Some Series and Sequence questions (1 Viewer)

Najji · Jul 5, 2013

Hello guys, I need help with these questions;
1. For the series function defined by Sn = 3n^2 -11n, find tn and hence show that the sequence is arithmetic
2. The first and second terms of an arithmetic sequence are a and b respectively. If the nth term is c, express i terms of a b and c and hence find the sum of these n terms.
3. Find the sum of a) the first odd positive integers b) the first n even positive integers c) the first n positive integers and find this value of n if the sum is 210.
Any help will be appreciated, Thanks.

pokka · Jul 5, 2013

Najji · Jul 5, 2013

pokka said:
$1.\quad\\\begin{align*}t_n&=S_n-S_{n-1}\\\\&=3n^2-11n-[3(n-1)^2-11(n-1)]\\\\&=3n^2-3(n-1)^2-11n+11(n-1)\\\\&=3(n^2-(n-1)^2)-11(n-n+1)\\\\&=3(n^2-n^2+2n-1)-11\\\\&=3(2n-1)-11\\\\&=6n-14\end{align*}\\\\\\$To check if it is arithmetic you check if $t_n-t_{n-1}=t_{n+1}-t_n$ such that both sides give a constant since each consecutive term has a common difference in an arithmetic sequence.$

I've seen this question posted on the net twice, but I couldn't understand how it was done until you posted your working, Thanks a lot.

Najji · Jul 6, 2013

Need help on this one as well; Find the first three terms of a geometric sequence given that the sum of the first four terms is 65/3 and the sum to infinity is 27

Drongoski · Jul 6, 2013

asalaam alaikum

$\frac {a(1-r^4 )}{1-r} = \frac {65}{3} \\ \frac {a}{1-r} = 27 \\ \\ \therefore (1-r^4) \times 27 = \frac {65}{3} \implies 1- r^4 = \frac {65}{3 \times 27} \implies r^4 = \frac {16}{81} = (\frac {2}{3} )^4} \implies r = \pm \frac {2}{3} \\ \\ \therefore a = 27 \times (1- \frac {2}{3}) = 9 $ or $ a = 27 \times (1+ \frac {2}{3}) = 45$

.: 1st 3 terms are: 9, 6, 4

or 45, -30, 20

Najji · Jul 6, 2013

Drongoski said:
asalaam alaikum

$\frac {a(1-r^4 )}{1-r} = \frac {65}{3} \\ \frac {a}{1-r} = 27 \\ \\ \therefore (1-r^4) \times 27 = \frac {65}{3} \implies 1- r^4 = \frac {65}{3 \times 27} \implies r^4 = \frac {16}{81} = (\frac {2}{3} )^4} \implies r = \pm \frac {2}{3} \\ \\ \therefore a = 27 \times (1- \frac {2}{3}) = 9 $ or $ a = 27 \times (1+ \frac {2}{3}) = 45$

.: 1st 3 terms are: 9, 6, 4

or 45, -30, 20

Walaykum Salaam, thanks for the answer mate

Some Series and Sequence questions (1 Viewer)

Najji

New Member

pokka

d'ohnuts

Najji

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Najji

New Member

Drongoski

Well-Known Member

Najji

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