# Arithmetic progression

1. May 4, 2016

### kaspis245

1. The problem statement, all variables and given/known data
A finite arithmetic progression is given such that $S_n>0$ and $d>0$. If the first member of the progression remains the same but $d$ increases by 2, then $S_n$ increases 3 times. If the first member of the progression remains the same but $d$ increases 4 times, then $S_n$ increases 5 times. Find $d$.

2. Relevant equations
$S_n=\frac{2a_1+d(n-1)}{2}\cdot n$

3. The attempt at a solution
$\frac{2a_1+2d(n-1)}{2}\cdot n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 3$
$\frac{2a_1+4d(n-1)}{2}\cdot n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 5$

When I try to solve this I get $a_1=0$ which is clearly not possible. Can somebody explain me what am I doing wrong?

Last edited: May 4, 2016
2. May 4, 2016

### Samy_A

I also find $a_1=0$.

Are you sure about the way you interpret the exercise?

Maybe it should be: "If $d$ is increased by 2, then $S_n$ is multiplied by 3", and similarly for the second one.

3. May 4, 2016

### kaspis245

Sorry for the confusion. I've updated my problem statement. The first part of the problem is: "If the first member of the progression remains the same but $d$ increases by 2, then $S_n$ increases 3 times." Everything else is the same.

Now I got $d=\frac{4}{3}$.

4. May 4, 2016

So do I.