1. The sum of the first three terms of a geometric progression is 65 and their product is 3375. Find the three terms. (6 m)

Let x be the first number. Since these numbers are in geometric progression, there is some constant r such that the second number is xr and the third number is xr•r = xr ².

Their sum is 65 :

x + xr + xr ² = 65

Their product is 3375 :

x • xr • xr ² = 3375

We can simplify the first equation as

x (1 + r + r ²) = 65

and the second one as

x ³ r ³ = (xr )³ = 3375

Since 15³ = 3375, it follows that xr = 15. Then x = 15/r. Substitute this into the second equation and solve for r :

15/r (1 + r + r ²) = 15/r + 15 + 15r = 65

15 + 15r + 15r ² = 65r

15 - 50r + 15r ² = 0

3 - 10r + 3r ² = 0

I'll complete the square:

3r ² - 10r = -3

3 (r ² - 10/3 r ) = -3

3 (r ² - 10/3 r + 100/36) = -3 + 300/36

3 (r - 10/6)² = 16/3

(r - 10/6)² = 16/9

r - 10/6 = ± √(16/9) = ± 4/3

r = 10/6 ± 4/3

r = 3   or   r = 1/3

Both cases give the same numbers:

If r = 3, then x = 15/3 = 5. Then the 3 numbers are 5, 15, 45.

If r = 1/3, then x = 15/(1/3) = 45. Then the 3 numbers are the same but in reverse order: 45, 15, 3.

answer: i believe c. is the answer sorry if i am wrong

step-by-step explanation: