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

An arithmetic progression (AP) is a sequence where each term increases by the same fixed amount called the common difference.

The nth term of an AP = a + (n-1)d where a = first term, d = common difference, n = term number

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Let's identify our terms:

First term = a

Common difference = d (given as positive)

3rd term = a + (3-1)d = a + 2d

7th term = a + (7-1)d = a + 6d

17th term = a + (17-1)d = a + 16d

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The problem states: (7th term)² = (3rd term) × (17th term)

Substituting our expressions:

$(a + 6d)^2 = (a + 2d)(a + 16d)$

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Left side: $(a + 6d)^2$

Using $(x + y)^2 = x^2 + 2xy + y^2$:

$(a + 6d)^2 = a^2 + 12ad + 36d^2$

Right side: $(a + 2d)(a + 16d)$

Using FOIL method:

First: $a \times a = a^2$

Outer: $a \times 16d = 16ad$

Inner: $2d \times a = 2ad$

Last: $2d \times 16d = 32d^2$

So: $(a + 2d)(a + 16d) = a^2 + 18ad + 32d^2$

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Setting left side = right side:

$a^2 + 12ad + 36d^2 = a^2 + 18ad + 32d^2$

Subtracting $a^2$ from both sides:

$12ad + 36d^2 = 18ad + 32d^2$

$36d^2 - 32d^2 = 18ad - 12ad$

$4d^2 = 6ad$

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We want to find $\tfrac{a}{d}$, so let's divide both sides by d.

Since $d > 0$ (given that common difference is positive), we can safely divide by d.

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

$4d = 6a$

Dividing both sides by $6d$:

$\dfrac{4d}{6d} = \dfrac{6a}{6d}$

$\dfrac{4}{6} = \dfrac{a}{d}$

$\dfrac{a}{d} = \dfrac{2}{3}$

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The ratio of the first term to the common difference is $\dfrac{2}{3}$.

This means that if the common difference is 3, then the first term would be 2, giving us an AP like: 2, 5, 8, 11, 14, 17, 20...