Lesson Explainer: Finding the Arithmetic Sequence Mathematics

In this explainer, we will learn how to find arithmetic sequences given information about their terms and relationships between them.

We recall that an arithmetic sequence or arithmetic progression is a sequence of numbers such that the difference between any two consecutive terms in the sequence is constant. We call this constant difference the โ€œcommon difference.โ€

If the sequence has a finite number of terms, then we call it a finite arithmetic sequence; otherwise, we refer to it as just an arithmetic sequence.

For example, if the first term in our arithmetic sequence is 100 and the common difference between any two terms in the sequence is โˆ’5, then we can find the next term in the sequence.

Denoting the second term in the sequence as ๐‘‡๏Šจ, the difference between the second and first terms must be โˆ’5. Therefore, ๐‘‡โˆ’100=โˆ’5๐‘‡=โˆ’5+100๐‘‡=95.๏Šจ๏Šจ๏Šจ

We can continue this to find an infinite number of terms of the sequence.

This is equivalent to saying that we find the next term in the sequence by adding the common difference to the previous term.

We can also use this same process on any arithmetic sequence. If we denote the ๐‘›th term in an arithmetic sequence as ๐‘‡๏Š and the common difference as ๐‘‘, so that the first term in the sequence is ๐‘‡๏Šง, then we can find a formula for the ๐‘›th term in this arithmetic sequence.

Since the difference between any two consecutive terms must be the common difference ๐‘‘, the difference between the first and second terms in the sequence must be ๐‘‘: ๐‘‡โˆ’๐‘‡=๐‘‘.๏Šจ๏Šง

Rearranging this gives us ๐‘‡=๐‘‡+๐‘‘.๏Šจ๏Šง

The same will be true if we take the difference between the second and third terms: ๐‘‡โˆ’๐‘‡=๐‘‘๐‘‡=๐‘‘+๐‘‡.๏Šฉ๏Šจ๏Šฉ๏Šจ

Then, we can substitute in our expression for ๐‘‡๏Šจ: ๐‘‡=๐‘‘+๐‘‡+๐‘‘๐‘‡=๐‘‡+2๐‘‘.๏Šฉ๏Šง๏Šฉ๏Šง

In other words, to construct the third term in our arithmetic sequence, we take the first term and add the common difference twice. This process will continue for any number of terms, such as ๐‘›, giving us the following expression for the ๐‘›th term in our arithmetic sequence: ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We can summarize this as follows.

Definition: Arithmetic Sequences

An arithmetic sequence or progression is a sequence of numbers where the difference between any two consecutive terms is constant.

The ๐‘›th term of an arithmetic sequence with common difference ๐‘‘ and first term ๐‘‡๏Šง is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We can use this formula to determine information about arithmetic sequences by using the terms of the sequence. Letโ€™s see some examples of finding an arithmetic sequence given information about its terms.

Example 1: Finding an Arithmetic Sequence given Two Nonconsecutive Terms

Find the finite sequence ๐‘‡๏Š given that ๐‘‡=โˆ’82๏Šง, ๐‘‡=โˆ’203๏Šง๏Šจ, and the twelfth-to-last term is โˆ’115.

Answer

We want to find an expression for a finite arithmetic sequence by using information about its terms. We recall that an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant, called the common difference.

To find an expression for this finite arithmetic sequence, we are going to need to find the first term, the last term, and the common difference of the sequence. We start by recalling the formula for the ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘: ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

The question tells us ๐‘‡=โˆ’82๏Šง and ๐‘‡=โˆ’203๏Šง๏Šจ, and we can substitute ๐‘›=12 into this formula: ๐‘‡=๐‘‡+(12โˆ’1)๐‘‘=๐‘‡+11๐‘‘.๏Šง๏Šจ๏Šง๏Šง

Then, we can use the values of ๐‘‡๏Šง and ๐‘‡๏Šง๏Šจ: โˆ’203=โˆ’82+11๐‘‘.

Now, we can rearrange for ๐‘‘: โˆ’203+82=11๐‘‘โˆ’121=11๐‘‘๐‘‘=โˆ’11.

We now know the first term of this finite arithmetic sequence, ๐‘‡=โˆ’82๏Šง, and the common difference, ๐‘‘=โˆ’11. We still need to determine the last term in this arithmetic sequence. To determine the last term in this sequence, we will use the fact that the twelfth-to-last term is โˆ’115.

