Finding the average of the following sequence?

Question

Given the sequence $5,-10,15,-20,25,-30,...$ find the average of the first 200 terms.

Is the formula that I am supposed to use is $t_n=a+(n-1)d$ ?

I know that $a = 5$, but I'm confused about the common difference and how to find the average of $200$ terms.

Answer 1

Notice:

There are 2 AP's here

$$5,15,25..$$

$$-10, -20, -30 ..$$

In the first, $a= 5$, $d = 10$

In the second $a = -10$, $d= -10$

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

so $(\frac{100}{2})(2\times5 + (99)10 + 2\times-10 + (99)-10)$ and divide the entire thing by 200

Answer 2

Hint

Take $5,15,25,\dots$ and $-10,-20,-30,\dots$ as two different arithmetic series. Then average the first $100$ terms of each.

Answer 3

Given $$5, -10, 15, -20, 25, -30, ... $$ taking two terms at a time we have, $$-5, -5, -5,...$$ So sum of n terms can be given as: $$S_{\ n\ (is\ even)} = -(\frac{n}{2})5$$ $$S_{\ n\ (is\ odd)} = S_{n-1}+T_n = -(\frac{n-1}{2})5+5+(\frac{n-1}{2})10$$$$=(\frac{n+1}{2})5$$ Note: $T_n$ of sequence$(5,-10,15,...)=t_{\frac{n+1}{2}}$ of sequence$(5,15,25,...)$ when n is odd.