The third term of a geometric progression is 63/4 and the sixth term is 1701/32. Find the sum of the first 12 terms of the sequence.

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) = , where

a is the first termr is the common ratio between each two consecutive terms

The sum of n terms of the geometric progression is s(n) =

∵ The third term of a geometric progression is

∴ a(3) =

∵ a(3) = ar²

→ Equate its two right sides

∴ ar² =  ⇒ (1)

∵ The sixth term is

∴ a(6) =

∵ a(6) = a

→ Equate its two right sides

∴ a =  ⇒ (2)

→ Divide (2) by (1)

∴ r³ =

→ Take ∛  for both sides

∴ r = = 1.5

→ Substitute the value of r in (1) to find a

∵ a(1.5) =

∴ 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) =

∴ s(12) = 2703.673096

∴ The sum of the first 12 terms of the sequence is 2703.673096

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm

Hope i can you!