How to find the common difference in sequences - SAT Math

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Question

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

Answer

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.

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Question

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?

Answer

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

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Question

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

Answer

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

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Question

$2^{0}, 2^{\frac{1}{2}}, 2^{1}$

Find the $11^{th}$ term of the sequence above.

Answer

The sequence is geometric with a common ratio of $2^{\frac{1}{2}}$ .

The formula for finding the $n^{th}$ term of the sequence is

$a_{n} = a_{1} * r^{n-1}$ .

So

$a_{11} = 2^{0} * (2^{\frac{1}{2}})^{11-1} = 2^{5} = 32$ .

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Question

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

Answer

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

Compare your answer with the correct one above

Question

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

Compare your answer with the correct one above

Question

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

Answer

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

Compare your answer with the correct one above

Question

$2^{0}, 2^{\frac{1}{2}}, 2^{1}$

Find the $11^{th}$ term of the sequence above.

Answer

The sequence is geometric with a common ratio of $2^{\frac{1}{2}}$ .

The formula for finding the $n^{th}$ term of the sequence is

$a_{n} = a_{1} * r^{n-1}$ .

So

$a_{11} = 2^{0} * (2^{\frac{1}{2}})^{11-1} = 2^{5} = 32$ .

Compare your answer with the correct one above

Question

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

Answer

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

Compare your answer with the correct one above

Question

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

Compare your answer with the correct one above

Question

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

Answer

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

Compare your answer with the correct one above

Question

$2^{0}, 2^{\frac{1}{2}}, 2^{1}$

Find the $11^{th}$ term of the sequence above.

Answer

The sequence is geometric with a common ratio of $2^{\frac{1}{2}}$ .

The formula for finding the $n^{th}$ term of the sequence is

$a_{n} = a_{1} * r^{n-1}$ .

So

$a_{11} = 2^{0} * (2^{\frac{1}{2}})^{11-1} = 2^{5} = 32$ .

Compare your answer with the correct one above

Question

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

Answer

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

Compare your answer with the correct one above

Question

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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

Compare your answer with the correct one above

Question

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

Answer

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

Compare your answer with the correct one above

Question

$2^{0}, 2^{\frac{1}{2}}, 2^{1}$

Find the $11^{th}$ term of the sequence above.

Answer

The sequence is geometric with a common ratio of $2^{\frac{1}{2}}$ .

The formula for finding the $n^{th}$ term of the sequence is

$a_{n} = a_{1} * r^{n-1}$ .

So

$a_{11} = 2^{0} * (2^{\frac{1}{2}})^{11-1} = 2^{5} = 32$ .

Compare your answer with the correct one above

Question

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

Answer

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

Compare your answer with the correct one above

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