1. ## Arithmetic progression

d>0
a2+a5=1
a1a3=7
how many terms are negative?

1)3 2)4 3)2 4)1

2. ## Re: Arithmetic progression

solve the system to find a=-7 and d = 3 ..it is easy...

3. ## Re: Arithmetic progression

Originally Posted by parmis
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: $a_n = a + nd$. So your relations give
$(a + 2d) + (a + 5d) = 1$

and
$(a + d)(a + 3d) = 7$

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

4. ## Re: Arithmetic progression

Hint: if $a_1 \times a_3 = 7$ since 7 is prime that means these terms must be either -7 and -1 or +1 and +7. And if $a_2 + a_5=1$ 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.

6. ## Re: Arithmetic progression

Ibdutt : Pls verify your calculations

7. ## Re: Arithmetic progression

Originally Posted by MINOANMAN
Ibdutt : Pls verify your calculations
I am sorry for i mixed up in fact i should have got a = -7 and d = 3
Thanks for pointing out