1)tentukan suku ke-5 barisan aritmatika ,jika a=-3 dan b = -4 2)diketahui deret aritmatika 3+6+9+12+....hitunglah jumlah S16 3) diketahui barisan geometri denga

1) u5 = -3 + (5-1)-4
= -3 + -16
=-19
2) s16 = 16/2 (2.3 + (16-1)3
= 8 (6+(15)3)
= 8(51)
= 408
3) u10 = 1.rn-1
u7= 64
ar6=64
a=1
1.r6 = 64
r6= 64
r= 2
u10 = 1.rn-1
= 1. 2^10-1
=2^9
= 512
4) u5=ar4=64
u2= ar =8
a=8/r
ar4 = 64
8/r.r4=64
r3 = 64/8
r= 2
u1, u2, u3, u4, u5 = a, ar, ar2, ar3, ar4
8/2, 4. 2, 4.2^2, 4.2^3, 4.2^4
4, 8, 16, 32, 64

1.]
a = -3 , b = -4

U₅ = a + 4b
U₅ = -3 + 4(-4)
U₅ = -3 - 16 = -19

2.]
3 + 6 + 9 + ...
a = 3 , b = 6 - 3 = 3

U₁₆ = a + 15b
U₁₆ = 3 + 15(3)
U₁₆ = 3 + 45 = 48

Maka
S₁₆ = ¹⁶/₂ (a + U₁₆)
S₁₆ = 8 (3 + 48)
S₁₆ = 8 × 51
S₁₆ = 408

3.]
a = 1

U₇ = ar⁶
64 = r⁶
2⁶ = r⁶
r = 2

U₁₀ = ar⁹
U₁₀ = 2⁹
U₁₀ = 512

4.]
U₂ = 8
U₅ = 64

r = ⁵⁻²√(U₅ / U₂)
r = ∛(64 / 8)
r = ∛8
r = 2

U₂ = ar
8 = 2a
a = 4

S₅ = a(r⁵ - 1) / (r - 1)
S₅ = 4(2⁵ - 1) / (2 - 1)
S₅ = 4(31)
S₅ = 124