Question

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

Answer 1

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