To find the last term, letโ€™s recall what is meant by the twelfth-to-last term. If we let ๐‘‡๏ˆ be the last term, then the arithmetic sequence is given by ๐‘‡,๐‘‡,โ€ฆ,๐‘‡,๐‘‡,๐‘‡.๏Šง๏Šจ๏ˆ๏Šฑ๏Šจ๏ˆ๏Šฑ๏Šง๏ˆ

๐‘‡๏ˆ๏Šฑ๏Šง will be the second-to-last term, since it is the penultimate term.

๐‘‡๏ˆ๏Šฑ๏Šจ will be the third-to-last term; we can continue this to see ๐‘‡๏ˆ๏Šฑ๏Šง๏Šง will be the twelfth-to-last term. Therefore, ๐‘‡=๐‘‡+(๐‘™โˆ’11โˆ’1)๐‘‘=๐‘‡+(๐‘™โˆ’12)๐‘‘.๏ˆ๏Šฑ๏Šง๏Šง๏Šง๏Šง

Substituting in our values for ๐‘‡๏Šง, ๐‘‘, and ๐‘‡๏ˆ๏Šฑ๏Šง๏Šง gives us โˆ’115=โˆ’82+(๐‘™โˆ’12)(โˆ’11)โˆ’115+82=(๐‘™โˆ’12)(โˆ’11)โˆ’33=(๐‘™โˆ’12)(โˆ’11)3=๐‘™โˆ’12๐‘™=15.

Thus, the sequence has 15 terms and we can find the last term: ๐‘‡=โˆ’82+(15โˆ’1)(โˆ’11)=โˆ’82โˆ’154=โˆ’236.๏Šง๏Šซ

Hence, the first term in our sequence is โˆ’82, the last term is โˆ’236, and the common difference is โˆ’11, giving us the sequence (โˆ’82,โˆ’93,โˆ’104,โ€ฆ,โˆ’236).

Example 2: Finding an Arithmetic Sequence given Sums of Its Terms

Find the arithmetic sequence in which the sum of the first and third terms is โˆ’142 and the sum of the third and fourth terms is โˆ’151.

Answer

We want to find an expression for a finite arithmetic sequence by using information about its terms. We recall that an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant, called the common difference.

The ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We are told in the question that ๐‘‡+๐‘‡=โˆ’142๏Šง๏Šฉ and ๐‘‡+๐‘‡=โˆ’151๏Šฉ๏Šช. We can use our formula to write ๐‘‡๏Šฉ and ๐‘‡๏Šช in terms of ๐‘‡๏Šง and ๐‘‘: ๐‘‡=๐‘‡+2๐‘‘๐‘‡=๐‘‡+3๐‘‘.๏Šฉ๏Šง๏Šช๏Šง

Substituting our expression for ๐‘‡๏Šฉ into the first sum gives us ๐‘‡+๐‘‡=โˆ’142๐‘‡+(๐‘‡+2๐‘‘)=โˆ’1422๐‘‡+2๐‘‘=โˆ’142๐‘‡+๐‘‘=โˆ’71.๏Šง๏Šฉ๏Šง๏Šง๏Šง๏Šง

We can do the same with our expression for ๐‘‡๏Šช: ๐‘‡+๐‘‡=โˆ’151(๐‘‡+2๐‘‘)+(๐‘‡+3๐‘‘)=โˆ’1512๐‘‡+5๐‘‘=โˆ’151.๏Šฉ๏Šช๏Šง๏Šง๏Šง

We have two linear equations in two variables; we can solve this system by eliminating a variable. Rearranging the first equation gives us ๐‘‡+๐‘‘=โˆ’71๐‘‡=โˆ’71โˆ’๐‘‘.๏Šง๏Šง

We then substitute this into our second equation: 2๐‘‡+5๐‘‘=โˆ’1512(โˆ’71โˆ’๐‘‘)+5๐‘‘=โˆ’151โˆ’142โˆ’2๐‘‘+5๐‘‘=โˆ’1513๐‘‘=โˆ’9๐‘‘=โˆ’3.๏Šง

We can then use this to find the value of ๐‘‡๏Šง: ๐‘‡+๐‘‘=โˆ’71๐‘‡+(โˆ’3)=โˆ’71๐‘‡=โˆ’68.๏Šง๏Šง๏Šง

Hence, the first term in our sequence is โˆ’68, and the common difference is โˆ’3, giving us the sequence (โˆ’68,โˆ’71,โˆ’74,โ€ฆ).

Letโ€™s now see an example where we are given the sum and product of distinct terms in an arithmetic sequence.

Example 3: Finding an Arithmetic Sequence given a Sum and a Product of Its Terms

Find the arithmetic sequence in which ๐‘‡+๐‘‡=โˆ’28๏Šจ๏Šช and ๐‘‡ร—๐‘‡=140๏Šฉ๏Šซ.

Answer

In an arithmetic sequence, the difference between any two consecutive terms is constant; this is called the common difference.

We recall that the ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We can use this to find expressions for the four terms given to us in the question: ๐‘‡=๐‘‡+(2โˆ’1)๐‘‘=๐‘‡+๐‘‘,๐‘‡=๐‘‡+(3โˆ’1)๐‘‘=๐‘‡+2๐‘‘,๐‘‡=๐‘‡+(4โˆ’1)๐‘‘=๐‘‡+3๐‘‘,๐‘‡=๐‘‡+(5โˆ’1)๐‘‘=๐‘‡+4๐‘‘.๏Šจ๏Šง๏Šง๏Šฉ๏Šง๏Šง๏Šช๏Šง๏Šง๏Šซ๏Šง๏Šง

We can then substitute these expressions into the two equations given to us.

First, ๐‘‡+๐‘‡=โˆ’28(๐‘‡+๐‘‘)+(๐‘‡+3๐‘‘)=โˆ’282๐‘‡+4๐‘‘=โˆ’28๐‘‡+2๐‘‘=โˆ’14.๏Šจ๏Šช๏Šง๏Šง๏Šง๏Šง

Second, ๐‘‡ร—๐‘‡=140(๐‘‡+2๐‘‘)ร—(๐‘‡+4๐‘‘)=140๐‘‡+4๐‘‡๐‘‘+2๐‘‡๐‘‘+8๐‘‘=140๐‘‡+6๐‘‡๐‘‘+8๐‘‘=140.๏Šฉ๏Šซ๏Šง๏Šง๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ

Therefore, we have two equations in two variables; we can solve these by eliminating a variable. We will start by rearranging an equation to find ๐‘‡๏Šง in terms of ๐‘‘: ๐‘‡+2๐‘‘=โˆ’14๐‘‡=โˆ’14โˆ’2๐‘‘.๏Šง๏Šง

We can then substitute this into our other equation: ๐‘‡+6๐‘‡๐‘‘+8๐‘‘=140(โˆ’14โˆ’2๐‘‘)+6(โˆ’14โˆ’2๐‘‘)๐‘‘+8๐‘‘=140196+56๐‘‘+4๐‘‘โˆ’84๐‘‘โˆ’12๐‘‘+8๐‘‘=140โˆ’28๐‘‘=โˆ’56๐‘‘=2.๏Šง๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

We can then substitute ๐‘‘=2 to find the value of ๐‘‡๏Šง: ๐‘‡+2๐‘‘=โˆ’14๐‘‡+2(2)=โˆ’14๐‘‡=โˆ’18.๏Šง๏Šง๏Šง

Hence, the first term in our sequence is โˆ’18, and the common difference is 2, giving us the sequence (โˆ’18,โˆ’16,โˆ’14,โ€ฆ).

In our next example, we are told two terms in an arithmetic sequence are additive inverses of each other. We need to use this information and the value of another term in the sequence to find the entire sequence.

Example 4: Finding an Arithmetic Sequence Given Relations between Its Terms

Find the arithmetic sequence for which ๐‘‡=279๏Šช๏Šฉ and ๐‘‡๏Šง๏Šง is the additive inverse of ๐‘‡๏Šง๏Šฉ.

Answer

In an arithmetic sequence, the difference between any two consecutive terms is constant; this is called the common difference. We can determine any arithmetic sequence from its first term and common difference.

