Find the terms of a geometric progression given two partial sums.

Question

Given an increasing geometric progression such that: $$a_1+a_2+a_3+a_4+a_5=93\qquad\text{ and }\qquad a_2+a_4=30,$$ find $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ without using the sum formula.

What I did: Let $q$ denote the common ratio of the geometric progression, so that $$a_1q+a_1q^3=30\qquad\text{ and }\qquad a_1+a_1q+a_1q^2+a_1q^3+a_1q^4=93.$$

Dividing the two equations yields $$\frac{1+q+q^2+q^3+q^4}{q+q^3}=\frac{93}{30},$$

but I don't know how to solve this equation.

Edit: I edited the question so that $a_1+a_2+a_3+a_4+a_5=93$ and NOT $a_1+a_2+a_3=93$.

Answer 1

Hint:

\begin{align} a_1+a_2+a_3+a_4+a_5&=93\\ a_1+(a_2+a_4)(1+q)&=93\\ a_1+30(1+q)&=93\\ a_1&=63-30q \end{align}

Then:

\begin{align} a_2+a_4&=30\\ a_1q(1+q^2)&=30\\ (63-30q)q(1+q^2)&=30\\ (21-10q)q(1+q^2)&=10\\ \iff 10q^4-21q^3+10q^2-21q+10&=0 \end{align}

Now try with the Rational Root Theorem.

Answer 2

This answer is a thorough explanation of my comment on Mario G's answer; I could not fit a clear explanation into the comments, and I felt like writing something up. His answer is a quick and clean solution, which I like very much. This answer uses the simpler quadratic formula in stead of the rational root theorem. All credit for the middle part of my answer (from equations $(1)$ to $(3)$) goes to Mario G, as it is essentialy the same argument.

---

Suppose $(a_1,a_2,a_3,a_4,a_5)$ is such a geometric progression. It is clear that $a_1\neq0$ as otherwise all terms in the progression equal $0$, in which case their sum cannot equal $93$. Let $a:=a_1$ the base factor and $q:=\tfrac{a_2}{a_1}$ the common ratio of the sequence, and note that $q>0$ because the progression is increasing. Moreover, if $q<1$ then $a_1<0$ for the same reason, but then all terms in the progression are negative, in which case their sum again cannot equal $93$. Therefore $q>1$.

We can rewrite the given equations as \begin{eqnarray} a+aq+aq^2+aq^3+aq^4&=&\ 93,\tag{1}\\ aq+aq^3&=&\ 30.\tag{2} \end{eqnarray} We can take out the left hand side of $(2)$ from the left hand side of $(1)$ as follows: \begin{eqnarray*} a+aq+aq^2+aq^3+aq^4 &=&\ a+(aq+aq^3)+(aq^2+aq^4)\\ &=&\ a+(aq+aq^3)+q(aq+aq^3)\\ &=&\ a+(aq+aq^3)(1+q). \end{eqnarray*} Then plugging in the numbers from $(1)$ and $(2)$ shows us that $93=a+30(1+q)$, or equivalently $$a=93-30(1+q)=63-30q.$$ Substituting this back into equation $(2)$ yields $$30=(63-30q)q+(63-30q)q^3=63q-30q^2+63q^3-30q^4,$$ and now rearranging the terms and dividing by $3$ we see that $q$ satisfies the quartic equation $$10q^4-21q^3+10q^2-21q+10=0.\tag{3}$$ Solving a quartic is generally not fun, but in this case it is${}^☺$ because it's a palindromic polynomial! This allows us to use a neat trick; we can divide equation $(3)$ by $q^2$ (because we already saw $q>0$) to get a quadratic equation in terms of $q+q^{-1}$: \begin{eqnarray*} q^{-2}(10q^4-21q^3+10q^2-21q+10) &=&10q^2-21q+10-21q^{-1}+10q^{-2}\\ &=&10(q^2+q^{-2})-21(q+q^{-1})+10\\ &=&10(q+q^{-1})^2-21(q+q^{-1})-10. \end{eqnarray*} Be careful to note what happens with the squares in the last step! Now we see that $q+q^{-1}$ is a root of the quadratic polynomial $10x^2-21x-10$ with discriminant $29^2$, so by the quadratic formula $$q+q^{-1}=\frac{21}{20}\pm\frac{1}{20}\cdot29,$$ where $q+q^{-1}>0$ because $q>0$, so we must have the '$+$'-sign with $q+q^{-1}=\tfrac{21}{20}+\tfrac{29}{20}=\tfrac52$. Multiplying through by $q$ and rearranging terms shows that $q$ is a root of the quadratic equation $$x^2-\tfrac52x+1=0,$$ with discriminant $\left(\tfrac{3}{2}\right)^2$, so by the quadratic formula $q=\tfrac54\pm\tfrac12\cdot\tfrac32$. Because $q>1$ we get $$q=\tfrac54+\tfrac12\cdot\tfrac32=2\qquad\text{ and }\qquad a=63-30q=3,$$ so the unique such geometric progression is $$(a_1,a_2,a_3,a_4,a_5)=(3,6,12,24,48).$$

---

☺: This may depend on your definition of fun.