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 May 3rd, 2016, 10:37 PM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 Arithmetic progression A finite arithmetic progression is given such that $\displaystyle S_n>0$ and $\displaystyle d>0$. If the first member of the progression remains the same but $\displaystyle d$ increases 2 times, then $\displaystyle S_n$ increases 3 times. If the first member of the progression remains the same but $\displaystyle d$ increases 4 times, then $\displaystyle S_n$ increases 5 times. Find $\displaystyle d$. My try: $\displaystyle S_n=\frac{2a_1+d(n-1)}{2}\cdot n$ $\displaystyle \frac{2a_1+2d(n-1)}{2}\cdot n=S_n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 3$ $\displaystyle \frac{2a_1+4d(n-1)}{2}\cdot n=S_n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 5$ When I try to solve this I get $\displaystyle a_1=0$ which is clearly not possible. Can somebody explain me what am I doing wrong?
 May 4th, 2016, 03:51 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 There is no solution. Was the problem originally in English? If not, what was the original wording in full?
 May 4th, 2016, 07:33 AM #3 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 I am sorry. I made a mistake in the problem statement. The first part of the problem is: "If the first member of the progression remains the same but $\displaystyle d$ increases by 2, then $\displaystyle S_n$ increases 3 times." Everything else is the same. I solved the equations and got $\displaystyle d=\frac{4}{3}$. Again, I am deeply sorry for the confusion.
 May 14th, 2016, 10:06 AM #4 Newbie   Joined: May 2016 From: India Posts: 2 Thanks: 1 You may find hints regarding Arithmetic Progression on Welcome to prep4paper.com

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