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If the 7th term of a series is 18 and the 9th term is 22 what is 21st term.

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x32 | Student | eNoter

Posted August 4, 2013 at 3:28 PM via web

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If the 7th term of a series is 18 and the 9th term is 22 what is 21st term.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 4, 2013 at 5:06 PM (Answer #1)

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If this is an arithmetic series then the terms form an arithmetic sequence. Since the 7th term is 18 and the 9th term is 22 the common difference is 2. (We can find teh 8th term as it is the arithmetic mean of the 7th and 9th terms, so the 8th term is 20 giving us the common difference.)

We know the formula for the nth term of an arithmetic sequence is `a_n=a_1+(n-1)d` where `a_n` is the nth term, `a_1` is the first term, n the number of terms, and d the common difference. Since `a_7=18,d=2,n=7` we can find the first term:

`18=a_1+6(2)==> a_1=6`

Then the 21st term is `a_21=6+(20)(2)=46`

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The 21st term of an arithmetic series whose 7th term is 18 and whose 9th term is 22 is 46

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If this is a geometric series the answer is much uglier. We find that the common ratio is `r=sqrt(22/18)` and the first term is approximately `a_1~~9.858752817` which gives the 21st term as

`a_(21)~~9.858752817sqrt(22/18)^20` whic is approximately 73.337.

If this is any other type of series, then you cannot find the answer from the given information.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 4, 2013 at 5:07 PM (Answer #2)

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You may use the general formula that helps you to evaluate each member of an arithmetic series, such that:

`a_n = a_1 + (n - 1)*d`

The problem provides the information that `a_7 = 18` and `a_9 = 22` , hence, you may write the following equations, such that:

`a_7 = a_1 + 6d => 18 = a_1 + 6d`

`a_9 = a_1 + 8d => 22 = a_1 + 8d`

Subtracting the equation `18 = a_1 + 6d` from `22 = a_1 + 8d` yields:

`22 - 18 = a_1 - a_1 + 8d - 6d`

`4 = 2d => d = 4/2 => d = 2`

`a_1 = 18 - 6*2 => a_1 = 18 - 12 => a_1 = 6`

You may evaluate the member `a_21` , such that:

`a_21 = a_1 + 20d => a_21 = 6 + 20*2 => a_21 = 46`

Hence, evaluating `a_21` , under the given conditions, yields `a_21 = 46` .

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 4, 2013 at 5:12 PM (Answer #3)

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I need to mention that since you have tagged the label "arithmetic series", the problem is solved only with respect to this request.

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