# WITH SOLUTIONS PLEASE! 1. Show that the sequence tn: {-0.9,-0.7,0.5,...} is an arithmetic sequence 2. Find the rule for the arithmetic sequence tn: {1/8,3/8,5/8,...} 3. Find the 10th term of the arithmetic sequence tn: {-10,-6,-2,...} 4. The three consecutive terms of an arithmetic sequence are 3.6,y,8.2. Find the value of y. 5. Insert three evenly spaced numbers between -2 and 10. 6. Find the !0th term of an arithmetic sequence where the first term is 5 and the 4th term is 17. 7. Find the difference between 4th term and the 10th term of the arithmetic sequence tn:{2,-1,-4,...} 8. The 10th term in an arithmetic sequence is 8 and the 4th term is -4. Determine the first term A. 9. Fnd the sum to first 8 terms of the arithmetic sequence tn:{-14,-12,-10,...}

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by soulPsiioniics

2016-06-17T17:40:38+08:00
1. t2 - t1 = -0.7 - (-0.9) = 0.2
t3 - t2 = -0.5 - (0.7) = 0.2 The common difference is 0.2. Therefore, it is an arithmetic sequence

2. a = 1/8, d = 2/8 = 1/4
tn = 1/8 + (n-1)1/4
tn = 1/8 + (n/4 - 1/4)
tn = n/4 - 1/8 This is the required rule

3. a = -10, d = 4
t10 = -10 + (10-1)4
t10 = -10 + 36
t10 = 26

4. y = (3.6 + 8.2) / 2
y = 5.9

5. a = -2, t5 = 10
-2 + (5-1)d =10
-2 + 4d = 10
4d = 12
d = 12/4
d = 3 Thus, we have -2, 1, 4, 7, 10 evenly spaced

6. a =5, t4 = 17
5 + (4-1)d = 17
5 + 3d = 17
3d = 12
d = 12/3
d = 4

t10 = 5 + (10-1)4
t10 = 5 + 36
t10 = 41

7. a = 2, d = 3
t4 = 2 +(4-1)-3
t4 = 2 + (-9)
t4 = -7

t10 = 2 + (10-1)-3
t10 = 2 + (-27)
t10 = -25
Difference = -7 - (-25)
Difference = 18

8. t10 = a + (10-1)d
8 = a + 9d -----equation 1

-4 = a + (4-1)d
-4 = a + 3d -----equation 2

Subtract eq. 2 from eq. 1

6d = 12
d = 12/6
d = 2

Now, replace d in eq. 1
a + 9(2) = 8
a +18 = 8
a = 8-18
a= -10

9. a = -14, d= 2
s8 = 8/2[2(-14) + (8-1)2]
s8 = 4[-28+14]
s8 = 4(-14)
s8 = -56