The two questions are here, they're probably easy but I'm having a brain freeze lol

http://img29.imageshack.us/img29/5626/file.png

The 4(c) sequence is 5, 7, 9, 11....

The 4(d) sequence is 5, -10, 20, -40.....

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- May 13th 2010, 03:00 PMbrumby_3Find a formula for the sequence....
The two questions are here, they're probably easy but I'm having a brain freeze lol

http://img29.imageshack.us/img29/5626/file.png

The 4(c) sequence is 5, 7, 9, 11....

The 4(d) sequence is 5, -10, 20, -40..... - May 13th 2010, 03:38 PMArchie Meade
Hi brumby_3,

1st one

$\displaystyle 5,\ 7,\ 9,\ 11,......$

is an ARITHMETIC SEQUENCE.

$\displaystyle T_1=5,\ T_2=5+(1)2,\ T_3=5+(2)2,\ T_4=5+(3)2,\ T_5=5+4(2)...$

$\displaystyle T_n=5+(n-1)2$

To sum all the terms of the sequence, the sum is

$\displaystyle S_n=\frac{T_1+T_n}{2}(n)$

which is the average value multiplied by the number of terms.

$\displaystyle S_n=\frac{5+5+(n-1)2}{2}(n)$

2nd one

$\displaystyle 5,\ -10,\ 20,\ -40,.....$

The next term is the previous one multiplied by $\displaystyle -2$

This is a GEOMETRIC SEQUENCE.

$\displaystyle T_1=5,\ T_2=(-2)5,\ T_3=(-2)^25,\ T_4=(-2)^35,...$

$\displaystyle T_n=5(-2)^{n-1}$

The sum of these terms is

$\displaystyle S_n=\frac{5\left(1-(-2)^{n}\right)}{1-(-2)}$

which is $\displaystyle \frac{a\left(1-r^n\right)}{1-r}$

where "a"=1st term, "r"=common ratio.