# Arithmetic progression

## 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?

Last edited:

Samy_A
Homework Helper

## 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}##.

Samy_A