Question

the Sum of three numbers in AP is 27 and their product is 405,find the numbers​

Answer 1

$Let \: \: the \: \: 3 \: \: numbers \: \: be : \: a-d, \: \: a, \: \: a+d \\ \\ (a - d) + (a) + (a + d) = 27 \\ a \cancel{- d} + a + a \cancel{+ d} = 27 \\ 3a = 27 \\ a = \frac{27}{3} \\ \large{ \boxed{a = \underline{ \underline{ \bf9}}}} \\ \\ (a - d)(a)(a + d) = 405 \\ ( {a}^{2} - {d}^{2} )a = 405 \\ ( {9}^{2} - {d}^{2} )9 = 405 \\ 81 - {d}^{2} = \frac{405}{9} \\ 81 - {d}^{2} = 45 \\ 81 - 45 = {d}^{2} \\ 36 = {d}^{2} \\ \sqrt{36} = d \\ \large {\boxed{ \underline{ \underline{ \bf6}} = d}} \\ \\ Three \: \: numbers \: \: are : \\ a - d, \: \: a, \: \: a + d \\ 9 - 6, \: \: 9, \: \: 9 + 6 \\ \large \bf 3, \: \: 9, \: \: 15 \\ \\ \large{ \underline{ \bf \: Let's \: \: Verify} : }\\ (a - d) + (a) + (a + d) = 27 \\ (9 - 6) + (9) + (9 + 6) = 27 \\ 3 + 9 + 15 = 27 \\ \large{ \underline{ \underline{ \bf27 = 27}}} \\ \\ (a - d)(a)(a + d) = 405 \\ (9 - 6)(9)(9 + 6) = 405 \\ 3 \times 9 \times 15 = 405 \\ \large{ \underline{ \underline{ \bf405 = 405}}} \\ \\ \large{ \bf Verified}$