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

-11, -8, -5 . . .

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 $\displaystyle S_n = \frac{n}{2}[2a+(n-1)d]$

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

4. So it'd be 171?

So it'd be 171?
what? how did you get 171? please look at what plato has told you to do

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

$\displaystyle = -209 + 513$

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!

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.