## Babynini one year ago A partial sum of an arithmetic sequence is given. find the sum. 1+5+9+...+401

1. anonymous

ok, we have $$a_1=1$$ and common difference is $$d=4$$, last term is $$a_n=401$$ and you sure know that nth term of the sequence (a_n) is given by:$a_n=a_1+(n-1)d$find $$n$$$401=1+4(n-1) \\ n-1=100 \\ n=101$and for sum use the formula$S=\frac{n(a_1+a_n)}{2}=\frac{101(1+401)}{2}=101\times 201$

2. Babynini

What if it is a geometric sum?

3. Babynini

For example: 1+3+9+...+2187 (thanks for the help on the other one by the way :) )

4. Babynini

@mukushla :)

5. Babynini

from that we get, a=, r=3, n = ?

6. Babynini

Plugging some numbers here and there I get that a_8=2187 now what?

7. Babynini

nvm, I got it now xD it = 3280