In a geometric sequence, the second term is $\frac{-4}{5}$ and the sum of the first three terms is $\frac{38}{25}$ . What is the value of the sequence's first term?

for some reason I keep getting a decimal as my answer which I'm pretty sure it can't be because it was a math homework problem:

My Steps

if we call the first term to be the number $n$ and $r$ to be the multiplyer

we have the following:


and then we multiply r on both sides to get the following


and then we substitute nr with $\frac{-4}{5}$

and we get the following:


but then when solving for the number $r$ I do not get a pretty number and I am sure that I am making some sort of mistake. I was wondering what I was doing wrong?

Also I understand that because this is a quadratic, I was wondering which "answer" to choose from the zero's?

  • 5
    $\begingroup$ WA gets a pretty answer $\endgroup$ – user137731 Jun 4 '17 at 3:19
  • 3
    $\begingroup$ To the edited in question: why not try them both and see if you can find an $n$ consistent with each? You might end up with $2$ distinct solutions for $n$ and $r$ or maybe one won't work. Give it a go. $\endgroup$ – user137731 Jun 4 '17 at 3:24

Beginning from $$\frac{-4}{5}+\frac{-4r}{5}+\frac{-4r^2}{5}=\frac{38r}{25}$$ we clear denominators by multiplying by 25:

$$-20 + -20r + -20r^2 = 38r$$ $$20r^2+58r+20=0$$

Now divide by $2$: $$10r^2+29r+10=0$$ This factors as $$(5r + 2 )(2r + 5) = 0$$ which gives $r=-2/5$ and $r = -5/2$ as two solutions. Both are valid so there is no reason to choose one over the other.

Once you've got $r$, it should be straightforward to find $n$.


\begin{align} -\dfrac 45\left( \dfrac 1r + 1 + r \right) &= \dfrac{38}{25} \\ \dfrac 1r + 1 + r &= -\dfrac{19}{10}\\ \dfrac 1r + \dfrac{29}{10} + r &= 0 \\ 10+29r +10r^2 &= 0 \\ (2r+5)(5r+2) &= 0 \\ r &\in \left\{-\dfrac 52, -\dfrac 25 \right\} \end{align} There are two solutions because the sequence and the sequence reversed will both have the same sum and will both be geometric sequences.

So the first term is either $\left(-\dfrac 45\right)\left(-\dfrac 52 \right) = 2$ or $\left(-\dfrac 45\right)\left(-\dfrac 25 \right) = \dfrac{8}{25}$


Following what you wrote Mr. John: $$1+r+r^2=-\frac{19}{10}r.$$ Then $$r^2+2.9r+1=0$$ We can easily find the two roots $r_1 = -2.5$ and $r_2 = -0.4$.

Now, since $nr = -\frac{4}{5}$ then $r_1 = -2.5$ will give $n = 0.32 = \frac{8}{25}$.

And $r_2 = -0.4$ will give $n = 2$.


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