Question

The first term of a GP is 27 and its common ration is 4/3. Find the least number of terms the sequence can have if its sum exceeds 550.

Answer 1

Answer:

a = 27

r = 4/3

Sum of GP = 550

S = a*(r^n-1)/(r-1)

S = 27*((4/3)^n - 1)/(4/3 - 1)

S = 27*((4/3)^n - 1)/(1/3)

550 = 81*((4/3)^n - 1)

550/81 = ((4/3)^n - 1)

6.79 = ((4/3)^n - 1)

7.79 = (4/3)^n

Then,

n = 7