A polynomial $p(x)$ with real coefficients is said to be [i]almeriense[/i] if it is of the form:

$$ 
p(x) = x^3+ax^2+bx+a
$$

And its three roots are positive real numbers in arithmetic progression. Find all [i]almeriense[/i] polynomials such that $p\left(\frac{7}{4}\right) = 0$
1. Let the roots of the polynomial \( p(x) \) be \( \{c, c+d, c+2d\} \) where \( c \) and \( d \) are positive real numbers. Since the roots are in arithmetic progression, we can use Vieta's formulas to express the coefficients of the polynomial in terms of \( c \) and \( d \).

2. By Vieta's formulas, the sum of the roots is equal to the coefficient of \( x^2 \) with a negative sign:
   \[
   c + (c + d) + (c + 2d) = 3c + 3d = -a
   \]
   The product of the roots taken one at a time is equal to the constant term:
   \[
   c(c + d)(c + 2d) = -a
   \]

3. Given that \( p\left(\frac{7}{4}\right) = 0 \), one of the roots must be \( \frac{7}{4} \). We consider two cases: \( c = \frac{7}{4} \) and \( c + d = \frac{7}{4} \).

4. **Case 1: \( c = \frac{7}{4} \)**
   \[
   3c + 3d = -a \implies 3\left(\frac{7}{4}\right) + 3d = -a \implies \frac{21}{4} + 3d = -a
   \]
   \[
   c(c + d)(c + 2d) = -a \implies \left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right)\left(\frac{7}{4} + 2d\right) = -a
   \]

5. **Case 2: \( c + d = \frac{7}{4} \)**
   \[
   c + (c + d) + (c + 2d) = 3c + 3d = -a \implies 3c + 3\left(\frac{7}{4} - c\right) = -a \implies \frac{21}{4} = -a
   \]
   \[
   c(c + d)(c + 2d) = -a \implies c\left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right) = -a
   \]

6. Solving for \( d \) in both cases:
   - For \( c = \frac{7}{4} \):
     \[
     3\left(\frac{7}{4}\right) + 3d = -a \implies \frac{21}{4} + 3d = -a
     \]
     \[
     \left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right)\left(\frac{7}{4} + 2d\right) = -a
     \]
     Solving these equations, we find \( d = \frac{1}{6} \) and \( c = \frac{3}{2} \).

   - For \( c + d = \frac{7}{4} \):
     \[
     c + d = \frac{7}{4} \implies c = \frac{7}{4} - d
     \]
     \[
     c\left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right) = -a
     \]
     Solving these equations, we find \( d = \frac{1}{96} \) and \( c = \frac{12}{7} \).

7. Constructing the polynomials for each case:
   - For \( c = \frac{3}{2} \) and \( d = \frac{1}{6} \):
     \[
     p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4}
     \]

   - For \( c = \frac{12}{7} \) and \( d = \frac{1}{96} \):
     \[
     p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}
     \]

The final answer is \( \boxed{ p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4} } \) or \( p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56} \).