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a_10 = 58 and a_5 =28 so subtract \[\huge\rm \frac{ 58-28 }{ 10-5 }\]

Ok, so you get the common difference of 6.

and i just noticed we don't have the first term hm

yes we do- its 4

we can use the formula to find term \[\huge\rm a_n =(r)^{n-k}\]

opps sorry i was looking at the wrong page thats for geometric sequence

btw you can use 10th term doesn't matter u will get the same answer :=)

oh!! you already found it yes it's 4

can i see the answer for last part
little confused on that part

The last part? You mean the first question that I didn't include?

So I got help yesterday for the first one and now I m stuck on how to find the answers for (ii)

ohh i see..

444**

Ok I will try that quickly

try to work on it
ive to go to eat something BRB

sure

Ok I also actually have to be away for a little while. Will come back to this as soon as I can.

alright tag me when u r ready :=)

Trying to solve it now

888 ?O_*

Yeah that didn't work

I can send a picture of my working out

ohh it supposed to be a_1 `+` a_n

sorry about taht ..

What do you mean?

Oooooh ok let me try that quickly

i'll try it let's see what we get

Now the quadratic I get is \[n ^{2} + \frac{ 1 }{ 3 } -148=0\]

1/3 ? how did you get that ? o.O

Oh dear. Umm I had -6n^2 -2n +888 and I divided everything by -6

ohh so 1/3n*

Oh yes, sorry

\[n ^{2} + \frac{ 1 }{ 3 }n -148=0\] can you solve for n ?

Yup doing that now-almost done

alright let me know what you get

Ok done- the positive value I got was 12.00333333. Do I round that down to 12?

good job! :=)

np & you too!