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

Dec 13, 2017

1

#### Explanation:

In this GP,

a = 1, r = 7, N = 365

${S}_{N} = a 1 \cdot \frac{{r}^{N} - 1}{r - 1}$

${S}_{N} = 1 \cdot \frac{{7}^{365} - 1}{7 - 1}$

${S}_{N} = \frac{{7}^{365} - 1}{6}$

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