The Sequence Sum

Let the first term be $a$ and the common ratio be $r$ . We have:

$S_{10} = 4 \implies \color{#D61F06}{\dfrac{a(r^{10}-1)}{r-1} = 4}\\ S_{11-30} = S_{30} - S_{10} = \dfrac{a(r^{30}-1)}{r-1} - 4 = 48\\ \color{#3D99F6}{\dfrac{a(r^{30}-1)}{r-1} =52}$

Now, we can use the identity $x^3-1=(x-1)(x^2+x+1)$ :

$r^{30}-1=(r^{10})^3 - 1 = (r^{10}-1)(r^{20}+r^{10}+1)$

Substitute this in:

$\dfrac{a(r^{10}-1)(r^{20}+r^{10}+1)}{r-1} = 52\\ \left(\color{#D61F06}{\dfrac{a(r^{10}-1)}{r-1}}\right)(r^{20}+r^{10}+1)=52\\ \color{#D61F06}{4}(r^{20}+r^{10}+1)=52\\ r^{20}+r^{10}+1=13\\ r^{20}+r^{10}-12=0\\ (r^{10}+4)(r^{10}-3)=0\\ r^{10}=-4,\;r^{10}=3$

The left factor does not give a real value of $r$ , so we reject it. The value of $r=\pm\sqrt[10]{3}$

We are now looking for this:

$S_{31-60} = S_{60} - S_{30}\\ =\dfrac{a(r^{60}-1)}{r-1} - \color{#3D99F6}{\dfrac{a(r^{30}-1)}{r-1}}\\ =\dfrac{a(r^{30}-1)(r^{30}+1)}{r-1} - \color{#3D99F6}{52}\\ =\left(\color{#3D99F6}{\dfrac{a(r^{30}-1)}{r-1}}\right)(r^{30}+1)-52\\ =\color{#3D99F6}{52}\left((r^{10})^3+1\right)-52\\ =52(3^3+1-1)\\ =52(27)\\ =\boxed{1404}$

The information that we are given is:

$\displaystyle \sum_{n=1}^{10}a \times r^n = 4$

$\displaystyle \sum_{n=11}^{30}a \times r^n = 48$

What we need to solve is:

$\displaystyle \sum_{n=31}^{60}a \times r^n = \boxed{?}$

In order to solve the sum, we do not need to know what $a$ or $r$ are equal to. In any geometric series $a \times r^n$ , the series

$\displaystyle \sum_{n=1}^{10}a \times r^n, \displaystyle \sum_{n=11}^{20}a \times r^n, \displaystyle \sum_{n=21}^{30}a \times r^n, \ldots$

is also geometric. In other words, the series {the sum of the first 10 terms, the sum of the next 10 terms, the sum of the next 10 terms . . .} is geometric.

Let us use the following sequence notation to represent each sum as a term in the sequence $A_n$ :

$t_x = \displaystyle \sum_{n = 10x-9}^{10x}a \times r^n$

We know that

$t_1 = \displaystyle \sum_{n=1}^{10}a \times r^n = 4$

and we know that

$t_2 + t_3 = (\displaystyle \sum_{n=11}^{20}a \times r^n) + (\displaystyle \sum_{n=21}^{30}a \times r^n) = \displaystyle \sum_{n=11}^{30}a \times r^n = 48$

Since the sequence is geometric, we can represent it as

$A_n = (4, 4r, 4r^2, \ldots)$

So we can use this information to solve for r.

$t_2 + t_3 = 48$

$4r + 4r^2 = 48$

$4r^2 + 4r - 48 = 0$

$r^2 + r - 12 = 0$

$(r+4)(r-3) = 0$

$r = -4, 3$

We know $r$ cannot be negative, because all of the terms are positive.

So, $r = 3$

Now we can complete our sequence and solve the problem.

$A_n = (4, 12, 36, 108, 324, 972, \ldots)$

$\displaystyle \sum_{n=31}^{60}a \times r^n = \displaystyle \sum_{n=31}^{40}a \times r^n, \displaystyle \sum_{n=41}^{50}a \times r^n, \displaystyle \sum_{n=51}^{60}a \times r^n$

$= t_4 + t_5 + t_6$

$= 108 + 324 + 972$

$= \boxed {1404}$

The formula for the sum of a Geometric Sequence is:

$\sum_{k=1}^{n} a\times {r}^{k} = \frac {a({r}^{n}-1)}{r-1}$

[ $n$ represents the number of terms, $r$ represents the common ratio, and $a$ represents the first term]

We do not need to know $a$ or $r$ , so we can convert the given sum formula into what was given: $\frac {a({r}^{10}-1)}{r-1}=4$ $\frac {a({r}^{30}-1)}{r-1}=48+4=52$

We can divide the bottom equation by the top equation to give:

$\frac {(\frac {a({r}^{30}-1)}{r-1})}{(\frac {a({r}^{10}-1)}{r-1})}=13 \rightarrow {r}^{20}+{r}^{10}+1=13$

Simplifying:

${r}^{20}+{r}^{10}=12 \rightarrow {r}^{10}\times ({r}^{10}+1)=12$

We can solve for ${r}^{10}$ to get that ${r}^{10}=3,-4$ . ${r}^{10}$ cannot equal -4, so it must equal 3. Now, we can use the formula for the sum of the first 60 terms of a geometric sequence:

$\frac {a({r}^{60}-1)}{r-1}=x$

We can divide this by $\frac {a({r}^{30}-1)}{r-1}=52$ to get ${r}^{30}+1=\frac{x}{52}$ .Then, we can solve for $x$ by plugging in ${r}^{10}=3$ into the equation.

${r}^{30}+1=\frac{x}{52} \quad \rightarrow \quad ({{r}^{10}})^{3}+1=\frac{x}{52}\quad \rightarrow \quad 28=\frac{x}{52} \quad \rightarrow \quad x=1456$

We're not done here yet! Subtract $\frac {a({r}^{30}-1)}{r-1}=52$ from 1456 to get your final answer:

$=\boxed {1404}$