How to find the common difference in sequences - PSAT Math

Card 0 of 42

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

The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n - m}$

Setting $m = 1, n = 10, a_{1} = 20, a_{10} = 100$ , the common difference is found to be

$d = \frac{100 - 20}{10-1} = \frac{80}{9} = 8 \frac{8}{9}$

The $n\textup{th}$ term of the sequence $a_{n}$ is

$a_{n} = a_{1} + (n-1)d$

Setting $a_{1} = 20, n = 16, d = 8 \frac{8}{9}$ , this term is found to be

$a_{16} = 20 + (16-1) \cdot 8\frac{8}{9}$

$a_{16} = 20 + 15 \cdot 8\frac{8}{9}$

$a_{16} = 20 + 133 \frac{1}{3}$

$a_{16} = 153 \frac{1}{3}$

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Question

An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?

Answer

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

$a_{2} = a _{1}+ d = 17 + d$

and

$a_{10} = a _{9}+ d$ , or

$a _{9} = a_{10} - d = X- d$

The sum of the second and ninth terms is given by the expression

$a_{2} +a _{9} = 17 + d+ X - d = X + 17$

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Question

The first and fourth terms of an arithmetic sequence are, respectively,

$a _{1} = x^{2}- x+ 17$

and

$a _{4} = 3 x^{2}- 5x+ 81$ .

Which of the following is the common difference of the sequence?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n-m}$ .

Set :

$d = \frac{a_{4} - a_{1}}{4-1}$

$d = \frac{\left (3 x^{2}- 5x+ 81 \right ) - \left (x^{2}- x+ 17 \right )}{3}$

$d = \frac{3 x^{2}- x^{2} - 5x+ x + 81 - 17 }{3}$

$d = \frac{2 x^{2} - 4x + 64 }{3}$

$d = \frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

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

The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n - m}$

Setting $m = 1, n = 10, a_{1} = 20, a_{10} = 100$ , the common difference is found to be

$d = \frac{100 - 20}{10-1} = \frac{80}{9} = 8 \frac{8}{9}$

The $n\textup{th}$ term of the sequence $a_{n}$ is

$a_{n} = a_{1} + (n-1)d$

Setting $a_{1} = 20, n = 16, d = 8 \frac{8}{9}$ , this term is found to be

$a_{16} = 20 + (16-1) \cdot 8\frac{8}{9}$

$a_{16} = 20 + 15 \cdot 8\frac{8}{9}$

$a_{16} = 20 + 133 \frac{1}{3}$

$a_{16} = 153 \frac{1}{3}$

Compare your answer with the correct one above

Question

An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?

Answer

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

$a_{2} = a _{1}+ d = 17 + d$

and

$a_{10} = a _{9}+ d$ , or

$a _{9} = a_{10} - d = X- d$

The sum of the second and ninth terms is given by the expression

$a_{2} +a _{9} = 17 + d+ X - d = X + 17$

Compare your answer with the correct one above

Question

The first and fourth terms of an arithmetic sequence are, respectively,

$a _{1} = x^{2}- x+ 17$

and

$a _{4} = 3 x^{2}- 5x+ 81$ .

Which of the following is the common difference of the sequence?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n-m}$ .

Set :

$d = \frac{a_{4} - a_{1}}{4-1}$

$d = \frac{\left (3 x^{2}- 5x+ 81 \right ) - \left (x^{2}- x+ 17 \right )}{3}$

$d = \frac{3 x^{2}- x^{2} - 5x+ x + 81 - 17 }{3}$

$d = \frac{2 x^{2} - 4x + 64 }{3}$

$d = \frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

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

The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n - m}$

Setting $m = 1, n = 10, a_{1} = 20, a_{10} = 100$ , the common difference is found to be

$d = \frac{100 - 20}{10-1} = \frac{80}{9} = 8 \frac{8}{9}$

The $n\textup{th}$ term of the sequence $a_{n}$ is

$a_{n} = a_{1} + (n-1)d$

Setting $a_{1} = 20, n = 16, d = 8 \frac{8}{9}$ , this term is found to be

$a_{16} = 20 + (16-1) \cdot 8\frac{8}{9}$

$a_{16} = 20 + 15 \cdot 8\frac{8}{9}$

$a_{16} = 20 + 133 \frac{1}{3}$

$a_{16} = 153 \frac{1}{3}$

Compare your answer with the correct one above

Question

An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?

Answer

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

$a_{2} = a _{1}+ d = 17 + d$

and

$a_{10} = a _{9}+ d$ , or

$a _{9} = a_{10} - d = X- d$

The sum of the second and ninth terms is given by the expression

$a_{2} +a _{9} = 17 + d+ X - d = X + 17$

Compare your answer with the correct one above

Question

The first and fourth terms of an arithmetic sequence are, respectively,

$a _{1} = x^{2}- x+ 17$

and

$a _{4} = 3 x^{2}- 5x+ 81$ .

Which of the following is the common difference of the sequence?

Answer

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, , the common difference is

$d = \frac{a_{n} - a_{m}}{n-m}$ .

Set :

$d = \frac{a_{4} - a_{1}}{4-1}$

$d = \frac{\left (3 x^{2}- 5x+ 81 \right ) - \left (x^{2}- x+ 17 \right )}{3}$

$d = \frac{3 x^{2}- x^{2} - 5x+ x + 81 - 17 }{3}$

$d = \frac{2 x^{2} - 4x + 64 }{3}$

$d = \frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

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

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