Question

the sum of three term of an a.pis 21 and product of first and third term exceed the second term by 6 find three terms

Answer 1

let the three consecutive terms are (a-d),(a),(a+d).
from the first condition
$a - d + a + a + d = 21 \\ 3a = 21 \\ a = 7$
from the second condition
$(a - d)(a + d) = a + 6 \\ (7 - d)(7 + d) = 13 \\ 49 - {d}^{2} = 13 \\ 49 - 13 = {d}^{2} \\ d = \sqrt{36} \\ d = 6 \: or \: - 6$
If a = 7 and d= 6 then
The A.P.is 1,7,13.....
and
if a=7 and d=-6 then
The A.P is 13,7,1......
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