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Mathematics 10 Online
OpenStudy (anonymous):

What is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13 ?

OpenStudy (anonymous):

Can you tell me the formula for the nth term of an arithmetic sequence?

OpenStudy (anonymous):

Its an act question. Thats all it gives me

OpenStudy (anonymous):

What do you think the answer is?

OpenStudy (anonymous):

@debins33, I know that's all they give you but in order to solve it you're going to need to know a few formulas - the first of which being the formula for the nth term of an arithmetic sequence. Look it up and tell me when you find it.

OpenStudy (anonymous):

a+(n-1)d

OpenStudy (anonymous):

All of them seem to involve knowing the first term

OpenStudy (anonymous):

The ACT says the answer is 14.5

OpenStudy (anonymous):

That's it! Great job. So now we look at the information in the question: "What is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13 ?" From this, we know that: \[a_6=8=a+(6-1)d=a+5d\]\[a_{10}=13=a+(10-1)d=a+9d\] As you said, we need to find the first term. To do this, we can just solve these two equations simultaneously: \[8=a+5d\]\[13=a+9d\] Can you solve these for a and d?

OpenStudy (anonymous):

Oh okay I see how that works. I'm a bit rusty with my algebra would I put it in terms of d or a to solve?

OpenStudy (anonymous):

Great, glad you understand. There's two ways you could solve those: rearranging and just subtracting. Rearranging: \[8=a+5d, a=8-5d\]\[13=a+9d, a=13-9d\]\[8-5d=13-9d\]\[4d=5\]\[d=1.25\]\[a=8-5*1.25=1.75\] Subtracting: \[8=a+5d, 13=a+9d\] \[13-8=9d-5d\]\[5=4d\]\[d=1.25\]\[a=8-5*1.25=1.75\]

OpenStudy (anonymous):

Now we have the first term and the common difference, that's all we need to know! Can you tell me the equation for the sum of n terms in an arithmetic sequence?

OpenStudy (anonymous):

(n/2)(a1+an)

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

Yep, so we can work out a4 using the a and d we just worked out, then it's simply a case of subbing those numbers into that equation :)

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