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Answer:
1. 675
2. 11
3. 5
4. 34
5. 13
6. 1600
Step-by-step explanation:
1. 10, 15, 20, 25
to find the sum of the first 15 terms:
Given:
a1 = 10
a2 = 15
n = 15
to find d:
d = a2-a1
d = 15-10
d = 5
formula for the sum of Arithmetic Series:
Sn = n/2 [2a1+(n-1)d]
S15 = 15/2 [2(10)+(15-1)5
S15 = 15/2 [20+(14)5]
S15 = 15/2 [20+70]
S15 = 15/2 [90]
S15 = 1350/2
S15 = 675
The sum of the first 15 terms is 675 :D
2. s8 = 172 d = 3
to find the first term:
Given:
s8 = 172
d = 3
n = 8
a1 = ?
formula for the sum of Arithmetic Series:
Sn = n/2 [2a1+(n-1)d]
172 = 8/2 [2a1+(8-1)3]
172 = 8/2 [2a1+(7)3]
172 = 8/2 [2a1+21]
2 • 172 / 8 = 8/2 [2a1+21]
344 / 8 = [2a1+21] / 8
43 = 2a1+21
43-21 = 2a1
22 = 2a1
11 = a1
The first term is 11 :D
3. s9 = 216 a1 = 4
to find the common difference:
Given:
s9 = 216
a1 = 4
formula for the sum of Arithmetic Series:
Sn = n/2 [2a1+(n-1)d]
216 = 9/2 [2(4)+(9-1)d]
216 = 9/2 [8+(8)d]
2 • 216 / 9 = 9/2 [8+(8)d]
432 / 9 = [8+(8)d] / 9
48 = 8+8d
48-8 = 8d
40 = 8d
40 ÷ 8 = d
5 = d
The common difference is 5 :D
4. s12 = 606 an = 67 a1 = ?
to find the first term:
Given:
s12 = 606
an = 67
n = 12
the formula for the sum of Arithmetic Series if the last term is given:
Sn = n/2 (a1+an)
606 = 12/2 (a1+67)
2 • 606 / 12 = 12/2 (a1+67)
1212 / 12 = (a1+67) /12
101 = a1+67
101-67 = a1
34 = a1
The first term is 34 :D
5. a1 = 8 an = 56 Sn = 416 n=?
to find the terms:
Given:
a1 = 8
an = 56
Sn = 416
the formula for the sum of Arithmetic Series if the last term is given:
Sn = n/2 (a1+an)
416 = n/2 (8+56)
416 = n/2 (64)
416 • 2 = 64n
832 = 64n
13 = n
The number of terms of 56 and 416 is 13 :D
6. 4, 9, 14, 19
to get the sum of the first 25 terms:
Given:
a1 = 4
a2 = 9
a3 = 14
a4 = 19
n = 25
to find d:
d = a2-a1
d = 9-4
d = 5
formula for the sum of Arithmetic Series:
Sn = n/2 [2a1+(n-1)d]
S25 = 25/2 [2(4)+(25-1)5]
S25 = 25/2 [8+(24)5]
S25 = 25/2 [8+120]
S25 = 25/2 [128]
S25 = 3200/2
S25 = 1600
The sum of the first 25 terms is 1600 :D