Question

K , 3k , (8k + 4) ..... adalah deret geometri ...tentukan suku ke 10

Answer 1

U₂² = U₁.U₃
(3k)² = k(8k + 4)
9k² = 8k² + 4k
k² - 4k = 0
k(k - 4) = 0
k = 0 atau k = 4
ambil k = 4 amak:
4, 12, 36
a = 4
r = 12/4 = 3
U₁₀ = ar⁹
= 4.3⁹
= 78732