Math / Algebra

Question7 The third term of an arithmetic sequence is twice the first term. The second term is 45 . a) Find the value of the first term and the common difference. b) Determine whether or not 310 is a term of the sequence.

Studdy Solution

STEP 1

What is this asking? We're exploring an arithmetic sequence where the third term is double the first, the second term is 45, and we need to find the starting term, the common difference between terms, and if 310 is part of the sequence! Watch out! Don't mix up the terms of the sequence with the position of the terms.
Also, remember that an arithmetic sequence has a constant difference between consecutive terms.

STEP 2

1. Define Variables
2. Express Terms
3. Solve for First Term and Common Difference
4. Check if 310 is a Term

STEP 3

Let $a$ be the first term of our arithmetic sequence.
Let $d$ be the common difference between consecutive terms.
This means that to get from one term to the next, we add $d$ .

STEP 4

The first term is simply $a$ .

STEP 5

The second term is $a + d$ , which we know is equal to 45.
So, $a + d = 45$ .

STEP 6

The third term is $a + 2d$ , and we're told it's twice the first term, which is $2a$ .
So, $a + 2d = 2a$ .

STEP 7

From $a + 2d = 2a$ , we can subtract $a$ from both sides to get $2d = a$ .
This tells us that the first term ( $a$ ) is twice the common difference ( $d$ ).
Cool, right?

STEP 8

Now, let's substitute $2d$ for $a$ in the equation for the second term: $a + d = 45$ .
Replacing $a$ with $2d$ , we get $2d + d = 45$ , which simplifies to $3d = 45$ .

STEP 9

To find $d$ , we divide both sides of $3d = 45$ by 3.
This gives us $d = \frac{45}{3} = 15$ .
So, our common difference ( $d$ ) is 15!

STEP 10

Since $a = 2d$ , and we know $d = 15$ , we can find $a$ by calculating $a = 2 \cdot 15 = 30$ .
Our first term ( $a$ ) is 30!

STEP 11

Any term in an arithmetic sequence can be written as $a + (n-1)d$ , where $n$ is the term number.
We want to see if there's an integer $n$ for which $a + (n-1)d = 310$ .

STEP 12

We know $a = 30$ and $d = 15$ , so we substitute those values into the equation: $30 + (n-1) \cdot 15 = 310$ .

STEP 13

Subtract 30 from both sides: $(n-1) \cdot 15 = 280$ .

STEP 14

Divide both sides by 15: $n-1 = \frac{280}{15} = \frac{56}{3}$ .

STEP 15

Add 1 to both sides: $n = \frac{56}{3} + 1 = \frac{56}{3} + \frac{3}{3} = \frac{59}{3}$ .
Since $\frac{59}{3}$ is not an integer, 310 is not a term in the sequence.

STEP 16

The first term is $a = 30$ , the common difference is $d = 15$ , and 310 is not a term in the sequence.

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Question7 The third term of an arithmetic sequence is twice the first term. The second term is 45 . a) Find the value of the first term and the common difference. b) Determine whether or not 310 is a term of the sequence.

Studdy Solution

STEP 1

STEP 2

1. Define Variables
2. Express Terms
3. Solve for First Term and Common Difference
4. Check if 310 is a Term

STEP 3

Let $a$ be the first term of our arithmetic sequence.
Let $d$ be the common difference between consecutive terms.
This means that to get from one term to the next, we add $d$ .

STEP 4

The first term is simply $a$ .

STEP 5

The second term is $a + d$ , which we know is equal to 45.
So, $a + d = 45$ .

STEP 6

The third term is $a + 2d$ , and we're told it's twice the first term, which is $2a$ .
So, $a + 2d = 2a$ .

STEP 7

From $a + 2d = 2a$ , we can subtract $a$ from both sides to get $2d = a$ .
This tells us that the first term ( $a$ ) is twice the common difference ( $d$ ).
Cool, right?

STEP 8

Now, let's substitute $2d$ for $a$ in the equation for the second term: $a + d = 45$ .
Replacing $a$ with $2d$ , we get $2d + d = 45$ , which simplifies to $3d = 45$ .

STEP 9

To find $d$ , we divide both sides of $3d = 45$ by 3.
This gives us $d = \frac{45}{3} = 15$ .
So, our common difference ( $d$ ) is 15!

STEP 10

Since $a = 2d$ , and we know $d = 15$ , we can find $a$ by calculating $a = 2 \cdot 15 = 30$ .
Our first term ( $a$ ) is 30!

STEP 11

Any term in an arithmetic sequence can be written as $a + (n-1)d$ , where $n$ is the term number.
We want to see if there's an integer $n$ for which $a + (n-1)d = 310$ .

STEP 12

We know $a = 30$ and $d = 15$ , so we substitute those values into the equation: $30 + (n-1) \cdot 15 = 310$ .

STEP 13

Subtract 30 from both sides: $(n-1) \cdot 15 = 280$ .

STEP 14

Divide both sides by 15: $n-1 = \frac{280}{15} = \frac{56}{3}$ .

STEP 15

Add 1 to both sides: $n = \frac{56}{3} + 1 = \frac{56}{3} + \frac{3}{3} = \frac{59}{3}$ .
Since $\frac{59}{3}$ is not an integer, 310 is not a term in the sequence.

