Question

3. Find the minimum value of n such that the sum of the first n terms of a G.P. 1, 2, 22, 23, ... .is greater
than or equal to 2000.
у у
Chow
0 in 121​

Answer 1

Answer:

n=6

Sn = a(r^n-1) / r-1 (as r>1)

here a=1 r=3 and Sn < 365

=> 1{(3^n)-1} / 3-1 < 365

=>   (3^n)-1 <  730

=>  3^n <  731

Now, 3^6 = 729 which is less than 731, therefore n=6