Question -
Answer -
Given,
Third term, a3┬а= 16
As we know,
a┬а+(3тИТ1)d┬а=16
a+2d┬а= 16тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж.┬а(i)
It is given that, 7th┬аtermexceeds the 5th┬аterm by 12.
a7┬атИТ┬аa5┬а= 12
[a+(7тИТ1)d]тИТ[a┬а+(5тИТ1)d]= 12
(a+6d)тИТ(a+4d) = 12
2d┬а= 12
d┬а= 6
From equation┬а(i), we get,
a+2(6) = 16
a+12 = 16
a┬а= 4
Therefore, A.P. will be4, 10, 16, 22, тАж