Arithmetic progression

  • Thread starter kaspis245
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  • #1
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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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Answers and Replies

  • #2
Samy_A
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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 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.
 
  • #3
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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
Samy_A
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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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