Tentukan banyak suku dan jumlah deret aritmetika berikut .... A. 4 +9+ 14+ 19+... 104 b. 72+ 66+ 60.... Question from @Anugrahh

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Tentukan banyak suku dan jumlah deret aritmetika berikut ....
A. 4 +9+ 14+ 19+... 104
b. 72+ 66+ 60+ 54+... 12
c. -12- 8- 4- 0- ... -128
d. -3 -7 -11 -15 ... -107

iipmilan A. a = 4, b = 5
Un = a + (n-1)b
104 = 4 + (n-1) 5 --> n = 21
Sn = n/2 (a+Un) = 21/2 (4+104) = 1134
B. a = 72, b = -6
Un = a + (n-1)b
12 = 72 + (n-1) (-6) --> n = 11
Sn = n/2 (a+Un) = 11/2 (72+12) = 462
C. -
D. a = -3, b = -4
Un = a + (n-1)b
-107 = -3 + (n-1) (-4) --> n = 27
Sn = n/2 (a+Un) = 27/2 ((-3)+(-107)) = 1485

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Takamori37 A.
a = 4
b = 5
Un = 104
104 = a+(n-1)b
104 = 4+(n-1)5
100 = 5n-5
5n = 105
n = 21
$S_{21}=\frac{21}{2}(2.4+(21-1)5) \\ S_{21}=\frac{21}{2}(8+100) \\ S_{21}=\frac{21}{2}\times 108 \\ S_{21}=21\times 54 \\ S_{21}=1134$

B.
a = 72
b = -6
Un = 12
12 = a+(n-1)b
12 = 72+(n-1)(-6)
-60 = -6n+6
-6n = -66
n = 11
$S_{11}=\frac{11}{2}(2.72+(11-1)(-6)) \\ S_{11}=\frac{11}{2}(144-60) \\ S_{11}=\frac{11}{2}\times 84 \\ S_{11}=11\times 42 \\ S_{21}=462$

C.
Sepertinya ada kesalahan pengetikan,
a = -12
b = 4
Un = 128
128 = -12+(n-1)4
140 = 4n-4
4n = 144
n = 36
$S_{36}=\frac{36}{2}(2.(-12)+(36-1)4) \\ S_{36}=18(-24+140) \\ S_{36}=18\times116 \\ S_{36}=2088$

D.
a = -3
b = -4
Un = -107
-107 = -3+(n-1)(-4)
-104 = -4n+4
4n = 108
n = 27
$S_{27}=\frac{27}{2}(2.(-3)+(27-1)(-4)) \\ S_{27}=\frac{27}{2}(-6-104) \\ S_{27}=\frac{27}{2}\times (-110) \\ S_{11}=27\times (-55) \\ S_{21}=-1485$

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