We recall that the ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

By substituting ๐‘›=43 into this formula, we can use this to find an expression for ๐‘‡๏Šช๏Šฉ: ๐‘‡=๐‘‡+(43โˆ’1)๐‘‘=๐‘‡+42๐‘‘.๏Šช๏Šฉ๏Šง๏Šง

In the question, we are told ๐‘‡=279๏Šช๏Šฉ, so 279=๐‘‡+42๐‘‘.๏Šง

Next, we want to use the fact that ๐‘‡๏Šง๏Šง is the additive inverse of ๐‘‡๏Šง๏Šฉ. We recal that two numbers are additive inverses of each other if they add to give zero. Therefore, ๐‘‡+๐‘‡=0.๏Šง๏Šง๏Šง๏Šฉ

We can find expressions for these terms by substituting ๐‘›=11 and ๐‘›=13 into our formula: ๐‘‡=๐‘‡+10๐‘‘,๐‘‡=๐‘‡+12๐‘‘.๏Šง๏Šง๏Šง๏Šง๏Šฉ๏Šง

Since these are additive inverses, adding these expressions together must give us zero: ๐‘‡+๐‘‡=0๐‘‡+10๐‘‘+๐‘‡+12๐‘‘=02๐‘‡+22๐‘‘=0๐‘‡+11๐‘‘=0.๏Šง๏Šง๏Šง๏Šฉ๏Šง๏Šง๏Šง๏Šง

Remember, we already showed 279=๐‘‡+42๐‘‘.๏Šง

This means we have two equations in two unknowns; we can solve these by subtracting one equation from the other: ๐‘‡+11๐‘‘=0โˆ’(๐‘‡+42๐‘‘=279)โˆ’31๐‘‘=โˆ’279๏Šง๏Šง

We can then solve for ๐‘‘: โˆ’31๐‘‘=โˆ’279๐‘‘=9.

We can use this to find ๐‘‡๏Šง: ๐‘‡+11๐‘‘=0๐‘‡+11(9)=0๐‘‡=โˆ’99.๏Šง๏Šง๏Šง

Hence, the first term in our sequence is โˆ’99, and the common difference is 9, giving us the sequence (โˆ’99,โˆ’90,โˆ’81,โ€ฆ).

In our next example, we are asked to find an arithmetic sequence given information about the terms in words rather than in mathematical notation.

Example 5: Finding the Arithmetic Sequence under a Certain Condition

Find the arithmetic sequence whose twentieth term is 28, given that the sum of its third and sixth terms is greater than its ninth term by 8.

Answer

We want to find an arithmetic sequence. To do this, we will start by recalling that, in an arithmetic sequence, the difference between any two consecutive terms is the same; this is called the common difference. We also know that, in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘, the ๐‘›th term can be found by the following formula: ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We are told in the question the twentieth term is equal to 28, and we can find this by substituting ๐‘›=20: 28=๐‘‡+(20โˆ’1)๐‘‘=๐‘‡+19๐‘‘.๏Šง๏Šง

We can also find expressions for the third, sixth, and ninth terms: ๐‘‡=๐‘‡+2๐‘‘,๐‘‡=๐‘‡+5๐‘‘๐‘‡=๐‘‡+8๐‘‘.๏Šฉ๏Šง๏Šฌ๏Šง๏Šฏ๏Šง

In the question, we are told ๐‘‡+๐‘‡=๐‘‡+8๏Šฉ๏Šฌ๏Šฏ, so we can substitute these expressions into this equation, giving us ๐‘‡+๐‘‡=๐‘‡+8(๐‘‡+2๐‘‘)+(๐‘‡+5๐‘‘)=(๐‘‡+8๐‘‘)+82๐‘‡+7๐‘‘=๐‘‡+8๐‘‘+8๐‘‡โˆ’๐‘‘=8.๏Šฉ๏Šฌ๏Šฏ๏Šง๏Šง๏Šง๏Šง๏Šง๏Šง

Remember, we have already shown 28=๐‘‡+19๐‘‘.๏Šง

This means we have a pair of simultaneous equations with two unknowns. We can solve these by eliminating a variable: ๐‘‡+19๐‘‘=28โˆ’(๐‘‡โˆ’๐‘‘=8)20๐‘‘=20๏Šง๏Šง

Hence, ๐‘‘=1. Substituting this into one of our simultaneous equations gives us ๐‘‡โˆ’1=8๐‘‡=9.๏Šง๏Šง

Therefore, the first term of this arithmetic sequence is 9 and the common difference is 1. This is the sequence (9,10,11,โ€ฆ).

