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Let $$a_n$$ be the n-th term of an arithmetic sequence, where the common difference is a non-zero constant. It is given that $$a_1 + a_3 + a_5 =\dfrac{72}{5}  $$ and $$a^2_1=   a^2_{11}$$

Uniteasy Admin · Accepted Answer

Since $$a_n$$ is an arithmetic sequence, let $$d$$ be the common difference of the sequence 
$$a_n=a_1+(n-1)d$$
$$a_3=a_1+2d$$
$$a_5=a_1+4d$$
$$a_{11}=a_1+10d$$
$$\because a_1+a_3+a_5=\dfrac{72}{5}   $$
$$	herefore  3a_1+6d=\dfrac{72}{5}   $$n
$$\because a^2_1=   a^2_{11} $$
$$a^2_1= (a_1+10d)^2$$
$$a_1+5d =0$$n
Solve the function system (1) and (2) 
$$ d = -\dfrac{8}{5}$$ and $$a_1 =8$$
Therefore 
$$a_n = 8-(n-1)\dfrac{8}{5} $$
$$=\dfrac{40-8n+8}{5}  $$

$$=\dfrac{48}{5}-\dfrac{8}{5}n    $$
 If $$a_n <0$$
$$\dfrac{48}{5}-\dfrac{8}{5}n< 0$$
$$n>6$$
So the first negative term happens when $$n=7$$
Plug in the formula for general terms
$$a_7 = \dfrac{48}{5}-\dfrac{8}{5}	imes 7 $$
$$=\dfrac{48-56}{5}  $$
$$=-\dfrac{8}{5}  $$
 Let $$h$$ be a positive integer such that $$5a_h = h^2$$, that is 
$$5( \dfrac{48}{5}-\dfrac{8}{5}h ) = h^2 $$
$$h^2+8h-48 = 0$$
$$(h+12)(h-4) = 0$$
$$h = 4 $$
So yes, there exists a positive integer  such that $$5a_h = h^2$$

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