determine the AP whose third term is 6 and 7th term exceeds the 5th term by 12
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Answered by
4
a+2d=6,a=6-2d
a+6d=a+4d+12
substitute a value
6-2d+6d=6-2d+4d+12
6+4d=18+2d
2d=12
d=6
a=?
so substitute d value in 6-2d
a=6-2(6)
a=-6
AP: -6,0,6.....
Answered by
0
Answer:
Step-by-step explanation:
third term = a+2d
6= a+2d
fifth term = a+4d
seventh term = a+6d
a/q,
a+6d-(a+4d)= 12
a+6d-a-4d=12
2d=12
d=6
put in third term
a+2*6=6
a=6-12
a = -6
- second term =a+d= -6+ 6 = 0
hence ap is
-6,0,6,12,18..............
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