Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
When it tells you to find h(2), that means you plug in 2 for x.
If 2x^3 (to the third power)
2(2)^3-3
1. PEMDAS, exponents first
2^3=8
2. multiply
8x2=16
3. subtract
16-3=13
Hope this helps!
I’ve attached my work...
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Let the # be n.
Then 8(1/n) = 4(1/6). Mult both sides by 6n to elim. the fractions:
6n(8) 6n(4)
------- = -------- => 48 = 4n, and so n = 12 (answer)
n 6
Which stem-and-leaf plot represents the data 80, 81, 91, 92, 66, 55, 54, 30, 55, 79, 78?
const2013 [10]
Convert the given data into ascending order.
30,54,55,55,66,78,79,80,81,91,92
Here the minimum value is 30 and maximum is 92.
Total counts=11
Now plot the data.
Stem | Leaf
3 | 0
5 | 4 5 5
6 | 6
7 | 8 9
8 | 0 1
9 | 1 2
This is the required stem leaf plot.