How to find the common difference in sequences - PSAT Math
Card 0 of 42
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
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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.
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.
In a sequence of numbers, each term is 
times larger than the one before it. If the 3rd term of the sequence is 12, and the 6th term is 96, what is the sum of all of the terms less than 250?
In a sequence of numbers, each term is
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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
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
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Tap to see back →
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
Tap to see back →
Given terms 
of an arithmetic sequence, 
, the common difference 
is
Setting 
, the common difference is found to be
The 
term of the sequence 
is
Setting 
, this term is found to be
Given terms
Setting
The
Setting
An arithmetic sequence has as its first term 17. Its tenth term is 
. What is the sum of its second and ninth terms?
An arithmetic sequence has as its first term 17. Its tenth term is
Tap to see back →
It is not necessary to determine the common difference or any other terms to answer this question. If we let 
be the common difference, then
and
, or
The sum of the second and ninth terms is given by the expression
It is not necessary to determine the common difference or any other terms to answer this question. If we let
and
The sum of the second and ninth terms is given by the expression
The first and fourth terms of an arithmetic sequence are, respectively,
and
.
Which of the following is the common difference of the sequence?
The first and fourth terms of an arithmetic sequence are, respectively,
and
Which of the following is the common difference of the sequence?
Tap to see back →
Given terms 
of an arithmetic sequence, 
, the common difference 
is
.
Set 
:
Given terms
Set
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
Tap to see back →
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.
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.
In a sequence of numbers, each term is times larger than the one before it. If the 3rd term of the sequence is 12, and the 6th term is 96, what is the sum of all of the terms less than 250?
In a sequence of numbers, each term is
Tap to see back →
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
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
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Tap to see back →
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
Tap to see back →
Given terms of an arithmetic sequence, , the common difference is
Setting , the common difference is found to be
The term of the sequence is
Setting , this term is found to be
Given terms
Setting
The
Setting
An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?
An arithmetic sequence has as its first term 17. Its tenth term is
Tap to see back →
It is not necessary to determine the common difference or any other terms to answer this question. If we let be the common difference, then
and
, or
The sum of the second and ninth terms is given by the expression
It is not necessary to determine the common difference or any other terms to answer this question. If we let
and
The sum of the second and ninth terms is given by the expression
The first and fourth terms of an arithmetic sequence are, respectively,
and
.
Which of the following is the common difference of the sequence?
The first and fourth terms of an arithmetic sequence are, respectively,
and
Which of the following is the common difference of the sequence?
Tap to see back →
Given terms of an arithmetic sequence, , the common difference is
.
Set :
Given terms
Set
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
Tap to see back →
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.
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.
In a sequence of numbers, each term is times larger than the one before it. If the 3rd term of the sequence is 12, and the 6th term is 96, what is the sum of all of the terms less than 250?
In a sequence of numbers, each term is
Tap to see back →
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
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
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Which of the following are not natural numbers?
I. 1
II. 0
III. 349010
IV. -2
V. 1/4
Tap to see back →
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?
Tap to see back →
Given terms of an arithmetic sequence, , the common difference is
Setting , the common difference is found to be
The term of the sequence is
Setting , this term is found to be
Given terms
Setting
The
Setting
An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?
An arithmetic sequence has as its first term 17. Its tenth term is
Tap to see back →
It is not necessary to determine the common difference or any other terms to answer this question. If we let be the common difference, then
and
, or
The sum of the second and ninth terms is given by the expression
It is not necessary to determine the common difference or any other terms to answer this question. If we let
and
The sum of the second and ninth terms is given by the expression
The first and fourth terms of an arithmetic sequence are, respectively,
and
.
Which of the following is the common difference of the sequence?
The first and fourth terms of an arithmetic sequence are, respectively,
and
Which of the following is the common difference of the sequence?
Tap to see back →
Given terms of an arithmetic sequence, , the common difference is
.
Set :
Given terms
Set
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
Tap to see back →
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.
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.
In a sequence of numbers, each term is times larger than the one before it. If the 3rd term of the sequence is 12, and the 6th term is 96, what is the sum of all of the terms less than 250?
In a sequence of numbers, each term is
Tap to see back →
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
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