Now, what do you do in a sequence when you have nonconsecutive terms?

Question

Now, I can't figure out how to solve these two questions:one is geometric, another arithmetic.

1. For a given sequence, a sub1=2/3 and a sub 5= 1/24 Find a sub 6 (How do I figure out the common ration without consecutive terms?)

2. Find a sub 8 for the following arithmetic sequence: a sub 5 = -5, a sub 7 = -2m-11

Many thanks for your direction and assistance!

Answer 1

1.)
an = a1 * r^(n - 1)

a5 = a1 * r^(5 - 1)
(1/24) = (2/3) * r^4
(3/24) = 2r^4
(1/8) = 2r^4
r^4 = (1/16)
r = (1/2)

an = (2/3)(1/2)^(n - 1)

a1 = (2/3)(1/2)^(1 - 1) = (2/3)(1/2)^0 = (2/3) * 1 = (2/3)
a2 = (2/3)(1/2)^(2 - 1) = (2/3)(1/2)^1 = (2/3)(1/2) = (1/3)
a3 = (2/3)(1/2)^(3 - 1) = (2/3)(1/2)^2 = (2/3)(1/4) = (1/6)
a4 = (2/3)(1/2)^(4 - 1) = (2/3)(1/2)^3 = (2/3)(1/8) = (1/12)
a5 = (2/3)(1/2)^(5 - 1) = (2/3)(1/2)^4 = (2/3)(1/16) = (1/24)
a6 = (2/3)(1/2)^(6 - 1) = (2/3)(1/2)^5 = (2/3)(1/32) = (1/48)

a6 = (1/48)

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2.)
an = a1 + (n - 1)d

a5 = a1 + (5 - 1)d
-5 = a1 + 4d
a1 = -4d - 5

a7 = a1 + (7 - 1)d
-2m - 11 = a1 + 6d
a1 = -2m - 11 - 6d

-4d - 5 = -2m - 11 - 6d
2d = -2m - 6
2d = 2(-m - 3)
d = -m - 3

a1 = -4(-m - 3) - 5 = 4m + 12 - 5 = 4m + 7

a1 = (4m + 7) + (1 - 1)(-m - 3)
a1 = 4m + 7 + 0(-m - 3)
a1 = 4m + 7

a2 = 4m + 7 + (2 - 1)(-m - 3)
a2 = 4m + 7 + 1(-m - 3)
a2 = 4m + 7 - m - 3
a2 = 3m + 4

a3 = 4m + 7 + (3 - 1)(-m - 3)
a3 = 4m + 7 + 2(-m - 3)
a3 = 4m + 7 - 2m - 6
a3 = 2m + 1

a4 = 4m + 7 + 3(-m - 3)
a4 = 4m + 7 - 3m - 9
a4 = m - 2

a5 = 4m + 7 + 4(-m - 3)
a5 = 4m + 7 - 4m - 12
a5 = -5

a6 = 4m + 7 + 5(-m - 3)
a6 = 4m + 7 - 5m - 15
a6 = -m - 8

a7 = 4m + 7 + 6(-m - 3)
a7 = 4m + 7 - 6m - 18
a7 = -2m - 11

a8 = 4m + 7 + 7(-m - 3)
a8 = 4m + 7 - 7m - 21
a8 = -3m - 14

ANS : a8 = -3m - 14

Answer 2

Geometric
if a2/a1 = r
then a3/a1 = r^2, right? because each new elment is obtained by multiplying by r...

so a5/a1 = 1/24*(3/2) = 3/48 = 1/16 = r^4
-> r =1/2

Arith, similar concept:

a(n+1) - a(n) = d, the common difference
a(n+2) -a(n) = 2d...

a(7) - a(5) = -2m-11-(-5) = -2m-6 = 2d
-> d = -m-3

a(8) = a(5) + 3/2(a(7)-a(5)) or a(5)+3d = -5+3(-m-3) = -3m-14

Now what? The trouble that I have is what to do about a(0) or a(1). I don't know if there's a way to get a handle on a numerical answer for a(8), perhaps not without some assumption or knowledge about a(0) or a(1).

Answer 3

the equation for geometric is A sub n=a1(r)^n-1
we know a1, position number and the fifth term. Sub those in the equation.
1/24=2/3(r)^5-1
1/16=(r)^4
r=1/2
since we know 1=1/2 we have:
A6=2/3(1/2)^6-1
A6=2/3(1/32)
A6=1/48

Answer 4

a1 = (2/3)
a2 = (2/3)r
a3 = (2/3)r^2
a4 = (2/3)r^3
a5 = (2/3)r^4 =1/24
a6 = (2/3)r^5

r^4 = (1/24)/(2/3) = 1/16
r = (1/2)
a6 = 1/48


a5 = -5
a6 = -5+k
a7 = -5 + 2k = -2m - 11
a8 = -5 + 3k

2k = (-2m-11) -(-5)
2k = -2m - 6
k = -m-3
a8 = a7 + k = -2m -11 -m -3 = -3m -14