Geometric Convergent Series

Question

I’ve been trying to understand and solve for this question but I don’t know where to start.

The sum of the first and second terms of a convergent series is $60$. The sum of all the terms is $108$. Determine the common ratio $r$ with $r > 0$, and the first term $a$.

Answer 1

The sum of the first and second terms of a convergent series is 60.

If the first term is called $a$ and the ratio $r$, then from this you have: $$a+ar=60 \tag{1}$$

The sum of all the terms is 108.

You probably know that the sum of a convergent geometric sequence is given by $\tfrac{a}{1-r}$; so you have: $$\frac{a}{1-r} = 108 \tag{2}$$ Now combine equations $(1)$ and $(2)$ and solve for $a$ and $r$; can you take it from there?

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Addition after comment; from $(2)$ you have $a=108(1-r)$ and plugging this into $(1)$ yields: $$108(1-r)+108(1-r)r=60 \iff \ldots \iff 9r^2=4 \iff r = \ldots$$ Now take the positive solution since it was given that $r>0$; then $a$ follows from $(2)$.

Answer 2

Take $a$ to be the initial value of the series and $r$ to be the common ratio. We have two equalities from the information given: $$a + ar = 60$$ $$\frac{a}{1 - r} = 108$$ The second statement comes from the formula for the value of an infinite geometric series. Manipulating the second equation yields: $a = 108(1-r)$. Plugging that expression into the first equation gives: $$108(1-r) + 108r(1-r) = 60$$ simplifying gives: $$108r^2 - 48 = 0 $$ This polynomial has two roots at $r = \pm \frac{2}{3}$. But we know the value of $r$ must be between 0 and 1, so $r= \frac{2}{3}$

To get the value of $a$, plug in for $r$ in the first equation to get $\frac{5}{3}a = 60$, which gives $a = 36$.

Answer 3

In a geometric series with $r>0$ you know that the sum of the first two terms is $a(1+r)=60$ and the sum of all terms is $ a/(1-r)=108.$

Upon dividing we get $ 1-r^2 = 5/9$ which implies $r=2/3.$

Solve $a(1+r)=60$ for $a$ and get $a=36.$