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Given that, the square of the $7^{\text{th}}$ term is equal to the products of the $3^{\text{rd}}$ and $17^{\text{th}}$ terms.

The $n^{\text{th}}$ term of an arithmetic progression $= a + (n-1)d$

Where, $a = $ first term, and $d = $common difference.

Now, ${(a+6d)}^{2} = (a + 2d)(a + 16d)$

$\Rightarrow a^{2} + 36d^{2} + 12ad = a^{2} + 16ad + 2ad + 32d^{2}$

$\Rightarrow 4d^{2} = 6ad$

$\Rightarrow 2d = 3a$

$\Rightarrow 3a = 2d$

$\Rightarrow \frac{a}{d} = \frac{2}{3}$

$\therefore$ The ratio of the first term to the common difference is $2:3.$

Correct Answer $:\text{A}$

The $n^{\text{th}}$ term of an arithmetic progression $= a + (n-1)d$

Where, $a = $ first term, and $d = $common difference.

Now, ${(a+6d)}^{2} = (a + 2d)(a + 16d)$

$\Rightarrow a^{2} + 36d^{2} + 12ad = a^{2} + 16ad + 2ad + 32d^{2}$

$\Rightarrow 4d^{2} = 6ad$

$\Rightarrow 2d = 3a$

$\Rightarrow 3a = 2d$

$\Rightarrow \frac{a}{d} = \frac{2}{3}$

$\therefore$ The ratio of the first term to the common difference is $2:3.$

Correct Answer $:\text{A}$