Consider a Geometric progression #1,7,49...7^364# . find the remainder when sum of this g.p is divided by 5.?

1 Answer

1

Explanation:

In this GP,

a = 1, r = 7, N = 365

#S_N = a1*(r^N-1)/(r-1)#

#S_N = 1*(7^365-1)/(7-1)#

#S_N = (7^365-1)/6#

Sum of digits of #7^365# is #7#, so #7^365-1# is divisible by #6#.
Last digit of #7^365# is #7#.
when divided by 5, #7^365# would leave a remainder 2 ,subract 1, the remainder would be 1