In our next example, we will see how to use information about the signs of the terms of our sequence to find an arithmetic sequence.

Example 6: Finding an Arithmetic Sequence given Relations between Products of Its Terms

Find the arithmetic sequence given that ๐‘‡๐‘‡=1708๏Šง๏Šง๏Šจ, ๐‘‡๐‘‡โˆ’๐‘‡๐‘‡=336๏Šฌ๏Šฉ๏Šง๏Šญ๏Šจ๏Šฌ, and all terms are positive.

Answer

We want to express an arithmetic sequence using information given about its terms. We recall that an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant, called the common difference.

We also know that the ๐‘›th term of an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We can use this to find expressions for all of the terms used in the equations given to us in the question: ๐‘‡=๐‘‡+11๐‘‘,๐‘‡=๐‘‡+5๐‘‘,๐‘‡=๐‘‡+30๐‘‘,๐‘‡=๐‘‡+6๐‘‘,๐‘‡=๐‘‡+25๐‘‘.๏Šง๏Šจ๏Šง๏Šฌ๏Šง๏Šฉ๏Šง๏Šง๏Šญ๏Šง๏Šจ๏Šฌ๏Šง

We substitute these expressions into each of the equations.

First, ๐‘‡๐‘‡=1708๐‘‡(๐‘‡+11๐‘‘)=1708๐‘‡+11๐‘‡๐‘‘=1708.๏Šง๏Šง๏Šจ๏Šง๏Šง๏Šง๏Šจ๏Šง

Second, ๐‘‡๐‘‡โˆ’๐‘‡๐‘‡=336(๐‘‡+5๐‘‘)(๐‘‡+30๐‘‘)โˆ’(๐‘‡+6๐‘‘)(๐‘‡+25๐‘‘)=336.๏Šฌ๏Šฉ๏Šง๏Šญ๏Šจ๏Šฌ๏Šง๏Šง๏Šง๏Šง

Expanding over the parentheses and simplifying gives us ๐‘‡+30๐‘‡๐‘‘+5๐‘‡๐‘‘+150๐‘‘โˆ’๐‘‡โˆ’25๐‘‡๐‘‘โˆ’6๐‘‡๐‘‘โˆ’150๐‘‘=3364๐‘‡๐‘‘=336๐‘‡๐‘‘=84.๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šง

We can substitute this value for ๐‘‡๐‘‘๏Šง into our first equation: ๐‘‡+11๐‘‡๐‘‘=1708๐‘‡+11(84)=1708๐‘‡=784.๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจ

Now, since we are told all terms of the sequence are positive, we can find ๐‘‡๏Šง by taking the positive square root: ๐‘‡=โˆš784=28.๏Šง

We can then substitute this into the equation ๐‘‡๐‘‘=84๏Šง to find the value of ๐‘‘: (28)๐‘‘=84๐‘‘=3.

Therefore, the first term of this arithmetic sequence is 28 and the common difference is 3. This is the sequence (28,31,34,โ€ฆ).

In our next example, we will find an arithmetic sequence given information about its terms in the form of a recurrence relation.

Example 7: Finding an Arithmetic Sequence given a Term and a Recurrence Relation

Find an arithmetic sequence given ๐‘‡=13๏Šง and ๐‘‡=18๐‘‡๏Šง๏Šฎ๏Š๏Š.

Answer

We want to find an arithmetic sequence. To do this, we can start by recalling that, in an arithmetic sequence, the difference between any two consecutive terms is the same and this is called the common difference. We also know that, in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘, the ๐‘›th term can be found by the following formula: ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

We can use this to find an expressions for ๐‘‡๏Šง๏Šฎ๏Š and ๐‘‡๏Š in terms of ๐‘‘, ๐‘‡๏Šง, and ๐‘›: ๐‘‡=๐‘‡+(18๐‘›โˆ’1)๐‘‘=๐‘‡+18๐‘‘๐‘›โˆ’๐‘‘.๏Šง๏Šฎ๏Š๏Šง๏Šง

In our case, ๐‘‡=13๏Šง; this gives us ๐‘‡=13+18๐‘‘๐‘›โˆ’๐‘‘.๏Šง๏Šฎ๏Š

Next, ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘=13+(๐‘›โˆ’1)๐‘‘=13+๐‘‘๐‘›โˆ’๐‘‘.๏Š๏Šง

