Answer:
Step-by-step explanation:
let the A.P. be a,a+d,a+2d,a+3d,a+4d,...
[tex]S_{n}=\frac{n}{2} [2a+(n-1)d] ,\\S_{4}=\frac{4}{2} [2a+(4-1)d]\\4=2[2a+3d]\\2a+3d=4~~~...(1)\\S_{12}=\frac{12}{2} [2a+(12-1)d]\\28=6[2a+11d]\\14=6a+33d\\6a+33d=14 ...(2)[/tex]
multiply (1) by 3
6a+9d=12 ...(3)
(2)-(3) gives
24d=2
d=2/24=1/12
from (1)
2a+3(1/12)=4
2a=4-1/4=15/4
a=15/8
a+d=15/8+1/12=(45+2)/24=47/24
a+2d=15/8+2(1/12)=15/8+1/6=(45+4)/24=49/24
so A.P. is 15/8,47/24,49/24,....
