Question

WILL GIVE MEDAL!!!!! what is the sum of the arithmetic sequence 3,9,15.....if there are 22 terms ???

Answer 1

Ok, you start from n=1, and go up by 6 each time.
For the sum you need the 22nd term.

\(\large { \displaystyle{\rm a}_{n}={\rm a}_1+{\rm d}({\rm n}-1) }\)

\(\large { \displaystyle{\rm a}_{22}=3+{\rm 6}({\rm 22}-1) }\)
\(\large { \displaystyle{\rm a}_{22}=3+{\rm 6}(21) }\)
\(\large { \displaystyle{\rm a}_{22}=129 }\)

Answer 2

Then the sum of the first 22 terms (of this arithmetic sequence) is:

\(\large { \displaystyle{\rm S}_{22}=\frac{ \left({\rm a}_1+{\rm a}_{22}\right) }{2} \times 22 }\)

Answer 3

\(\large { \displaystyle{\rm S}_{22}=\frac{ \left(3+129\right) }{2} \times 22 }\)