sequence and series c1

i am finding this chapter really hard to understand compared to the other chapters does any one know any good website which would make this all clear to me please thank you

i am finding this chapter really hard to understand compared to the other chapters does any one know any good website which would make this all clear to me please thank you

Hey I'm doing AS maths as well. I think sequences and series is my strongest chapter. Reading the textbook is really helpful, I learned most of it by reading it on the way to school (50min bus journey haha).

Post ques here if you need help!

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Upload some Qs here and I'll help. Just quote me

in Arithmetic progressions, u7=3u2 and u10=21. find U1

can show me step by step please

I think:
$u_7=u_1+6d=3u_2=3(u_1+d)=3u_1+3d$

$\therefore \ u_1+6d=3u_1+3d$

$\therefore \ 2u_1=3d \Rightarrow u_1=\dfrac{3d}{2}$

$u_{10}=21=u_1+9d$

$\dfrac{3d}{2}+9d=21$

$3d+18d=42$

$21d=42\Rightarrow d =2$


$u_1=\dfrac{3(2)}{2}=3$
Someone else will have to check. I'm rusty on this. Do you get what I've done though?

yh your right says so in my answer book, the problem with my book is that it gives be questions i has not explained as im self teaching, so how did you learn sequencing and series through a book or teacher as i need a bit more help and dont want to bother you. thanks alot tho

It's fine. I did C1 twice and the first time I did it in yr 11, I found series a tad hard. Particularly the new terminology $\displaystyle \sum_{r=0}^n \dfrac{r^2}{6}$ etc. The 2nd time, I just did every single example and got 100 as a consequence .

As I said, I don't mind helping with stuff. I know that usually full solutions aren't allowed but it can be helpful seeing the steps shown as the back only has answers. Did you follow my reasoning?

ow thx bro, also i was finding this problem hard as i keep getting 15 but the answer is supposed to be 6

$u_4=18$ and $u_7=30$ find u1

also are you a maths grad as you got an impressive score at C1

We'll apply a similar approach as we did the previous problem. I'll leave you to do the work though .
$u_1+3d=u_4=18$
$u_1+6d=u_7=30$

And no, I'd never be as good at maths as a maths grad. Compared to the majority of the regular posters on this forum, I'm an idiot, haha.
I get $u_1=6$

i done $3d-6d=-3d$
then $18-30=-12$
then $\dfrac{-12}{-3}=4$
then $4*3=12$
$18-12=6$

is this the right method thx whats your major btw? or are u in 6 form

I'm a sixth form student. That looks good although you haven't clearly stated what you're doing....

yh i know just want to get the concept,

also would you be able to help me with my last question for the day that would be great if you have time of course.

u4=9 and s4=21 calcualulate s8



and finding this one tricky

Un=4n-7. find S40

$s_n=\dfrac{n}{2}(a+l)$
$n=4$, $a=u_1$, $l=u_4=9$
$S_4=2(u_1+9)=21$
$u_1+9=10.5$
$u_1=0.5$ (I think)
$u_4=u_1+3d=9=0.5+3d$
$d=\dfrac{8.5}{3}$
$S_8=4(0.5+\dfrac{8.5}{3})=\dfrac{40}{3}$
There's a good possibility I'm wrong though as I'm tired

yh it was wrong but dont worry forget about doing the other one i have all ready learnt so much from you seriously like i only knew around 30%of sequencing and series know im up to 70% i will finish it tomorrow. your great help thx