SAT Math : Common Difference in Sequences
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Example Question #1 : How To Find The Common Difference In Sequences
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
9
10
12
13
11
12
The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.
Example Question #771 : Arithmetic
In a sequence of numbers, each term is 
378
381
192
384
372
381
Let's call the first term in the sequence a1 and the nth term an.
We are told that each term is r times larger than the one before it. Thus, we can find the next term in the sequence by multiplying by r.
a1 = a1
a2 = r(a1)
a3 = r(a2) = r(r(a1)) = r2(a1)
a4 = r(a3) = r(r2(a1)) = r3(a1)
an = r(n–1)a1
We can use this information to find r.
The problem gives us the value of the third and the sixth terms.
a3 = r2(a1) = 12
a6 = r5(a1) = 96
Let's solve for a1 in terms of r and a3.
a1 = 12/(r2)
Let's then solve for a1 in terms of r and a6.
a1 = 96/(r5)
Now, we can set both values equal and solve for r.
12/(r2) = 96/(r5)
Multiply both sides by r5 to get rid of the fraction.
12r5/r2 = 96
Apply the property of exponents which states that ab/ac = ab–c.
12r3 = 96
Divide by 12 on both sides.
r3 = 8
Take the cube root of both sides.
r = 2
This means that each term is two times larger than the one before it, or that each term is one half as large as the one after it.
a2 must equal a3 divided by 2, which equals 12/2 = 6.
a1 must equal a2 divided by 2, which equals 6/2 = 3.
Here are the first eight terms of the sequence:
3, 6, 12, 24, 48, 96, 192, 384
The question asks us to find the sum of all the terms less than 250. Only the first seven terms are less than 250. Thus the sum is equal to the following:
sum = 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381
Example Question #11 : Sequences
Find the 
The sequence is geometric with a common ratio of 
The formula for finding the 
So
Example Question #3 : How To Find The Common Difference In Sequences
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
II, III, IV, V
II, IV, V
I, IV, V
I, V
IV, V
II, IV, V
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
Example Question #12 : Sequences
An arithmetic sequence begins as follows:
Give the first integer in the sequence.
The sequence has no integers.
Subtract the first term 
Setting 
If 
Solving for 
Therefore, we seek the least positive integer value of 
which is an integer.
Therefore, 2 is the first integer value in the sequence.
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