### All ACT Math Resources

## Example Questions

### Example Question #1 : Nth Term Of An Arithmetic Sequence

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

**Possible Answers:**

Wednesday

Saturday

Monday

Tuesday

**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 #1 : 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 #1 : Nth Term Of An Arithmetic Sequence

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 #9 : Arithmetic Sequences

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 #10 : Arithmetic Sequences

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 #1 : 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

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