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+(n-1)1.5

(b)What is the sum of 4 terms between U10 and U13 inclusive?

2)Un=1/2^n-3

(a)What is the nature of this sequence?

(b)Calculate the sum of the first nine terms?

3)Un=12-0.5n

(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+(n-1)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)^(n-3)

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 + (n-1)d]

0 = n/2[23 - .5(n-1)]

solve for n

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