Terms in a Series

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Pre-Calculus › Terms in a Series

Questions 1 - 6

Given the terms of the sequence $\small \small 0,1,1,2,3,5,...$ , what are the next two terms after $\small 5$ ?

$\small 8, 13$

$\small 7, 10$

$\small 8, 12$

$\small 7, 13$

Explanation

The next two terms are $\small 8$ and $\small 13$ . This is the Fibonacci sequence where you start off with the terms $\small 0$ and $\small 1$ , and the next term is the sum of two previous terms. So then

$\small a_3=0+1=1$

$\small a_4=1+1=2$

$\small a_5=1+2=3$

$\small a_6 = 2+3=5$

$\small a_7=3+5=8$

$\small a_8 = 5+8=13$

$\small \small a_{9}=8+13=21$

and so on.

What is the sum of the first terms of an arithmetic series if the first term is , and the last term is ?

$\frac{315}{2}$

$\frac{181}{15}$

$\frac{271}{15}$

$\frac{317}{4}$

Explanation

Write the formula to find the arithmetic sum of a series where is the number of terms, is the first term, and is the last term.

$S=\frac{n(a_1+a_n)}{2}$

Substitute the given values and solve for the sum.

$S=\frac{15(1+20)}{2}=\frac{15 \cdot21}{2}= \frac{315}{2}$

What is the fifth term of the series

$\frac{1}{32}$

$\frac{1}{16}$

Explanation

Let's try to see if this series is a geometric series.

We can divide adjacent terms to try and discover a multiplicative factor.

Doing this it seems the series proceeds with a common multiple of $\frac{1}{2}$ between each term.

Rewriting the series we get,

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\left(\frac{1}{2}\right)^n$ .

When

$n=5\rightarrow \frac{1}{2^5}=\frac{1}{32}$ .

What is the 10th term in the series:

1, 5, 9, 13, 17....

Explanation

The pattern in this arithmetic series is that each term is created by adding 4 to the previous one. You can then continue the series by continuing to add 4s until you've gotten to the tenth term:

1, 5, 9, 13, 17, 21, 25, 29, 33, 37

The correct answer, then, is 37.

What is the 9th term of the series that begins 2, 4, 8, 16...

144

256

512

488

1024

Explanation

In this geometric series, each number is created by multiplying the previous number by 2. You may also see that, because the first number is 2, it also becomes a list of powers of 2. The list is 2, 4, 8, 16, 32, 64, 128, 256, 512, where you can see that the 9th term is 512.

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$