Question

30 points!!
What is the sum of the first six terms of the series?
48 - 12 + 3 - 0.75 +...

Answer 1

Answer:

The sum of the first six terms is 38.39

Step-by-step explanation:

This is a geometric sequence since the common difference between each term is $-\frac{1}{4}$

Thus, $r=-\frac{1}{4}$

To find the sum of first six terms, we need to find the fifth and sixth term of the sequence.

To find the fifth term:

The general form of geometric sequence is $a_{n}=a_{1} \cdot r^{n-1}$

To find the fifth term, substitute $n=5$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{5} &=(48) \cdot\left(-\frac{1}{4}\right)^{5-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{4} \\&=(48)\left(\frac{1}{256}\right) \\a_{5} &=0.1875\end{aligned}$

To find the sixth term, substitute $n=6$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{6} &=(48) \cdot\left(-\frac{1}{4}\right)^{6-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{5} \\&=(48)\left(-\frac{1}{1024}\right) \\a_{5} &=-0.046875\end{aligned}$

To find the sum of the first six terms:

The general formula to find Sn for $|r|<1$ is $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\begin{aligned}S_{6} &=\frac{48\left(1-\left(-\frac{1}{4}\right)^{6}\right)}{1-\left(-\frac{1}{4}\right)} \\&=\frac{48\left(1-\frac{1}{4096}\right)}{1+\frac{1}{4096}} \\&=\frac{48(0.95)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{47.9904}{5} \\&=38.39\end{aligned}$

Thus, the sum of first six terms is 38.39

Answer 2

You can use the formula for finding the sum of an infinite geometric series to get the needed sum.

The sum of the first six term of the series is 38.390625

What is  a geometric sequence?

There are three parameters which differentiate between which geometric sequence we're talking about.

• The first parameter is the initial value of the sequence.
• The second parameter is the quantity by which we multiply previous term to get the next term.
• The third parameter is the length of the sequence. It can be finite or infinite.

Suppose the initial term of a geometric sequence is  $a$   and the term by which we multiply the previous term to get the next term is   $r$

Then the sequence would look like

$a, ar, ar^2, ar^3, ...$

 (till the terms to which it is defined)

What is a geometric series?

When we do addition of all the terms of a geometric sequence, we the expression we get is called geometric series.

For the above discussed geometric sequence, its geometric series would look like:

$a + ar + ar^2 + ...$

How to find the sum of a geometric series till n terms?

If the initial term is 'a', and the multiplication is done by 'r' term by term, then we have the sum till nth term as:

$S_n = \dfrac{a(1-r^n)}{(1-r)}$

Using the above formula to find the sum of the given series till nth term

You can see that, to get the next term, you multiply -1/4 to the previous term.

This is the reason why it is geometric series.

The first term is a = 48

The multiplication is done by r =   -1/4 on previous term to get the next terms.

Thus, as we need sum til 6 terms, thus, n = 6.

Putting that in the above specified formula, we get:

$S_n = \dfrac{a(1-r^n)}{(1-r)}\\\\S_6 = \dfrac{48 \times (1 - (\dfrac{-1}{4})^6)}{(1 - \dfrac{-1}{4})} = \dfrac{48 \times \dfrac{4095}{4096}}{\dfrac{5}{4}} = \dfrac{48 \times 819}{1024} = 38.390625$

Thus,

The sum of the first six term of the series is 38.390625

Learn more about geometric series here:

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