the fourth term of an arithmetic progression is equal to 3 times the first term and the seventh term exceeds twice the third term by 1 find the first term and the common differrence

1 answer

a+3d = 3a
a+6d = 2(a+2d)+1

a=3
d=2

chec:
3+6 = 3*3
3+12 = 2(7)+1