d>0
a2+a5=1
a1a3=7
how many terms are negative?
1)3 2)4 3)2 4)1
To expand upon MINOANMAN's comment, since this is an arithmetic sequence we know that all terms are in the form: $\displaystyle a_n = a + nd$. So your relations give
$\displaystyle (a + 2d) + (a + 5d) = 1$
and
$\displaystyle (a + d)(a + 3d) = 7$
I'd solve the first equation for a, then plug that into the second equation. That will give you a quadratic equation in d.
-Dan
Hint: if $\displaystyle a_1 \times a_3 = 7$ since 7 is prime that means these terms must be either -7 and -1 or +1 and +7. And if $\displaystyle a_2 + a_5=1$ that means that one of these terms must be negative, or perhaps 0. Given these constraints, and given that as an arithmetic progression the difference between successive terms is constant and positive, there is only one sequence that fits. Work on it a bit and post back with what you get for an answer.