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#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;
}
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Log2aa=x then, a=(2a)x ......(1)
Log3a2a=y then,2a=(3a)y ......(2)
Log4a 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.
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routine of nikas banda