Answer:
There are 7 terms in the GP
Step-by-step explanation:
In the geometric progression, there is a constant ratio between each two consecutive terms
The rule of the nth term of the geometric progression is
[tex]a(n)=a(r)^{n-1}[/tex], where
- n is the position of the term in the sequence
∵ The first term = 5[tex]\frac{1}{3}[/tex]
∴ a = 5[tex]\frac{1}{3}[/tex]
∵ The last term = [tex]\frac{243}{256}[/tex]
∴ a(n) = [tex]\frac{243}{256}[/tex]
∵ The common ratio = [tex]\frac{3}{4}[/tex]
→ Substitute these values in the rule above to find n
∵ [tex]\frac{243}{256}[/tex] = 5[tex]\frac{1}{3}[/tex] . [tex][\frac{3}{4}]^{n-1}[/tex]
→ Divide both sides by 5[tex]\frac{1}{3}[/tex]
∴ [tex]\frac{729}{4096}[/tex] = [tex][\frac{3}{4}]^{n-1}[/tex]
→ Let us find how many 3 in 729
∵ 729 ÷ 3 = 243 ÷ 3 = 81 ÷ 3 = 27 ÷ 3 = 9 ÷ 3 = 3 ÷ 3 = 1
∴ There are 6 three in 729
∴ 729 = [tex]3^{6}[/tex]
→ Let us find how many 4 in 4096
∵ 4096 ÷ 4 = 1024 ÷ 4 = 256 ÷ 4 = 64 ÷ 4 = 16 ÷ 4 = 4 ÷ 4 = 1
∴ There are 6 four in 4096
∴ 4096 = [tex]4^{6}[/tex]
∵ [tex]\frac{729}{4096}[/tex] = [tex]\frac{3^{6} }{4^{6}}[/tex] = [tex][\frac{3}{4}]^{6}[/tex]
∴ [tex][\frac{3}{4}]^{6}[/tex] = [tex][\frac{3}{4}]^{n-1}[/tex]
∵ The bases are equal
∴ Their exponents are equal
∴ 6 = n - 1
→ Add 1 to both sides
∴ 6 + 1 = n - 1 + 1
∴ 7 = n
∴ There are 7 terms in the GP
