Question

Let An be the sum of first n terms of GP 704+704/2+704/4.... And
Bn be the sum of first n terms of GP
1984-1984/2+1984/4-1984/8....
If An=Bn then the value of n is

Answer 1

For A(n), a = 704 , r = 1/2
A(n) = 704((1 - 1/2)^n )/(1 - 1/2)
= 1408( 1 - (1/2)^n )

For B(n), a = 1984 , r = -1/2
B(n) = 1984( 1 - (-1/2)^n)/( 1 + 1/2)
= (2/3)(1984)(1 - (-1/2)^n)

but 1408( 1 - (1/2)^n ) = (2/3)(1984)(1 - (-1/2)^n)
4224( 1 - (1/2)^n ) = 3968(1 - (-1/2)^n)
33( 1 - (1/2)^n ) = 31(1 - (-1/2)^n)

if n is even:
33( 1 - (1/2)^n ) = 31(1 - (1/2)^n)
33 - 33(1/2)^n = 31 - 31(1/2)^n
- 2(1/2)^n = -2
this can only be true if n = 0 , but n has to be a whole number, so n can't be even.

if n is odd:
33( 1 - (1/2)^n ) = 31(1 + (1/2)^n)
33 - 33(1/2)^n = 31 + 31(1/2)^n
2 = 64(1/2)^n
(1/2)^n = 1/32
n = 5

check:
A(5) = 704( 1 - (1/2)^5)/(1 - 1/2)
= 704(31/32) / (1/2) = 1364

B(5) = 1984( 1 - (-1/32) / (1 + 1/2)
= 1984(33/32) / (3/2) = 1364

n=5 is the correct answer