Adding terms

May 2010
43
0
Can someone help me find the sum of the first 19 terms of the following sequence?



-11, -8, -5 . . .
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
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West Midlands, England
Can someone help me find the sum of the first 19 terms of the following sequence?



-11, -8, -5 . . .
It is an arithmetic sequence with first term (a) -11 and common difference (d) 3 and you want to find \(\displaystyle S_{19}\)


The sum of an arithmetic sequence is given by \(\displaystyle S_n = \frac{n}{2}[2a+(n-1)d]\)
 

Plato

MHF Helper
Aug 2006
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Can someone help me find the sum of the first 19 terms of the following sequence?
-11, -8, -5 . . .
\(\displaystyle \sum\limits_{k = 0}^{18} {\left( { - 11 + 3k} \right)} = \left( { - 11} \right)\left( {19} \right) + 3 \cdot \frac{{18 \cdot 19}}{2}\)
 
Feb 2010
1,036
386
Dirty South
So it'd be 171?
what? how did you get 171? please look at what plato has told you to do

\(\displaystyle (-11 \times 19 ) + \left(3 \times \frac{18 \times 19}{2} \right)\)

\(\displaystyle = -209 + 513\)
 
Last edited:
Jun 2008
292
87
here is the general solution

use arithmetic progression
 
Last edited:
Feb 2010
1,036
386
Dirty South
let get the general solution by which particular solution can be obtained.
let there b n terms with t(n) as the nth term. let
s=(-11)+(-8)+(-5)+....................................+t(n) (1)
s= ____(-11)+(-8)+(-5)+..............................+t(n) (2)(shift right by 1 term)
subtracting 2 from 1 we have
0=-11+3+3+3+..................................-t(n) or
t(n)=-11+3+3+....n-1 times therefor
t(n)=-11+3(n-1) or
t(n)=3n-14
now we are suppose to add t(1),t(2) till t(n) therefor
S(t(n))=S(3n)-S(14) S stands for summation
S(t(n))=3[(n(n+1))/2]-14n.................(after operating summation)
now for particular solution put n=19
S(t(19))=3(19)(10)-126=444
therefor solution is 444
Nikhil,

your notation is a little hard to understand, but 444 is not the correct answer!
 
Jun 2008
292
87
two ways

final equation is
3[(n(n+1))/2]-14n
put n=19
so answer is
3(19)(10)- 14(19)
570-266=304
also
using arithmatic progression
s=(n/2)[2a+(n-1)d]
here d=3,n=19,a=-11
s=(19)[-11+27]=19x16=304
I did calculation mistake(put 14(9) instead of 14(19)) that I realised after posting. I edited it but you being so fast quoted my previous answer which had a mistake
I gave the complex method because it can also be used to find sum to n terms of series like 1+3+6+10 and so on...