Answer:
The sum of the first 12 terms of the sequence is 2703.673096
Step-by-step explanation:
The nth term of a geometric progression is a(n) = [tex]ar^{n-1}[/tex], where
- r is the common ratio between each two consecutive terms
The sum of n terms of the geometric progression is s(n) = [tex]\frac{a(r^{n}-1)}{r-1}[/tex]
∵ The third term of a geometric progression is [tex]\frac{63}{4}[/tex]
∴ a(3) = [tex]\frac{63}{4}[/tex]
∵ a(3) = ar²
→ Equate its two right sides
∴ ar² = [tex]\frac{63}{4}[/tex] ⇒ (1)
∵ The sixth term is [tex]\frac{1701}{32}[/tex]
∴ a(6) = [tex]\frac{1701}{32}[/tex]
∵ a(6) = a[tex]r^{5}[/tex]
→ Equate its two right sides
∴ a[tex]r^{5}[/tex] = [tex]\frac{1701}{32}[/tex] ⇒ (2)
→ Divide (2) by (1)
∴ r³ = [tex]\frac{27}{8}[/tex]
→ Take ∛ for both sides
∴ r = [tex]\frac{3}{2}[/tex] = 1.5
→ Substitute the value of r in (1) to find a
∵ a(1.5) = [tex]\frac{63}{4}[/tex]
∴ 1.5a = 15.75
→ Divide both sides by 1.5
∴ a = 10.5
→ Let us find the sum of the first 12 terms
∵ n = 12, a = 10.5, and r = 1.5
→ Substitute them in the rule of the sum above
∵ s(12) = [tex]\frac{10.5(1.5^{12}-1)}{1.5-1}[/tex]
∴ s(12) = 2703.673096
∴ The sum of the first 12 terms of the sequence is 2703.673096