STEP 16

The first term is $a = 30$ , the common difference is $d = 15$ , and 310 is not a term in the sequence.

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Question7 The third term of an arithmetic sequence is twice the first term. The second term is 45 . a) Find the value of the first term and the common difference. b) Determine whether or not 310 is a term of the sequence.

Studdy Solution

STEP 1

STEP 2

1. Define Variables2. Express Terms3. Solve for First Term and Common Difference4. Check if 310 is a Term

STEP 3

Let aaa be the **first term** of our arithmetic sequence.Let ddd be the **common difference** between consecutive terms.This means that to get from one term to the next, we *add* ddd.

STEP 4

The **first term** is simply aaa.

STEP 5

The **second term** is a+da + da+d, which we know is equal to **45**.So, a+d=45a + d = 45a+d=45.

STEP 6

The **third term** is a+2da + 2da+2d, and we're told it's *twice* the first term, which is 2a2a2a.So, a+2d=2aa + 2d = 2aa+2d=2a.

STEP 7

From a+2d=2aa + 2d = 2aa+2d=2a, we can subtract aaa from both sides to get 2d=a2d = a2d=a.This tells us that the **first term** (aaa) is *twice* the **common difference** (ddd).Cool, right?

STEP 8

Now, let's substitute 2d2d2d for aaa in the equation for the **second term**: a+d=45a + d = 45a+d=45.Replacing aaa with 2d2d2d, we get 2d+d=452d + d = 452d+d=45, which simplifies to 3d=453d = 453d=45.

STEP 9

To find ddd, we divide both sides of 3d=453d = 453d=45 by 3.This gives us d=453=15d = \frac{45}{3} = 15d=345​=15.So, our **common difference** (ddd) is **15**!

STEP 10

Since a=2da = 2da=2d, and we know d=15d = 15d=15, we can find aaa by calculating a=2⋅15=30a = 2 \cdot 15 = 30a=2⋅15=30.Our **first term** (aaa) is **30**!

STEP 11

Any term in an arithmetic sequence can be written as a+(n−1)da + (n-1)da+(n−1)d, where nnn is the term number.We want to see if there's an integer nnn for which a+(n−1)d=310a + (n-1)d = 310a+(n−1)d=310.

STEP 12

We know a=30a = 30a=30 and d=15d = 15d=15, so we substitute those values into the equation: 30+(n−1)⋅15=31030 + (n-1) \cdot 15 = 31030+(n−1)⋅15=310.

STEP 13

Subtract 30 from both sides: (n−1)⋅15=280(n-1) \cdot 15 = 280(n−1)⋅15=280.

STEP 14

Divide both sides by 15: n−1=28015=563n-1 = \frac{280}{15} = \frac{56}{3}n−1=15280​=356​.

STEP 15

Add 1 to both sides: n=563+1=563+33=593n = \frac{56}{3} + 1 = \frac{56}{3} + \frac{3}{3} = \frac{59}{3}n=356​+1=356​+33​=359​.Since 593\frac{59}{3}359​ is *not* an integer, 310 is *not* a term in the sequence.

STEP 16

The **first term** is a=30a = 30a=30, the **common difference** is d=15d = 15d=15, and **310 is not a term** in the sequence.

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1. Define Variables
2. Express Terms
3. Solve for First Term and Common Difference
4. Check if 310 is a Term

Let $a$ be the first term of our arithmetic sequence.
Let $d$ be the common difference between consecutive terms.
This means that to get from one term to the next, we add $d$ .

The first term is simply $a$ .

The second term is $a + d$ , which we know is equal to 45.
So, $a + d = 45$ .

The third term is $a + 2d$ , and we're told it's twice the first term, which is $2a$ .
So, $a + 2d = 2a$ .

From $a + 2d = 2a$ , we can subtract $a$ from both sides to get $2d = a$ .
This tells us that the first term ( $a$ ) is twice the common difference ( $d$ ).
Cool, right?

Now, let's substitute $2d$ for $a$ in the equation for the second term: $a + d = 45$ .
Replacing $a$ with $2d$ , we get $2d + d = 45$ , which simplifies to $3d = 45$ .

To find $d$ , we divide both sides of $3d = 45$ by 3.
This gives us $d = \frac{45}{3} = 15$ .
So, our common difference ( $d$ ) is 15!

Since $a = 2d$ , and we know $d = 15$ , we can find $a$ by calculating $a = 2 \cdot 15 = 30$ .
Our first term ( $a$ ) is 30!

Any term in an arithmetic sequence can be written as $a + (n-1)d$ , where $n$ is the term number.
We want to see if there's an integer $n$ for which $a + (n-1)d = 310$ .

We know $a = 30$ and $d = 15$ , so we substitute those values into the equation: $30 + (n-1) \cdot 15 = 310$ .

Subtract 30 from both sides: $(n-1) \cdot 15 = 280$ .

Divide both sides by 15: $n-1 = \frac{280}{15} = \frac{56}{3}$ .

Add 1 to both sides: $n = \frac{56}{3} + 1 = \frac{56}{3} + \frac{3}{3} = \frac{59}{3}$ .
Since $\frac{59}{3}$ is not an integer, 310 is not a term in the sequence.

The first term is $a = 30$ , the common difference is $d = 15$ , and 310 is not a term in the sequence.