Posted by Anonymous on Saturday, April 19, 2008 at 3:14am.
1)AP has U6=12.5 and U19=32
a+5d=12.5 (&)
a+18d=32 (2)
(2)(1), d=1.5
therefore, a=5
(a)What is the nth term?Tn=5+(n1)1.5
(b)What is the sum of 4 terms between U10 and U13 inclusive?
2)Un=1/2^n3
(a)What is the nature of this sequence?
(b)Calculate the sum of the first nine terms?
3)Un=120.5n
(a)Solve SUMn=0
(b)If Wm=2+5m, when is Wm first a three digit number?
(c)Find m for the case where Wm=U10

Math  Reiny, Saturday, April 19, 2008 at 9:02am
for 1. a) I would simplify a bit further.
Tn=5+(n1)1.5
= 5 + 1.5 n  1.5
= 1.5n + 3.5
for b)
find the sum of the first 13 terms, then
find the sum of the first 9 terms.
2. Clearly a geometric sequence if the expression is (1/2)^(n3)
The problem is that you did not specify the starting value of n
e.g. if you start with n = 4, term 1 is 1/2
if you start with n = 1 then the first term is 4
In any case r = 1/2
Use you sum formula to find the sum of 9 terms.
3. obviously an arithmetic sequence with a= 11.5 , d= .5
Sum(n) = n/2[2a + (n1)d]
0 = n/2[23  .5(n1)]
solve for n
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