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Example Questions
Example Question #31 : Integers
The first term in a sequence of integers is 2 and the second term is 10. All subsequent terms are the arithmetic mean of all of the preceding terms. What is the 39th term?
6
5
300
600
1200
6
The first term and second term average out to 6. So the third term is 6. Now add 6 to the preceding two terms and divide by 3 to get the average of the first three terms, which is the value of the 4th term. This, too, is 6 (18/3)—all terms after the 2nd are 6, including the 39th. Thus, the answer is 6.
Example Question #1 : Sequences
Consider the following sequence of integers:
5, 11, 23, 47
What is the 6th element in this sequence?
None of the other answers
93
191
95
189
191
First, consider the change in each element. Notice that in each case, a given element is twice the preceding one plus one:
11 = 2 * 5 + 1
23 = 11 * 2 + 1
47 = 23 * 2 + 1
To find the 6th element, continue following this:
The 5th: 47 * 2 + 1 = 95
The 6th: 95 * 2 + 1 = 191
Example Question #1 : Arithmetic Sequences
The sequence begins with the numbers
and has the
term defined as
, for
.
What is the value of the term of the sequence?
The first term of the sequence is , so here
, and we're interested in finding the 20th term, so we'll use n = 20.
Plugging these values into the given expression for the nth term gives us our answer.
and
Example Question #1 : Sequences
In a sequence of numbers, the first two values are 1 and 2. Each successive integer is calculated by adding the previous two and mutliplying that result by 3. What is fifth value in this sequence?
None of the other answers
Our sequence begins as 1, 2.
Element 3: (Element 1 + Element 2) * 3 = (1 + 2) * 3 = 3 * 3 = 9
Element 4: (Element 2 + Element 3) * 3 = (2 + 9) * 3 = 11 * 3 = 33
Element 5: (Element 3 + Element 4) * 3 = (9 + 33) * 3 = 42 * 3 = 126
Example Question #1 : Sequences
Let Z represent a sequence of numbers wherein each term is defined as seven less than three times the preceding term. If
, what is the first term in the sequence?
Let us first write the value of a consecutive term in a numerical format:
Consequently,
Using the first equation, we can define in terms of
:
This allows us to rewrite
as
Rearrangement of terms allows us to solve for :
Now, using our second equation, we can find , the first term:
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