Answer:
Explanation:
The way you have written the first question may be what is confusing you. It should be written as
bn = 3*b_(n-1) + 2
So b2 =
b2 =3*b_(2 -1) + 2
b2 = 3*b1 + 2
b2 = 3*5 + 2
b2 = 15 + 2
b2 = 17
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b3 = 3b_(n _1) + 2
b_2 = 17 (from the step above)
b3 = 3*17 + 2
b3 = 51 + 2
b3 = 53
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b_4 = 3*b_3 + 2
b_4 = 3*53 + 2
b_4 = 159 + 2
b_4 = 161
Do you see how this works? You take the previous term, multiply by 3 and add 2 to get the current term. This one builds up rather quickly.
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Next Question
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tn = a + (n - 1)*d
t6 = a + (6 - 1)*d
t4 = a + 5d
4 = a + 5d
t10 = a + 9d
Subtract t4 [4 = a + 3d ] from t10 written bellow
- 4 = a + 9d
4 = a + 5d
- 8 = 4d Divide by 4
-8/4 = 4d/4
-2 = d
t6 = a + 5d
4 = a + 5*(-2)
4 = a - 10 Add 10 to both sides.
4 + 10 = a - 10 + 10
14 = a
tn = 14 + (n - 1)*d
Answer: d
Please don't use red. It is really hard to read.