If ‘a’ is a natural number, then $$(7a^2 + 7a)$$ is always divisible by:

Given, 'a' is a natural number

$$7a^2+7a=7a\left(a+1\right) = 7k$$

$$=$$>  $$7a^2+7a$$ is always divisible by 7

Since a is a natural number,

When a is odd then a+1 is even

When a is even then a+1 is odd

$$\therefore\ a\left(a+1\right)$$ is always even

Let $$\ a\left(a+1\right)=2p$$

$$=$$>  $$7a^2+7a=7a\left(a+1\right)=7\left(2p\right)=14p$$

$$=$$>  $$7a^2+7a$$ is always divisible by 14

$$\therefore\ $$ $$7a^2+7a$$ is always divisible by both 7 and 14

Hence, the correct answer is Option C