### All ACT Math Resources

## Example Questions

### Example Question #1681 : Act Math

If the first day of the year is a Monday, what is the 295th day?

**Possible Answers:**

Monday

Wednesday

Tuesday

Saturday

**Correct answer:**

Monday

The 295th day would be the day after the 42nd week has completed. 294 days/7 days a week = 42 weeks. The next day would therefore be a monday.

### Example Question #2 : How To Find The Nth Term Of An Arithmetic Sequence

If the first two terms of a sequence are and , what is the 38th term?

**Possible Answers:**

**Correct answer:**

The sequence is multiplied by each time.

### Example Question #1686 : Act Math

Find the term of the following sequence:

**Possible Answers:**

**Correct answer:**

The formula for finding the term of an arithmetic sequence is as follows:

where

= the difference between consecutive terms

= the number of terms

Therefore, to find the term:

### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence

What is the rd term of the following sequence:?

**Possible Answers:**

**Correct answer:**

Notice that between each of these numbers, there is a difference of ; however the first number is , the second , and so forth. This means that for each element, you know that the value must be , where is that number's place in the sequence. Thus, for the rd element, you know that the value will be or .

### Example Question #4 : Nth Term Of An Arithmetic Sequence

What is the th term in the following series of numbers: ?

**Possible Answers:**

148

**Correct answer:**

Notice that between each of these numbers, there is a difference of . This means that for each element, you will add . The first element is or . The second is or , and so forth... Therefore, for the th element, the value will be or .

### Example Question #2 : How To Find The Nth Term Of An Arithmetic Sequence

Find the sum of the first fifteen terms in an arithmetic sequence whose sixth term is and whose ninth term is .

**Possible Answers:**

**Correct answer:**

Use the formula *a*_{n} = *a*_{1} + (*n* – 1)*d*

*a*_{6} = *a*_{1} + 5*d*

*a*_{9} = *a*_{1} + 8*d*

Subtracting these equations yields

*a*_{6 }– *a*_{9} = –3*d*

–7 – 8 = –3*d*

*d* = 5

*a*_{1} = 33

Then use the formula for the series; = –30