Expression and value of Sn

Arithmetic and Geometric progressions.

Expression and value of Sn

Postby teraslintu » Sat Sep 21, 2019 2:08 pm

a) Write an expression for Sn, for the series 1 + 7 + 13 + ...

b) Hence, find the value of n for which Sn=833

Please help me.... I don't even know where to begin.
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Re: Expression and value of Sn

Postby Guest » Sun Sep 22, 2019 3:37 pm

$\div1,7,13,...$

$a_1=1$

$a_2=7=1+1.6$

$a_3=13=1+2.6$

$a_4=19=1+3.6$

$a_5=25=1+4.6$

$\cdots$

$a_n=1+(n-1).6$ - general term of arithmetic progression with difference $d=6$

$a_1+a_2+a_3+\cdots+a_n=S_n=\frac{2a_1+(n-1).d}{2}\cdot n=\frac{2+(n-1).6}{2}\cdot n=(1+3(n-1)).n=(3n-2).n$

$S_n=833$

$n(3n-2)=833$

$3n^2-2n-833=0$ - quadratic equation for n with abbreviated discriminant $D_1=\left(\frac{-2}{2}\right)^2-3.(-833)=1+2499=2500=50^2$

$n=\frac{-\frac{b}{2}\pm\sqrt{D_1}}{a}=\frac{-\frac{-2}{2}\pm50}{3}=\frac{51}{3};\ -\frac{49}{3}$

$n>0\Rightarrow n=\frac{51}{3}=17$
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Re: Expression and value of Sn

Postby Guest » Wed Oct 02, 2019 2:11 pm

teraslintu wrote:a) Write an expression for Sn, for the series 1 + 7 + 13 + ...

b) Hence, find the value of n for which Sn=833

Please help me.... I don't even know where to begin.

You are asking us to guess what you mean!
My first thought was that just three terms is not enough to identify a sequence. Here it happens that 7= 1+ 6 and 13= 7+ 6 so we can guess that this is an arithmetic sequence with initial value of 1 and common difference 6. But you understand, I hope, that there are infinitely many different sequences that start "1, 7, 13, …". Also, while [tex]S_n[/tex] is commonly used for the "nth sum" it would have been better to actually say that!

Assuming this is an arithmetic sequence then we can write the nth term as [tex]a_n= 1+ 6(n-1)= 6n- 5[/tex] so that [tex]a_1= 6- 5= 1[/tex], [tex]a_2= 12- 5= 7[/tex], [tex]a_3= 18- 5= 13[/tex]. An arithmetic sequence has the nice property that the "average value" is equal to the average of the first and last terms. Here the first term is 1 and the nth term is, as I said above, [tex]6n- 5[/tex]. The average of those two numbers is (1+ 6n- 5)/2= 3n- 2 and that is the average of the entire n term sequence. So the sum is [tex]n(3n- 2)= 3n^2- 2n[/tex]. Again, you can check that for the three numbers given- if n= 1, that is 3- 2= 1, if n= 2, that is 12- 4= 8= 1+ 7, if n= 3, that is 27- 6= 21= 1+ 7+ 13. That sum will be 833 when [tex]3n^2- 2n= 833[/tex] or [tex]3n^2- 2n- 833= 0[/tex]. Solve that quadratic equation. (It has two roots but obviously we want the positive one.)
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