here are the first four terms of an arithmetic sequence
10 16 22 28
what is the 10th term of this sequence? is it 64?

Question

here are the first four terms of an arithmetic sequence
10  16  22  28
what is the 10th term of this sequence? is it 64?

jim_thompson5910 · Answer

You got it, the nth term is 6n+4, so the tenth term is 6(10)+4 = 60+4 =64

anonymous · Answer

yay:D

anonymous · Answer

also$$3\sqrt{7}$$ how do i rationalise the denominator?

jim_thompson5910 · Answer

Is that the whole fraction?

anonymous · Answer

its 3 over then |dw:1339356855249:dw|

jim_thompson5910 · Answer

ah gotcha

jim_thompson5910 · Answer

multiply top and bottom by the denominator sqrt(7) to rationalize this

jim_thompson5910 · Answer

$$\Large \frac{3}{\sqrt{7}}$$

$$\Large \frac{3\sqrt{7}}{\sqrt{7}*\sqrt{7}}$$

anonymous · Answer

i seeeee is that all of it?

jim_thompson5910 · Answer

no, there's more, you have to multiply the two sqrt(7) terms in the bottom, what do you get when you do that?

anonymous · Answer

i swear you still end up with 7 though?

jim_thompson5910 · Answer

yes you do, since sqrt(7)*sqrt(7) =sqrt(7*7) = sqrt(49) = 7

jim_thompson5910 · Answer

So the final answer is $$\Large \frac{3\sqrt{7}}{7}$$

anonymous · Answer

will it always be that with every equation?

jim_thompson5910 · Answer

what do you mean

anonymous · Answer

|dw:1339357466179:dw|

jim_thompson5910 · Answer

Do the same thing, but now multiply top and bottom by sqrt(8)

anonymous · Answer

|dw:1339357532573:dw|