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May 3rd, 2016, 10:37 PM   #1
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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?
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May 4th, 2016, 03:51 AM   #2
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There is no solution. Was the problem originally in English? If not, what was the original wording in full?
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May 4th, 2016, 07:33 AM   #3
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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.
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May 14th, 2016, 10:06 AM   #4
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You may find hints regarding Arithmetic Progression on Welcome to prep4paper.com
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