The second and ninth terms of an arithmetic sequence are 2 and 30, respectively. What is the fiftieth term?

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The second and ninth terms of an arithmetic sequence are 2 and 30, respectively. What is the fiftieth term?

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Answer 1 · 2018-07-21T20:33:10

$a_(50)=194$

Explanation:

Ok let's say:

$a_n= a_s+d(n-s)$

This is a characteristic of an arithmetic sequence, each term is separated by a common difference:

In this case:
$a_9= a_2+d(9-2)$

$30= 2+7d$

$28=7d$

$d=4$

So, the 5th term:
$a_(50)= a_2+d(50-2)$

$a_5= 2+4(48)$

$a_5=194$

Answer 2 · 2018-07-21T20:41:59

$color(blue)(194)$

Explanation:

The nth term of an arithmetic sequence is given by:

$a+(n-1)d$

Where:

$bba$ is the first term, $bbd$ is the common difference and $bbn$ is the nth term.

We have:

$a+(2-1)d=2 \ \ \ [1]$

and

$a+(9-1)d=30 \ \ \ [2]$

We need to find $bba$ and $bbd$ .

Solving $[1]$ and $[2]$ simultaneously:

Subtract $[1]$ from $[2]$

$a-a+8d-d=30-2$

$7d=28=>d=4$

Substituting in $[1]$

$a+4=2=>a=-2$

So our general term is:

$-2+(n-1)4$

50th term will therefore be:

$-2+(50-1)4=194$

Answer 3 · 2018-07-21T20:46:31

$a_(50)=194$

Explanation:

$"the n th term of an arithmetic sequence is"$

$•color(white)(x)a_n=a+(n-1)d$

$" where a is the first term and d the common difference"$

$a_2=a+d=2to(1)$

$a_9=a+8d=30 to(2)$

$(2)-(1)" gives"$

$7d=28rArrd=4$

$"substitute in "(1)a+4=2rArra=-4$

$rArra_(50)=-2+(49xx4)=-2+196=194$

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The second and ninth terms of an arithmetic sequence are 2 and 30, respectively. What is the fiftieth term?

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