What Is the Common Difference in This Arithmetic Progression?

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Homework Help Overview

The discussion revolves around a finite arithmetic progression where the sum of the first n terms, \( S_n \), is influenced by changes in the common difference, \( d \). Participants are tasked with finding the value of \( d \) given specific conditions regarding how changes in \( d \) affect \( S_n \).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants attempt to set up equations based on the relationship between \( S_n \) and changes in \( d \). There is a focus on interpreting the problem statement correctly, particularly regarding the conditions under which \( S_n \) increases.

Discussion Status

Some participants express confusion regarding the interpretation of the problem statement, leading to differing attempts at solving for \( d \). One participant suggests a possible misinterpretation of the conditions, while others confirm a derived value for \( d \) after clarifying the problem.

Contextual Notes

There is an emphasis on ensuring that the first term of the progression remains constant while varying \( d \). Participants note that \( S_n \) must remain positive, which adds constraints to their reasoning.

kaspis245
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Homework Statement


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##.

Homework Equations


##S_n=\frac{2a_1+d(n-1)}{2}\cdot n##

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?
 
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kaspis245 said:

Homework Statement


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 2 times, 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##.

Homework Equations


##S_n=\frac{2a_1+d(n-1)}{2}\cdot n##

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?
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.
 
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}##.
 
kaspis245 said:
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}##.
So do I.
 

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