GRE Math : Other Arithmetic Sequences

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Example Questions

Example Question #1 : How To Find The Answer To An Arithmetic Sequence

What is the sum of the odd integers $\displaystyle 1, 3, 5, ... ,97, 99$?

Possible Answers:

1975 $\displaystyle 1975$

2000 $\displaystyle 2000$

1000 $\displaystyle 1000$

None of the other answers

2500 $\displaystyle 2500$

Correct answer:

2500 $\displaystyle 2500$

Explanation:

Do NOT try to add all of these. It is key that you notice the pattern. Begin by looking at the first and the last elements: 1 and 99. They add up to 100. Now, consider 3 and 97. Just as 1 + 99 = 100, 3 + 97 = 100. This holds true for the entire list. Therefore, it is crucial that we find the number of such pairings.

1, 3, 5, 7, and 9 are paired with 99, 97, 95, 93, and 91, respectively. Therefore, for each 10s digit, there are 5 pairings, or a total of 500. To get all the way through our numbers, you will have to repeat this process for the 10s, 20s, 30s, and 40s (all the way to 49 + 51 = 100).

Therefore, there are 500 (per pairing) * 5 pairings = 2500.

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Example Question #2 : Sequences

A sequence is defined by the following formula:

$s_n = 3\cdot s_n_-_1+5$ $\displaystyle s_n = 3\cdot s_n_-_1+5$

$\displaystyle s_1 = 12$

What is the 4th element of this sequence?

Possible Answers:

389 $\displaystyle 389$

1172 $\displaystyle 1172$

145 $\displaystyle 145$

$\displaystyle 29$

$\displaystyle 41$

Correct answer:

389 $\displaystyle 389$

Explanation:

With series, you can always "walk through" the values to find your answer. Based on our equation, we can rewrite s_2 $\displaystyle s_2$ as :

$s_2 = 3\cdot s_1+5=3\cdot 12+5=36+5 = 41$ $\displaystyle s_2 = 3\cdot s_1+5=3\cdot 12+5=36+5 = 41$

You then continue for the third and the fourth element:

$s_3=3\cdot 41 + 5=128$ $\displaystyle s_3=3\cdot 41 + 5=128$

$s_4=3\cdot 128+5=389$ $\displaystyle s_4=3\cdot 128+5=389$

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Example Question #1 : Sequences

What is the sum of the 40th and the 70th elements of the series defined as:

$\displaystyle s_n = s_n_-_1 - 5$

s_1=281 $\displaystyle s_1=281$

Possible Answers:

$\displaystyle 45$

100 $\displaystyle 100$

$\displaystyle 55$

$\displaystyle 17$

$\displaystyle 22$

Correct answer:

$\displaystyle 22$

Explanation:

When you are asked to find elements in a series that are far into its iteration, you need to find the pattern. You absolutely cannot waste your time trying to calculate all of the values between $\displaystyle 1$ and $\displaystyle 70$. Notice that for every element after the first one, you subtract $\displaystyle 5$. Thus, for the second element you have:

$\displaystyle s_2=281-5$

For the third, you have:

$\displaystyle s_3 = 281 - (2)5$

Therefore, for the 40th and 70th elements, you will have:

$\displaystyle s_4_0 = 281 - (39)5 = 281 - 195 = 86$

$\displaystyle s_7_0 = 281 - (69)5 = 281-345=-64$

The sum of these two elements is:

$\displaystyle 86-64=22$

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GRE Math : Other Arithmetic Sequences

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #1 : How To Find The Answer To An Arithmetic Sequence

Example Question #2 : Sequences

Example Question #1 : Sequences

All GRE Math Resources

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DMCA Complaint

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