We can now substitute these expressions into the equation given to us in the question: ๐‘‡=18๐‘‡13+18๐‘‘๐‘›โˆ’๐‘‘=18(13+๐‘‘๐‘›โˆ’๐‘‘)13+18๐‘‘๐‘›โˆ’๐‘‘=234+18๐‘‘๐‘›โˆ’18๐‘‘0=221โˆ’17๐‘‘17๐‘‘=221๐‘‘=13.๏Šง๏Šฎ๏Š๏Š

Hence, the first term in our sequence is 13, and the common difference is 13, giving us the sequence (13,26,39,โ€ฆ).

In our final example, we will find an arithmetic progression by solving simultaneous equations.

Example 8: Finding the Arithmetic Sequence under a Certain Condition

Determine the arithmetic sequence in which ๐‘‡+๐‘‡=500๏Šซ๏Šฆ๏Šจ๏Šฎ and ๐‘‡+๐‘‡+๐‘‡=138๏Šฉ๏Šง๏Šง๏Šฉ๏Šซ.

Answer

We recall that, in an arithmetic sequence, the difference between any two consecutive terms is constant. We can determine any arithmetic sequence from its first term and common difference, where the ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ is given by ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง

By using this formula, we can construct expressions for each of the terms in the equations given to us in the question: ๐‘‡=๐‘‡+49๐‘‘,๐‘‡=๐‘‡+27๐‘‘,๐‘‡=๐‘‡+2๐‘‘,๐‘‡=๐‘‡+10๐‘‘,๐‘‡=๐‘‡+34๐‘‘.๏Šซ๏Šฆ๏Šง๏Šจ๏Šฎ๏Šง๏Šฉ๏Šง๏Šง๏Šง๏Šง๏Šฉ๏Šซ๏Šง

We can then substitute these expressions into the equations given to us in the question.

First, ๐‘‡+๐‘‡=500(๐‘‡+49๐‘‘)+(๐‘‡+27๐‘‘)=5002๐‘‡+76๐‘‘=500๐‘‡+38๐‘‘=250.๏Šซ๏Šฆ๏Šจ๏Šฎ๏Šง๏Šง๏Šง๏Šง

Second, ๐‘‡+๐‘‡+๐‘‡=138(๐‘‡+2๐‘‘)+(๐‘‡+10๐‘‘)+(๐‘‡+34๐‘‘)=1383๐‘‡+46๐‘‘=138.๏Šฉ๏Šง๏Šง๏Šฉ๏Šซ๏Šง๏Šง๏Šง๏Šง

This gives us two equations in two unknowns, we can solve these by eliminating one variable. We can rearrange the first equation: ๐‘‡+38๐‘‘=250๐‘‡=250โˆ’38๐‘‘.๏Šง๏Šง

Then, we substitute this expression for ๐‘‡๏Šง in our second equation: 3๐‘‡+46๐‘‘=1383(250โˆ’38๐‘‘)+46๐‘‘=138750โˆ’114๐‘‘+46๐‘‘=138612โˆ’68๐‘‘=0๐‘‘=9.๏Šง

Finally, we substitute this value for ๐‘‘ into one of our equations to find the first term in our sequence: ๐‘‡+38๐‘‘=250๐‘‡+38(9)=250๐‘‡+342=250๐‘‡=โˆ’92.๏Šง๏Šง๏Šง๏Šง

Therefore, the first term of this arithmetic sequence is โˆ’92 and the common difference is 9. This is the sequence (โˆ’92,โˆ’83,โˆ’74,โ€ฆ).

Letโ€™s finish by recapping some basic points.

Key Points

  • An arithmetic sequence or progression is a sequence in which the difference between any two consecutive terms is constant; this constant is called the common difference.
  • The ๐‘›th term in an arithmetic sequence with first term ๐‘‡๏Šง and common difference ๐‘‘ can be found by using the formula ๐‘‡=๐‘‡+(๐‘›โˆ’1)๐‘‘.๏Š๏Šง
  • We can use this formula to construct an expression for any term in our arithmetic sequence in terms of the initial term and common difference. If we can create two equations in these variables, then we can solve these as simultaneous equations.

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