If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Correct Option: A

Let the first term of the arithmetic progression be $a$ and the common difference be $d$ (where $d>0$).

The $n$-th term of an AP is given by:
$t_n=a+(n-1)d$

According to the problem:
${\left(t_7\right)}^2=t_3\times t_{17}$

Substitute the terms in terms of $a$ and $d$:
${(a+6d)}^2=(a+2d)(a+16d)$

Expand both sides:
$a^2+12ad+36d^2=a^2+18ad+32d^2$

Subtract $a^2$ from both sides and rearrange terms:
$36d^2-32d^2=18ad-12ad$
$4d^2=6ad$

Since the common difference $d$ is positive ($d>0$), we can divide both sides by $2d$:
$2d=3a$

Therefore, the ratio of the first term to the common difference is:
$\frac{a}{d}=\frac{2}{3}$

This gives the ratio as 2 : 3.