#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

int i, nextTerm;

int t1 = 2, t2 = 2;

nextTerm = t1+t2;

printf("%d, %d, ", t1, t2);

for (i = 3; i <= 10; ++i) {

printf("%d, ", nextTerm);

t1 = t2;

t2 = nextTerm;

nextTerm = t1 + t2;

}

return 0;

}

Log_2aa=x    then, a=(2a)^x  ......(1)

Log_3a2a=y    then,2a=(3a)^y ......(2)

Log_4a 3a=z  then, 3a=(4a)^z ......(3)

So, 

a=(2a)^x  [from (1)]

Or, a=(3a)^xy    [from(2)]

Or, a=(4a)^xyz     [from(3)]

Multiplying both sides by 4a,

4a.a=4a.(4a)^xyz  

Or,(2a)² =(4a)^{xyz + 1} 

Or,(3a)^2y =(4a)^xyz+1 

Or,(4a)^2yz =(4a)^xyz+1 

Or, 2yz = xyz+1 .proved.

No, if three vectors do not lie in a plane, they cannot give zero resultant.

Explanation:

Let A, B and C be three vectors. If they give zero resultant, then

     A+B+C=0

or, A= -(B+C)

Hence, they will produce zero resultant, if A is equal to negative of vector (B+C). The vector (B+C) lies in the plane of B and C. Hence, A will be equal to negative of (B+C) if A, B and C all lie in a plane.