1. ## Adding terms

Can someone help me find the sum of the first 19 terms of the following sequence?

-11, -8, -5 . . .

2. Originally Posted by tysonrss
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 $S_{19}$

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

3. Originally Posted by tysonrss
Can someone help me find the sum of the first 19 terms of the following sequence?
-11, -8, -5 . . .
$\sum\limits_{k = 0}^{18} {\left( { - 11 + 3k} \right)} = \left( { - 11} \right)\left( {19} \right) + 3 \cdot \frac{{18 \cdot 19}}{2}$

4. So it'd be 171?

5. Originally Posted by tysonrss
So it'd be 171?
what? how did you get 171? please look at what plato has told you to do

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

$= -209 + 513$

6. ## here is the general solution

use arithmetic progression

7. Originally Posted by nikhil
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!

8. 304 is the answer?

9. Originally Posted by tysonrss
304 is the answer?
Yes!!

10. ## two ways

final equation is
3[(n(n+1))/2]-14n
put n=19
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...

11. Originally Posted by harish21
Yes!!
Can you break it down how you got that equation, yours seem a bit more understandable than the others posted here and I need to know how to use that equation for the other problems.

12. Originally Posted by tysonrss
Can you break it down how you got that equation, yours seem a bit more understandable than the others posted here and I need to know how to use that equation for the other problems.
look at this to understand arithmetic sequences. Also look for arithmetic series in the same page. You should be able to understand how you can solve this problem.

13. Originally Posted by harish21
look at this to understand arithmetic sequences. Also look for arithmetic series in the same page. You should be able to understand how you can solve this problem.
Thanks, I already figuired it out