the sum of the first n terms of the AP is equal to half of the sum of the next n terms of the same AP.find the ratio of the sum of the first 3n terms of the progression to the sum of its first n terms.i)5 ii)6 iii)7 iv)8

Sum of first n terms is half of sum of next n terms. This means, sum of first 2n terms is 3 times the sum of first n terms. Let the first term be a and common difference be d.
This means that $$\frac{\frac{2n}{2}(2a+(2n-1)d)}{\frac{n}{2}(2a+(n-1)d}=3$$
=> 4a+4nd-2d=6a+3nd-3d
=>2a-nd-d=0
Now the ratio of sum of 3n terms to sum of n terms is:
$$\frac{\frac{3n}{2}(2a+(3n-1)d)}{\frac{n}{2}(2a+(n-1)d}$$
$$\frac{6a+9nd-3d}{2a+nd-d}$$
$$\frac{6a+3nd-3d+6nd}{2a+nd-d}$$
$$3+\frac{6nd}{2a+nd-d}$$
Now, 2a=nd+d
this means 2a+nd-d=2nd
Thus, answer is 3+3=6