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.....
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.....
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.